The Reimann-Stieltjes Integrals

William-Adolphe Bouguereau (1825-1905) was a French painter.

We have sections

Riemann-Stieltjes sum \ref{a}
Upper and Lower Riemann-Stieltjes Integrals
Step Functions as Integrators
Riemann-Stieltjes Integrals Depending on a Parameter

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

Riemann-Stieltjes sum.

The Riemann-Stieltjes integral is a more general concept than that of Riemann integral.

\begin{equation}{\label{mad22}}\tag{1}\mbox{}\end{equation}

Definition \ref{mad22}. Suppose that $f$ and $\alpha$ are two real-valued functions defined on the closed interval $[a,b]$.Let ${\cal P}=\{x_{0},x_{1},\cdots ,x_{n}\}$ be a partition of $[a,b]$, and let $t_{k}\in [x_{k-1},x_{k}]$. The following sum
\begin{align*} S({\cal P},f,\alpha ) & =\sum_{k=1}^{n}f(t_{k})\left [\alpha(x_{k})-\alpha(x_{k-1})\right ]\\ & \equiv\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}\end{align*}
is called a Riemann-Stieltjes sum of $f$ with respect to $\alpha$. We say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ when there exists a real number $A$ such that, given any $\epsilon >0$, there exists a partition ${\cal P}(\epsilon )$ of $[a,b]$ satisfying that, for any partition ${\cal P}$ that is finer than ${\cal P}(\epsilon )$ and for every choice $t_{k}\in [x_{k-1},x_{k}]$, we have
\[\left |S({\cal P},f,\alpha )-A\right |<\epsilon .\]

Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$. Then, the number $A$ is unique. In this case, we also write
\[A=\int_{a}^{b}fd\alpha =\int_{a}^{b}f(x)d\alpha (x).\]
The functions $f$ and $\alpha$ are called the integrand and integrator, respectively. If $\alpha (x)=x$, the Riemann-Stieltjes is reduced to the Reimann integral. There is an alternative definition of Reimann-Stieltjes integral.

\begin{equation}{\label{mad23}}\tag{2}\mbox{}\end{equation}

Definition \ref{mad23}. Suppose that $f$ and $\alpha$ are two real-valued functions defined on the closed interval $[a,b]$. We say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ when there exists a real number $A$ such that, given any $\epsilon >0$, there exists $\delta >0$ satisfying that, for any partition ${\cal P}$ with $\parallel {\cal P}\parallel <\delta$ and for every choice $t_{k}\in [x_{k-1},x_{k}]$, we have
\[\left |S({\cal P},f,\alpha )-A\right |<\epsilon .\]

\begin{equation}{\label{map24}}\tag{3}\mbox{}\end{equation}
Proposition \ref{map24}. Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ according to Definition \ref{mad23}. Then $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ according to Definition \ref{mad22}. In this case, the two integrals are equal. The converse is not true. $\sharp$

Proposition \ref{map24} shows that the concept of Riemann-Stieltjes integral in Definition \ref{mad22} is weaker than that of Definition \ref{mad23}. Here, we consider the Riemann-Stieltjes integral according to Definition \ref{mad22}. We denote by $f\in R(\alpha )$ on $[a,b]$ to mean that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$.

\begin{equation}{\label{ma7}}\tag{4}\mbox{}\end{equation}
Proposition \ref{ma7}. Let $\alpha (x)=\beta (x)+c$ for any $x\in [a,b]$ and for some constant $c$. Suppose that $f\in R(\alpha )$ on $[a,b]$. Then $f\in R(\beta )$ on $[a,b]$ and
\[\int_{a}^{b}f(x)d\alpha (x)=\int_{a}^{b}f(x)d\beta (x).\]

Proof. It is clear to see
\begin{align*} \Delta\alpha_{k} & =\alpha (x_{k})-\alpha (x_{k-1})\\ & =\beta (x_{k})+c-\beta (x_{k-1})-c\\ & =\beta (x_{k})-\beta (x_{k-1})\\ & =\Delta\beta_{k},\end{align*}
which also says $S({\cal P},f,\alpha )=S({\cal P},f,\beta )$ for any partition ${\cal P}$ of $[a,b]$. Since $f\in R(\alpha )$ on $[a,b]$, the proof is complete. $\blacksquare$

\begin{equation}{\label{map25}}\tag{5}\mbox{}\end{equation}
Proposition \ref{map25}. Suppose that $f\in R(\alpha )$ and $g\in R(\alpha )$ on $[a,b]$. Then $c_{1}f+c_{2}g\in R(\alpha )$ on $[a,b]$ for any constants $c_{1}$ and $c_{2}$. Moreover, we have
\[\int_{a}^{b}\left (c_{1}f+c_{2}g\right )d\alpha =c_{1}\int_{a}^{b}fd\alpha +c_{2}\int_{a}^{b}gd\alpha .\]

Proof. Let $h=c_{1}f+c_{2}g$. Given any partition ${\cal P}$ of $[a,b]$, we have
\begin{align*}
S({\cal P},h,\alpha ) & =\sum_{k=1}^{n}h(t_{k})\Delta\alpha_{k}\\ & =c_{1}\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}+c_{2}\sum_{k=1}^{n}g(t_{k})\Delta\alpha_{k}\\
& = c_{1}S({\cal P},f,\alpha )+c_{2}S({\cal P},g,\alpha ).
\end{align*}
Given $\epsilon >0$, there exists ${\cal P}'(\epsilon)$ such that ${\cal P}\supseteq {\cal P}'(\epsilon)$ implies
\[\left |S({\cal P},f,\alpha )-\int_{a}^{b}fd\alpha\right |<\frac{\epsilon}{2|c_{1}|},\]
and there exists ${\cal P}”(\epsilon)$ such that ${\cal P}\supseteq{\cal P}”(\epsilon)$ implies
\[\left |S({\cal P},g,\alpha )-\int_{a}^{b}gd\alpha\right |<\frac{\epsilon}{2|c_{2}|}.\]
We take ${\cal P}(\epsilon)={\cal P}'(\epsilon)\cup {\cal P}”(\epsilon)$. Then ${\cal P}\supseteq {\cal P}(\epsilon)$ implies ${\cal P}\supseteq {\cal P}'(\epsilon)$ and ${\cal P}\supseteq {\cal P}”(\epsilon)$. Now, for ${\cal P}\supseteq {\cal P}(\epsilon)$, we have
\begin{align*}
& \left |S({\cal P},h,\alpha )-c_{1}\int_{a}^{b}fd\alpha-c_{2}\int_{a}^{b}gd\alpha\right |\\
& \quad =\left |c_{1}S({\cal P},f,\alpha )+c_{2}S({\cal P},g,\alpha )-c_{1}\int_{a}^{b}fd\alpha-c_{2}\int_{a}^{b}gd\alpha\right |\\
& \quad \leq |c_{1}|\cdot\left |S({\cal P},f,\alpha )-\int_{a}^{b}fd\alpha\right |+|c_{2}|\cdot\left |S({\cal P},g,\alpha )-\int_{a}^{b}gd\alpha\right |\\
& \quad <|c_{1}|\cdot\frac{\epsilon}{2|c_{1}|}+|c_{2}|\cdot\frac{\epsilon}{2|c_{2}|}=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .
\end{align*}
This completes the proof. $\blacksquare$

\begin{equation}{\label{map40}}\tag{6}\mbox{}\end{equation}
Proposition \ref{map40}. Suppose that $f\in R(\alpha )$ and $f\in R(\beta )$ on $[a,b]$. Then $f\in R(c_{1}\alpha+c_{2}\beta )$ on $[a,b]$ for any constants $c_{1}$ and $c_{2}$. Moreover, we have
\[\int_{a}^{b}fd\left (c_{1}\alpha +c_{2}\beta\right )=c_{1}\int_{a}^{b}fd\alpha +c_{2}\int_{a}^{b}fd\beta .\]

Proof. The proof is similar to the argument of Proposition \ref{map25}, and is left as an exercise. $\blacksquare$

\begin{equation}{\label{map31}}\tag{7}\mbox{}\end{equation}
Proposition \ref{map31}. Given any $c\in (a,b)$, suppose that $f\in R(\alpha )$ on $[a,c]$ and $[c,b]$. Then $f\in R(\alpha )$ on $[a,b]$. Moreover, we have
\begin{equation}{\label{maeq26}}\tag{8}
\int_{a}^{c}fd\alpha +\int_{c}^{b}fd\alpha =\int_{a}^{b}fd\alpha .
\end{equation}

Proof. Let ${\cal P}$ be a partition of $[a,b]$ with $c\in {\cal P}$. Let ${\cal P}’={\cal P}\cap [a,c]$ and ${\cal P}”={\cal P}\cap [c,b]$ denote the corresponding partitions of $[a,c]$ and $[c,b]$, respectively. Therefore, we have
\begin{equation}{\label{maeq27}}\tag{9}
S({\cal P},f,\alpha )=S({\cal P}’,f,\alpha )+S({\cal P}”,f,\alpha ).
\end{equation}
Since $f\in R(\alpha )$ on $[a,c]$ and $[c,b]$, given any $\epsilon >0$, there exist partitions ${\cal P}'(\epsilon)$ of $[a,c]$ and ${\cal P}”(\epsilon)$ of $[c,b]$ such that ${\cal P}’\supseteq {\cal P}'(\epsilon)$ and ${\cal P}”\supseteq {\cal P}”(\epsilon)$ imply
\begin{equation}{\label{maeq28}}\tag{10}
\left |S({\cal P}’,f,\alpha )-\int_{a}^{c}fd\alpha\right |<\frac{\epsilon}{2}
\mbox{ and }\left |S({\cal P}”,f,\alpha )-\int_{c}^{b}fd\alpha\right |<\frac{\epsilon}{2}
\end{equation}
We take ${\cal P}(\epsilon)={\cal P}'(\epsilon)\cup {\cal P}”(\epsilon)$. Then ${\cal P}\supseteq {\cal P}(\epsilon)$ implies ${\cal P}’\supseteq {\cal P}'(\epsilon)$ and ${\cal P}”\supseteq {\cal P}”(\epsilon)$. Now, for ${\cal P}\supseteq {\cal P}(\epsilon)$, using (\ref{maeq27}) and (\ref{maeq28}), we obtain
\begin{align*}
& \left |S({\cal P},f,\alpha )-\int_{a}^{c}fd\alpha-\int_{c}^{b}fd\alpha\right |\\
& \quad =\left |S({\cal P}’,f,\alpha )+S({\cal P}”,f,\alpha )-\int_{a}^{c}fd\alpha -\int_{c}^{b}fd\alpha\right |\\
& \quad \leq\left |S({\cal P}’,f,\alpha )-\int_{a}^{c}fd\alpha\right |+\left |S({\cal P}”,f,\alpha )-\int_{c}^{b}fd\alpha\right |\\
& \quad <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
\end{align*}
This shows that $f\in R(\alpha )$ on $[a,b]$ and the desired equality (\ref{maeq26}) holds true. This completes the proof. $\blacksquare$

Using the mathematical induction, we can prove a similar result for a decomposition of $[a,b]$ into a finite number of sub-intervals. If $a<b$, we define
\[\int_{b}^{a}fd\alpha =-\int_{a}^{b}fd\alpha\]
whenever $f\in R(\alpha )$ on $[a,b]$. We also define
\[\int_{a}^{a}fd\alpha =0.\]
Therefore, the equality (\ref{maeq26}) can be rewritten as
\[\int_{a}^{b}fd\alpha +\int_{b}^{c}fd\alpha +\int_{c}^{a}fd\alpha =0.\]

\begin{equation}{\label{map61}}\tag{11}\mbox{}\end{equation}
Proposition \ref{map61}. Let $\alpha$ be increasing on $[a,b]$. Suppose that $f\in R(\alpha )$ and $g\in R(\alpha )$ on $[a,b]$, and that $f(x)\leq g(x)$ for all $x\in [a,b]$. Then, we have
\[\int_{a}^{b}f(x)d\alpha (x)\leq\int_{a}^{b}g(x)d\alpha (x).\]

Proof. Given any partition ${\cal P}$, since $\Delta\alpha_{k}\geq 0$ for all $k$, the corresponding Riemann-Stieltjes sums satisfy
\begin{align*} S({\cal P},f,\alpha ) & =\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}\\ & \leq\sum_{k=1}^{n}g(t_{k})\Delta\alpha_{k}\\ & =S({\cal P},g,\alpha ).\end{align*}
Let
\[A=\int_{a}^{b}f(x)d\alpha (x)\mbox{ and }B=\int_{a}^{b}g(x)d\alpha (x).\]
By definition, given any $\epsilon>0$, there exists partitions ${\cal P}'(\epsilon)$ and ${\cal P}”(\epsilon)$ satisfying
\[\left |S({\cal P},f,\alpha )-A\right |<\frac{\epsilon}{2}\mbox{ for }{\cal P}\supseteq {\cal P}'(\epsilon)\]
and
\[\left |S({\cal P}(\epsilon),g,\alpha )-B\right |<\frac{\epsilon}{2}\mbox{ for }{\cal P}\supseteq {\cal P}”(\epsilon).\]

Let ${\cal P}(\epsilon)={\cal P}'(\epsilon)\cup{\cal P}”(\epsilon)$. Then
\[\left |S({\cal P}(\epsilon),f,\alpha )-A\right |<\frac{\epsilon}{2}\mbox{ and }
\left |S({\cal P}(\epsilon),g,\alpha )-B\right |<\frac{\epsilon}{2},\]
i.e.,
\[A-\frac{\epsilon}{2}<S({\cal P}(\epsilon),f,\alpha )<A+\frac{\epsilon}{2}\] and
\[B-\frac{\epsilon}{2}<S({\cal P}(\epsilon),g,\alpha )<B+\frac{\epsilon}{2}.\]
Therefore, we obtain
\begin{align*} A-\frac{\epsilon}{2} & <S({\cal P}(\epsilon),f,\alpha )\\ & \leq S({\cal P}(\epsilon),g,\alpha )
\\ & <B+\frac{\epsilon}{2},\end{align*}
which says that $A<B+\epsilon$. Since $\epsilon$ can be any positive number, we must have $A\leq B$. This completes the proof. $\blacksquare$

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Upper and Lower Riemann-Stieltjes Integrals.

We are going to discuss the upper and lower integrals for the Riemann-Stietjes integrals. Let ${\cal P}$ be a partition of $[a,b]$, and let
\begin{equation}{\label{maeq58}}\tag{12}
M_{k}(f)=\sup_{x\in [x_{k-1},x_{k}]}f(x)\mbox{ and }m_{k}(f)=\inf_{x\in [x_{k-1},x_{k}]}f(x).
\end{equation}
The number
\[U({\cal P},f,\alpha )=\sum_{k=1}^{n}M_{k}(f)\Delta\alpha_{k}\]
is called the upper Riemann-Stieltjes sum of $f$ with respect to $\alpha$ for the partition ${\cal P}$. Also, the number
\[L({\cal P},f,\alpha )=\sum_{k=1}^{n}m_{k}(f)\Delta\alpha_{k}\]
is called the lower Riemann-Stieltjes sum of $f$ with respect to $\alpha$ for the partition ${\cal P}$.

\begin{equation}{\label{ma108}}\tag{13}\mbox{}\end{equation}

Remark \ref{ma108}. We have
\begin{align*} M_{k}(-f) & =\sup_{x\in [x_{k-1},x_{k}]}(-f(x))\\ & =-\inf_{x\in [x_{k-1},x_{k}]}f(x)=-m_{k}(f)\end{align*}
and
\begin{align*} m_{k}(-f) & =\inf_{x\in [x_{k-1},x_{k}]}(-f(x))\\ & =-\sup_{x\in [x_{k-1},x_{k}]}f(x)=-M_{k}(f).\end{align*}
It follows
\begin{align*} U({\cal P},-f,\alpha ) & =\sum_{k=1}^{n}M_{k}(-f)\Delta\alpha_{k}\\ & =-\sum_{k=1}^{n}m_{k}(f)\Delta\alpha_{k}\\ & =-L({\cal P},f,\alpha )\end{align*}
and
\begin{align*} L({\cal P},-f,\alpha ) & =\sum_{k=1}^{n}m_{k}(-f)\Delta\alpha_{k}\\ & =-\sum_{k=1}^{n}M_{k}(f)\Delta\alpha_{k}\\ & =-U({\cal P},f,\alpha )\end{align*}

\begin{equation}{\label{ma109}}\tag{14}\mbox{}\end{equation}

Remark \ref{ma109}. When the integrator $\alpha$ is increasing on $[a,b]$, we have $\Delta\alpha_{k}\geq 0$. Therefore, we have
\[m_{k}(f)\Delta\alpha_{k}\leq M_{k}(f)\Delta\alpha_{k},\]
which implies
\begin{align*} L({\cal P},f,\alpha ) & =\sum_{k=1}^{n}m_{k}(f)\Delta\alpha_{k}\\ & \leq\sum_{k=1}^{n}M_{k}(f)\Delta\alpha_{k}\\ & =U({\cal P},f,\alpha ).\end{align*}
On the other hand, for $t_{k}\in [x_{k-1},x_{k}]$, we also have
\[m_{k}(f)\leq f(t_{k})\leq M_{k}(f),\]
which implies
\begin{equation}{\label{maeq48}}\tag{15}
L({\cal P},f,\alpha )\leq S({\cal P},f,\alpha )\leq U({\cal P},f,\alpha ).
\end{equation}
The inequalities (\ref{maeq48}) does not necessarily hold true when $\alpha$ is not an increasing function on $[a,b]$. $\sharp$

\begin{equation}{\label{ma3}}\tag{16}\mbox{}\end{equation}
Proposition \ref{ma3}. Suppose that the real-valued function $\alpha$ is increasing on $[a,b]$. Given any two functions $f$ and $g$ defined on $[a,b]$, we have
\[U({\cal P},f+g,\alpha )\leq U({\cal P},f,\alpha )+U({\cal P},g,\alpha )\]
and
\[L({\cal P},f+g,\alpha )\geq L({\cal P},f,\alpha )+L({\cal P},g,\alpha ).\]
We also have
\[U({\cal P},f-g,\alpha )\leq U({\cal P},f,\alpha )-L({\cal P},g,\alpha )\]
and
\[L({\cal P},f-g,\alpha )\geq L({\cal P},f,\alpha )-U({\cal P},g,\alpha ).\]

Proof. We have
\begin{align*}
M_{k}(f+g) & =\sup_{x\in [x_{k-1},x_{k}]}\left (f(x)+g(x)\right )\\
& \leq\sup_{x\in [x_{k-1},x_{k}]}f(x)+\sup_{x\in [x_{k-1},x_{k}]}g(x)\\ & =M_{k}(f)+M_{k}(g)
\end{align*}
and
\begin{align*}
m_{k}(f+g) & =\inf_{x\in [x_{k-1},x_{k}]}\left (f(x)+g(x)\right )\\
& \geq\inf_{x\in [x_{k-1},x_{k}]}f(x)+\inf_{x\in [x_{k-1},x_{k}]}g(x)\\ & =m_{k}(f)+m_{k}(g).
\end{align*}
Since $\Delta\alpha_{k}\geq 0$ for all $k$, we obtain
\begin{align*}
U({\cal P},f+g,\alpha ) & =\sum_{k=1}^{n}M_{k}(f+g)\Delta\alpha_{k}\\
& \leq\sum_{k=1}^{n}M_{k}(f)\Delta\alpha_{k}+\sum_{k=1}^{n}M_{k}(g)\Delta\alpha_{k}\\ & =U({\cal P},f,\alpha )+U({\cal P},g,\alpha )
\end{align*}
and
\begin{align*}
L({\cal P},f+g,\alpha ) & =\sum_{k=1}^{n}m_{k}(f+g)\Delta\alpha_{k}\\
& \geq\sum_{k=1}^{n}m_{k}(f)\Delta\alpha_{k}+\sum_{k=1}^{n}m_{k}(g)\Delta\alpha_{k}\\ & =L({\cal P},f,\alpha )+L({\cal P},g,\alpha ).
\end{align*}
Now, using Remark \ref{ma108}, we also have
\begin{align*} U({\cal P},f-g,\alpha ) & \leq U({\cal P},f,\alpha )+U({\cal P},-g,\alpha )\\ & =U({\cal P},f,\alpha )-L({\cal P},g,\alpha )\end{align*}
and
\begin{align*} L({\cal P},f-g,\alpha ) & \geq L({\cal P},f,\alpha )+L({\cal P},-g,\alpha )\\ & =L({\cal P},f,\alpha )-U({\cal P},g,\alpha ).\end{align*}
This completes the proof. $\blacksquare$

\begin{equation}{\label{map51}}\tag{17}\mbox{}\end{equation}
Proposition \ref{map51}. Suppose that the real-valued function $\alpha$ is increasing on $[a,b]$. Then, we have the following properties.

(i) Suppose that ${\cal P}’$ is finer than ${\cal P}$. Then, we have $U({\cal P}’,f,\alpha )\leq U({\cal P},f,\alpha )$ and $L({\cal P}’,f,\alpha )\geq L({\cal P},f,\alpha )$.

(ii) Given any two partitions ${\cal P}_{1}$ and ${\cal P}_{2}$, we have $L({\cal P}_{1},f,\alpha )\leq U({\cal P}_{2},f,\alpha )$.

Proof. To prove part (i), it suffices to consider the partition ${\cal P}’$ that contains exactly one more point than ${\cal P}$. In this case, this point is said to be $c$. Suppose that $c$ is in the $i$th subinterval of ${\cal P}$. Then, we have
\[U({\cal P},f,\alpha )=M_{i}(f)\left [\alpha (x_{i})-\alpha (x_{i-1})\right ]+\sum_{k=1,k\neq i}^{n}M_{k}(f)\Delta\alpha_{k}\]
and
\[U({\cal P}’,f,\alpha )=M_{i}^{(1)}\left [\alpha (c)-\alpha (x_{i-1})\right ]+M_{i}^{(2)}\left [\alpha (x_{i})-\alpha (c)\right ]
+\sum_{k=1,k\neq i}^{n}M_{k}(f)\Delta\alpha_{k},\]
where
\[M_{i}^{(1)}(f)=\sup_{x\in [x_{i-1},c]}f(x)\mbox{ and }M_{i}^{(2)}(f)=\sup_{x\in [c,x_{i}]}f(x).\]
It follows
\[M_{i}^{(1)}(f)\leq M_{i}(f)\mbox{ and }M_{i}^{(2)}(f)\leq M_{i}(f).\]
Since
\[\alpha (c)-\alpha (x_{i-1})\geq 0\mbox{ and }\alpha (x_{i})-\alpha (c)\geq 0,\]
we obtain $U({\cal P}’,f,\alpha )\leq U({\cal P},f,\alpha )$. We can similarly prove $L({\cal P}’,f,\alpha )\geq L({\cal P},f,\alpha )$.

To prove part (ii), let ${\cal P}={\cal P}_{1}\cup {\cal P}_{2}$. Then, using part (i) and Remark \ref{ma109}, we have
\begin{align*} L({\cal P}_{1},f,\alpha ) & \leq L({\cal P},f,\alpha )\\ & \leq U({\cal P},f,\alpha )\\ & \leq U({\cal P}_{2},f,\alpha ).\end{align*}
This completes the proof. $\blacksquare$

Definition. Let $\alpha$ be a real-valued function defined on $[a,b]$. The upper Riemann-Stieltjes integral of $f$ with respect to $\alpha$ on $[a,b]$ is defined by
\[\overline{\int}_{a}^{b}fd\alpha =\inf_{{\cal P}\in\mathfrak{P}[a,b]}U({\cal P},f,\alpha )\]
and the lower Riemann-Stieltjes integral of $f$ with respect to $\alpha$ on $[a,b]$ is defined by
\[\underline{\int}_{a}^{b}fd\alpha =\sup_{{\cal P}\in\mathfrak{P}[a,b]}L({\cal P},f,\alpha ).\]

\begin{equation}{\label{map55}}\tag{18}\mbox{}\end{equation}
Proposition \ref{map55}. Suppose that the real-valued function $\alpha$ is increasing on $[a,b]$. Then, we have
\[\underline{\int}_{a}^{b}fd\alpha\leq\overline{\int}_{a}^{b}fd\alpha .\]

Proof. According to the concept of infimum, given any $\epsilon >0$, there exists a partition ${\cal P}_{1}$ satisfying
\[U({\cal P}_{1},f,\alpha )<\overline{\int}_{a}^{b}fd\alpha +\epsilon .\]
Using part (ii) of Proposition \ref{map51}, we also have
\[L({\cal P},f,\alpha )\leq U({\cal P}_{1},f,\alpha )<\overline{\int}_{a}^{b}fd\alpha +\epsilon\]

for any ${\cal P}\in\mathfrak{P}[a,b]$. By taking supremum on both sides, we obtain
\begin{align*} \underline{\int}_{a}^{b}fd\alpha & =\sup_{{\cal P}\in\mathfrak{P}[a,b]}L({\cal P},f,\alpha )\\ & \leq\overline{\int}_{a}^{b}fd\alpha +\epsilon .\end{align*}
Since $\epsilon$ can be any positive number, we obtain the desired inequality. $\blacksquare$

Proposition. Suppose that the real-valued function $\alpha$ is increasing on $[a,b]$. Then, we have the following properties.

(i) For $a<c<b$, we have
\[\overline{\int}_{a}^{b}fd\alpha =\overline{\int}_{a}^{c}fd\alpha+\overline{\int}_{c}^{b}fd\alpha\]
and \[\underline{\int}_{a}^{b}fd\alpha =\underline{\int}_{a}^{c}fd\alpha+\underline{\int}_{c}^{b}fd\alpha .\]

(ii) We have
\[\overline{\int}_{a}^{b}(f+g)d\alpha\leq\overline{\int}_{a}^{b}fd\alpha+\overline{\int}_{a}^{b}gd\alpha\] and
\[\underline{\int}_{a}^{b}(f+g)d\alpha\geq\underline{\int}_{a}^{b}fd\alpha+\underline{\int}_{a}^{b}gd\alpha .\]

Proof. The proofs are left as exercises by referring to Proposition \ref{ma3}.

Definition. Let $\alpha$ be a real-valued function defined on $[a,b]$. We say that $f$ satisfies the Riemann’s condition with respect to $\alpha$ on $[a,b]$ when, for every $\epsilon >0$, there exists a partition ${\cal P}(\epsilon )$ such that, for any partition ${\cal P}$ that is finer than ${\cal P}(\epsilon )$, we have $U({\cal P},f,\alpha )-L({\cal P},f,\alpha )<\epsilon$. $\sharp$

\begin{equation}{\label{mat62}}\tag{19}\mbox{}\end{equation}
Theorem \ref{mat62}. Suppose that $\alpha$ is increasing on $[a,b]$. Then, the following statements are equivalent:

(a) $f\in R(\alpha )$ on $[a,b]$;

(b) $f$ satisfying the Riemann’s condition with respect to $\alpha$ on $[a,b]$;

Moreover, if (c) holds true, then $f\in R(\alpha )$ on $[a,b]$ and
\[\overline{\int}_{a}^{b}fd\alpha =\underline{\int}_{a}^{b}fd\alpha =\int_{a}^{b}fd\alpha .\]

Proof. In order to prove the equivalence, we shall prove that (a) implies(b), (b) implies (c), and (c) implies(a). Suppose that (a) holds true. Since $\alpha$ is assumed to be increasing on $[a,b]$, if $\alpha (a)=\alpha (b)$ then $\alpha$ is a constant function on $[a,b]$. In this case, it is obvious that $f$ satisfies the Riemann’s condition. Now, we assume $\alpha (a)\neq\alpha (b)$, i.e., $\alpha (a)<\alpha (b)$. Given any $\epsilon>0$, there exists a partition ${\cal P}(\epsilon )$ such that, for any partition ${\cal P}$ finer than ${\cal P}(\epsilon )$ and any choices of $t_{k}, s_{k}\in [x_{k-1},x_{k}]$, we have
\[\left |\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}-\int_{a}^{b}fd\alpha\right |<\frac{\epsilon}{3}\mbox{ and }\left |\sum_{k=1}^{n}f(s_{k})\Delta\alpha_{k}-\int_{a}^{b}fd\alpha\right |<\frac{\epsilon}{3},\]
which implies
\begin{align}
\left |\sum_{k=1}^{n}\left [f(t_{k})-f(s_{k})\right ]\Delta\alpha_{k}\right |
& \leq\left |\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}-\int_{a}^{b}fd\alpha\right |
+\left |\int_{a}^{b}fd\alpha-\sum_{k=1}^{n}f(s_{k})\Delta\alpha_{k}\right |\nonumber\\
& <\frac{2}{3}\epsilon .\label{maeq53}\tag{20}
\end{align}
Since
\begin{align*}
M_{k}(f)-m_{k}(f) & =\sup_{z\in [x_{k-1},x_{k}]}f(z)-\inf_{z\in [x_{k-1},x_{k}]}f(z)\\
& =\sup_{z_{1}\in [x_{k-1},x_{k}]}f(z_{1})+\sup_{z_{2}\in [x_{k-1},x_{k}]}[-f(z_{2})]
\\ & =\sup_{z_{1},z_{2}\in [x_{k-1},x_{k}]}\left [f(z_{1})-f(z_{2})\right ],
\end{align*}
given any $\hat{\epsilon}$, we can take $t_{k}$ and $s_{k}$ satisfying
\begin{equation}{\label{maeq52}}\tag{21}
f(t_{k})-f(s_{k})>M_{k}(f)-m_{k}(f)-\hat{\epsilon}.
\end{equation}
Now, we take
\begin{equation}{\label{maeq54}}\tag{22}
\hat{\epsilon}=\frac{\epsilon}{3[\alpha (b)-\alpha (a)]}.
\end{equation}
Since $\Delta\alpha_{k}\geq 0$ for all $k$, using (\ref{maeq52}), (\ref{maeq53}) and (\ref{maeq54}) (in order), we obtain
\begin{align*}
U({\cal P},f,\alpha )-L({\cal P},f,\alpha ) & =\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta\alpha_{k}\\
& <\sum_{k=1}^{n}\left [f(t_{k})-f(s_{k})\right ]\Delta\alpha_{k}+\hat{\epsilon}\sum_{k=1}^{n}\Delta\alpha_{k}\\
& <\frac{2}{3}\epsilon +\hat{\epsilon}\cdot\left [\alpha (b)-\alpha (a)\right ]\\
& =\epsilon.
\end{align*}
This proves that (a) implies (b).

Suppose that (b) holds true. Given any $\epsilon >0$, there exists a partition ${\cal P}(\epsilon )$ such that, for any partition ${\cal P}$ finer than ${\cal P}(\epsilon )$, we have
\[U({\cal P},f,\alpha )<L({\cal P},f,\alpha )+\epsilon ,\]
which implies
\begin{align*}
\overline{\int}_{a}^{b}fd\alpha & =\inf_{{\cal P}\in\mathfrak{P}[a,b]}U({\cal P},f,\alpha )\leq U({\cal P},f,\alpha )<L({\cal P},f,\alpha )+\epsilon\\
& \leq\sup_{{\cal P}\in\mathfrak{P}[a,b]}L({\cal P},f,\alpha )=\underline{\int}_{a}^{b}fd\alpha +\epsilon .
\end{align*}
Since $\epsilon$ can be any positive number, we obtain
\[\overline{\int}_{a}^{b}fd\alpha\leq\underline{\int}_{a}^{b}fd\alpha .\]
Using Proposition \ref{map55}, we conclude that (b) implies (c).

Finally, we assume that (c) holds true. Let $A$ denote the common value
\begin{equation}{\label{maeq56}}\tag{23}
\overline{\int}_{a}^{b}fd\alpha =\underline{\int}_{a}^{b}fd\alpha =A.
\end{equation}
Since
\[\overline{\int}_{a}^{b}fd\alpha=\inf_{{\cal P}\in\mathfrak{P}[a,b]}U({\cal P},f,\alpha ),\]
according to the concept of infimum, given any $\epsilon >0$, there exists a partition ${\cal P}'(\epsilon )$ satisfying
\[U({\cal P}'(\epsilon ),f,\alpha )<\overline{\int}_{a}^{b}fd\alpha +\epsilon .\]
For any partition ${\cal P}$ finer than ${\cal P}'(\epsilon )$, using part (i) of Proposition \ref{map51}, we have
\[U({\cal P},f,\alpha )\leq U({\cal P}'(\epsilon ),f,\alpha )<\overline{\int}_{a}^{b}fd\alpha +\epsilon .\]
Since
\[L({\cal P},f,\alpha )=\sup_{{\cal P}\in\mathfrak{P}[a,b]}L({\cal P},f,\alpha ),\]
according to the concept of supremum, we can also take a partition ${\cal P}”(\epsilon )$ satisfying
\[L({\cal P}”(\epsilon ),f,\alpha )>\underline{\int}_{a}^{b}fd\alpha -\epsilon .\]
For any partition ${\cal P}$ finer than ${\cal P}”(\epsilon )$, using part (i) of Proposition \ref{map51}, we also have
\[L({\cal P},f,\alpha )\geq L({\cal P}”(\epsilon ),f,\alpha )>\underline{\int}_{a}^{b}fd\alpha -\epsilon .\]
Now, we take ${\cal P}(\epsilon )={\cal P}'(\epsilon )\cup {\cal P}”(\epsilon )$. For any partition ${\cal P}$ finer than ${\cal P}(\epsilon )$, we have
\begin{equation}{\label{maeq57}}\tag{24}
\underline{\int}_{a}^{b}fd\alpha -\epsilon <L({\cal P},f,\alpha )\leq
S({\cal P},f,\alpha )\leq U({\cal P},f,\alpha )<\overline{\int}_{a}^{b}fd\alpha+\epsilon .
\end{equation}
Combining (\ref{maeq56}) and (\ref{maeq57}), we obtain
\[\left |S({\cal P},f,\alpha )-A\right |<\epsilon\]
for any partition ${\cal P}$ finer than ${\cal P}(\epsilon )$. This proves that (c) implies (a) and
\[A=\overline{\int}_{a}^{b}fd\alpha=\underline{\int}_{a}^{b}fd\alpha =\int_{a}^{b}fd\alpha .\]
This completes the proof. $\blacksquare$

Let $\alpha$ be a function of bounded variation on $[a,b]$. Theorem \ref{mat18} says $\alpha =\alpha_{1}-\alpha_{2}$ for some increasing functions $\alpha_{1}$ and $\alpha_{2}$ on $[a,b]$. Suppose that $f\in R(\alpha_{1})$ and $f\in R(\alpha_{2})$. Then, by Proposition \ref{map40}, we have $f\in R(\alpha )$. Conversely, for $f\in R(\alpha )$ on $[a,b]$, since the difference $\alpha =\alpha_{1}-\alpha_{2}$ is not unique, it is quite possible to choose increasing functions $\alpha_{1}$ and $\alpha_{2}$ such that either $f\not\in R(\alpha_{1})$ or $f\not\in R(\alpha_{2})$. However, we can prove that there is at least one decomposition for which the converse is true, when $\alpha_{1}$ is the total variation of $\alpha$ and $\alpha_{2}=\alpha -\alpha_{1}$. This is shown below.

\begin{equation}{\label{map15}}\tag{25}\mbox{}\end{equation}
Lemma \ref{map15}. Suppose that the real-valued function $f$ is of bounded variation on $[a,b]$. Let $V$ be a real-valued function defined on $[a,b]$ by
\begin{equation}{\label{maeq19}}\tag{26}
V(x)=\left\{\begin{array}{ll}
V_{f}(a,x) & \mbox{if $a<x\leq b$}\\
0 & \mbox{if $x=a$}.
\end{array}\right .
\end{equation}
Then, we have the following properties.

(i) $V$ is an increasing function on $[a,b]$.

(ii) $V-f$ is an increasing function on $[a,b]$.

The above Lemma \ref{map15} can refer to the page Functions of Bounded Variation.

\begin{equation}{\label{map75}}\tag{27}\mbox{}\end{equation}
Proposition \ref{map75}. Let $\alpha$ be of bounded variation on $[a,b]$. We define
\begin{equation}{\label{maeq76}}\tag{28}
\beta (x)=\left\{\begin{array}{ll}
V_{\alpha}(a,x) & \mbox{if $a<x\leq b$}\\
0 & \mbox{if $x=a$}.
\end{array}\right .
\end{equation}
Let $f$ be bounded on $[a,b]$. Suppose that $f\in R(\alpha )$ on $[a,b]$. Then $f\in R(\beta )$ on $[a,b]$ and $f\in R(\beta -\alpha )$ on $[a,b]$, where $\beta$ and $\beta -\alpha$ are increasing functions on $[a,b]$.

Proof. Suppose that $\beta (b)=0$. Then $\beta$ is a constant function and the result is obvious. Now, we assume that $\beta (b)>0$. By the hypothesis, we also have $|f(x)|\leq M$ for $x\in [a,b]$. Since $\beta$ is increasing, in order to prove $f\in R(\beta )$ on $[a,b]$ by using Theorem \ref{mat62}, we only need to verify that $f$ satisfies the Riemann’s condition with respect to $\beta$ on $[a,b]$. Since $f\in R(\alpha )$ on $[a,b]$, given any $\epsilon >0$, there exists a partition ${\cal P}(\epsilon )$ such that, for any partition ${\cal P}$ finer than ${\cal P}(\epsilon )$ and any $t_{k},s_{k}\in [x_{k-1},x_{k}]$, we have
\[\left |\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}-\int_{a}^{b}fd\alpha\right |<\frac{\epsilon}{8}
\mbox{ and }\left |\sum_{k=1}^{n}f(s_{k})\Delta\alpha_{k}-\int_{a}^{b}fd\alpha\right |<\frac{\epsilon}{8},\]
which imply
\begin{align}
\left |\sum_{k=1}^{n}\left [f(t_{k})-f(s_{k})\right ]\Delta\alpha_{k}\right | & \leq
\left |\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}-\int_{a}^{b}fd\alpha\right |
+\left |\int_{a}^{b}fd\alpha-\sum_{k=1}^{n}f(s_{k})\Delta\alpha_{k}\right |\nonumber\\
& <\frac{\epsilon}{8}+\frac{\epsilon}{8}=\frac{\epsilon}{4}.\label{maeq80}\tag{29}
\end{align}
By the concept of supremum, since
\[0<\beta (b)=V_{\alpha}(a,b)=\sup_{{\cal P}\in\mathfrak{P}[a,b]}V_{\alpha}({\cal P},[a,b]),\]
there exists a partition ${\cal P}$ satisfying
\begin{equation}{\label{maeq71}}\tag{30}
\beta (b)<V_{\alpha}({\cal P},[a,b])+\frac{\epsilon}{4M}=\sum_{k=1}^{n}\left |\Delta\alpha_{k}\right |+\frac{\epsilon}{4M}.
\end{equation}
Therefore, using (\ref{maeq71}), we obtain
\begin{align}
& \sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\left (\Delta\beta_{k}-|\Delta\alpha_{k}|\right )
=\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta\beta_{k}
-\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]|\Delta\alpha_{k}|\nonumber\\
& \quad\leq 2M\sum_{k=1}^{n}\Delta\beta_{k}-2M\sum_{k=1}^{n}|\Delta\alpha_{k}|
=2M(\beta(b)-\beta(a))-2M\sum_{k=1}^{n}|\Delta\alpha_{k}|\label{maeq72}\tag{31}\\
& \quad=2M\left (\beta (b)-\sum_{k=1}^{n}\left |\Delta\alpha_{k}\right |\right )<\frac{\epsilon}{2}.\nonumber
\end{align}
Let
\[P({\cal P})=\{k:\Delta\alpha_{k}\geq 0\}\mbox{ and }N({\cal P})=\{k:\Delta\alpha_{k}<0\}.\]
Since
\[M_{k}(f)=\sup_{x\in [x_{k-1},x_{k}]}f(x)\mbox{ and }m_{k}(f)=\inf_{x\in [x_{k-1},x_{k}]}f(x).\]
For $k\in P({\cal P})$, by the concepts of supremum and infimum, given $h=\frac{1}{4}\epsilon /\beta (b)$, we can choose $t_{k},s_{k}\in [x_{k-1},x_{k}]$ satisfying
\[f(t_{k})+\frac{h}{2}>M_{k}(f)\mbox{ and }f(s_{k})<m_{k}(f)+\frac{h}{2},\]
which imply
\begin{equation}{\label{ma106}}\tag{32}
f(t_{k})-f(s_{k})+h>M_{k}(f)-m_{k}(f).
\end{equation}
For $k\in N({\cal P})$, we can also choose $s_{k},t_{k}\in [x_{k-1},x_{k}]$ satisfying
\[f(s_{k})+\frac{h}{2}>M_{k}(f)\mbox{ and }f(t_{k})<m_{k}(f)+\frac{h}{2},\]
which imply
\begin{equation}{\label{ma107}}\tag{33}
f(s_{k})-f(t_{k})+h>M_{k}(f)-m_{k}(f).
\end{equation}
Then, using (\ref{ma106}), (\ref{ma107}) and (\ref{maeq80}) (in order), we have
\begin{align}
& \sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\left |\Delta\alpha_{k}\right |\label{maeq74}\tag{34}\\
& \quad <\sum_{k\in P({\cal P})}\left [f(t_{k})-f(s_{k})\right ]\left |\Delta\alpha_{k}\right |+\sum_{k\in N({\cal P})}\left [f(s_{k})-f(t_{k})\right ]\left |\Delta\alpha_{k}\right |+h\sum_{k=1}^{n}\left |\Delta\alpha_{k}\right |\nonumber\\
& \quad =\sum_{k=1}^{n}\left [f(t_{k})-f(s_{k})\right ]\Delta\alpha_{k}+h\sum_{k=1}^{n}\left |\Delta\alpha_{k}\right |\nonumber\\
& \quad\leq\sum_{k=1}^{n}\left [f(t_{k})-f(s_{k})\right ]\Delta\alpha_{k}+h\cdot\sup_{{\cal P}\in\mathfrak{P}[a,b]}V_{\alpha}({\cal P},[a,b])\nonumber\\
& \quad =\sum_{k=1}^{n}\left [f(t_{k})-f(s_{k})\right ]\Delta\alpha_{k}+h\beta(b)<\frac{\epsilon}{4}+\frac{\epsilon}{4}=\frac{\epsilon}{2}.\nonumber
\end{align}
Now, using (\ref{maeq72}) and (\ref{maeq74}), we have
\begin{align*}
& U({\cal P},f,\beta )-L({\cal P},f,\beta )
=\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta\beta_{k}\\
& \quad =\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\left (\Delta\beta_{k}-|\Delta\alpha_{k}|
\right )+\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\left |\Delta\alpha_{k}\right |\\
& \quad <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
\end{align*}
This shows that $f$ satisfies the Riemann’s condition with respect to $\beta$ on $[a,b]$, i.e., $f\in R(\beta )$ on $[a,b]$. Now, we have $\alpha =\beta -(\beta -\alpha )$, where $\beta$ and $\beta -\alpha$ are increasing functions on $[a,b]$ by Lemma \ref{map15}. Since $f\in R(\beta )$ on $[a,b]$, Proposition \ref{map40} says that $f\in R(\beta -\alpha )$ on $[a,b]$. This completes the proof. $\blacksquare$

\begin{equation}{\label{mat95}}\tag{35}\mbox{}\end{equation}
Proposition \ref{mat95}. Let $\alpha$ be of bounded variation on $[a,b]$. Suppose that $f\in R(\alpha )$ on $[a,b]$. Then $f\in R(\alpha )$ on every sub-interval $[c,d]$ of $[a,b]$.

Proof. We first assume that $\alpha$ is increasing on $[a,b]$. Let ${\cal P}$ be a partition of $[a,x]$. We define
\[\Delta ({\cal P},x)=U({\cal P},f,\alpha )-L({\cal P},f,\alpha )\]
as the difference of upper and lower sums associated with the interval $[a,x]$. Since $f\in R(\alpha )$ on $[a,b]$, the Riemann’s condition holds true with respect to $\alpha$ on $[a,b]$. Therefore, given any $\epsilon >0$, there exists a partition ${\cal P}(\epsilon )$ of $[a,b]$ such that
\begin{equation}{\label{maeq79}}\tag{36}
{\cal P}\supseteq {\cal P}(\epsilon )\mbox{ implies }\Delta ({\cal P},b)<\epsilon .
\end{equation}
Assume that $a<c<b$ with $c\in {\cal P}(\epsilon )$. In this case, the points of ${\cal P}(\epsilon )$ in $[a,c]$ forms a partition ${\cal P}'(\epsilon )$ of $[a,c]$. Let ${\cal P}’$ be a partition of $[a,c]$, which is finer than ${\cal P}'(\epsilon )$. Then ${\cal P}={\cal P}’\cup {\cal P}(\epsilon )$ is a partition of $[a,b]$. Therefore, the sum defining $\Delta ({\cal P}’,c)$ contains only part of the terms in the sum defining $\Delta ({\cal P},b)$. Since each term is nonnegative and ${\cal P}$ is finer than ${\cal P}(\epsilon )$, using (\ref{maeq79}), we have
\[\Delta ({\cal P}’c)\leq\Delta ({\cal P},b)<\epsilon .\]
This says that
\[{\cal P}’\supseteq {\cal P}'(\epsilon )\mbox{ implies }\Delta ({\cal P}’,c)<\epsilon,\]
i.e., $f$ satisfies the Riemann’s condition on $[a,c]$. Theorem \ref{mat62} says that $f\in R(\alpha )$ on $[a,c]$. Similarly, we can show that $f\in R(\alpha )$ on $[a,d]$. Using Proposition \ref{map31}, we obtain $f\in R(\alpha )$ on $[c,d]$. Let $\beta$ be defined in (\ref{maeq76}). Proposition \ref{map75} says $f\in R(\beta )$ on $[a,b]$ and $f\in R(\beta -\alpha )$ on $[a,b]$, where $\beta$ and $\beta -\alpha$ are increasing on $[a,b]$ satisfying $\alpha =\beta -(\beta -\alpha )$. Using the previous results, we conclude that $f\in R(\beta )$ and $f\in R(\beta -\alpha )$ on $[c,d]$, which implies $f\in R(\alpha )$ on $[c,d]$ by Proposition \ref{map40}. This completes the proof. $\blacksquare$

\begin{equation}{\label{map414}}\tag{37}\mbox{}\end{equation}
Proposition \ref{map414}. Let $\alpha$ be increasing on $[a,b]$. Suppose that $f\in R(\alpha )$ on $[a,b]$. Then $|f|\in R(\alpha )$ on $[a,b]$. We also have
\begin{equation}{\label{maeq60}}\tag{38}
\left |\int_{a}^{b}f(x)d\alpha (x)\right |\leq\int_{a}^{b}|f(x)|d\alpha (x).
\end{equation}

Proof. From (\ref{maeq58}), we can write
\begin{align*} 0 & \leq M_{k}(f)-m_{k}(f)\\ & =\sup_{x,y\in [x_{k-1},x_{k}]}\left [f(x)-f(y)\right ]\\ & =\sup_{x,y\in [x_{k-1},x_{k}]}\left |f(x)-f(y)\right |.\end{align*}
Since $|f(x)|-|f(y)|\leq \left |f(x)-f(y)\right |$, we have
\begin{align}
M_{k}(|f|)-m_{k}(|f|) & =\sup_{x,y\in [x_{k-1},x_{k}]}\left [|f(x)|-|f(y)|\right ]\nonumber\\
& \leq\sup_{x,y\in [x_{k-1},x_{k}]}\left |f(x)-f(y)\right |\\ & =M_{k}(f)-m_{k}(f),\label{maeq59}\tag{39}
\end{align}
Since $\Delta\alpha_{k}\geq 0$, using (\ref{maeq59}), we obtain
\begin{align}
U({\cal P},|f|,\alpha )-L({\cal P},|f|,\alpha ) & =\sum_{k=1}^{n}\left [M_{k}(|f|)-m_{k}(|f|)\right ]\Delta\alpha_{k}\nonumber\\
& \leq\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta\alpha_{k}\nonumber\\
& =U({\cal P},f,\alpha )-L({\cal P},f,\alpha )\label{maeq63}\tag{40}
\end{align}
for every partition ${\cal P}$ of $[a,b]$. Since $f\in R(\alpha )$ on $[a,b]$, Theorem \ref{mat62} says that $f$ satisfies the Riemann’s condition. Therefore, from (\ref{maeq63}), we have
\[U({\cal P},|f|,\alpha )-L({\cal P},|f|,\alpha )\leq U({\cal P},f,\alpha )-L({\cal P},f,\alpha )<\epsilon .\]
Using Theorem \ref{mat62} again, we conclude that $|f|\in R(\alpha )$ on $[a,b]$. Finally, the inequality (\ref{maeq60}) follows immediately from Proposition \ref{map61} by considering $-|f|\leq f\leq |f|$. $\blacksquare$

\begin{equation}{\label{maeq70}}\tag{41}\mbox{}\end{equation}
Proposition \ref{maeq70}. Let $\alpha$ be of bounded variation on $[a,b]$. Suppose that $f$ is bounded and $f\in R(\alpha )$ on $[a,b]$. Then $f^{2}\in R(\alpha )$ on $[a,b]$.

Proof. We first assume that $\alpha$ is increasing on $[a,b]$. Let $M$ be an upper bound for $|f|$ on $[a,b]$, i.e., $|f(x)|\leq M$ for all $x\in [a,b]$. We have
\begin{align*} M_{k}(f^{2}) & =\sup_{x\in [x_{k-1},x_{k}]}f^{2}(x)\\ & =\left (\sup_{x\in [x_{k-1},x_{k}]}|f(x)|\right )^{2}\\ & =(M_{k}(|f|))^{2}\end{align*}
and
\begin{align*} m_{k}(f^{2}) & =\inf_{x\in [x_{k-1},x_{k}]}f^{2}(x)\\ & =\left (\inf_{x\in [x_{k-1},x_{k}]}|f(x)|\right )^{2}\\ & =(m_{k}(|f|))^{2}.\end{align*}
We also have
\[M_{k}(|f|)=\sup_{x\in [x_{k-1},x_{k}]}|f(x)|\leq M\] and
\[m_{k}(|f|)=\inf_{x\in [x_{k-1},x_{k}]}|f(x)|\leq M.\]
Therefore, we obtain
\begin{align*} M_{k}(f^{2})-m_{k}(f^{2}) & =\left [M_{k}(|f|)+m_{k}(|f|)\right ]\cdot \left [M_{k}(|f|)-m_{k}(|f|)
\right ]\\ & \leq 2M\cdot \left [M_{k}(|f|)-m_{k}(|f|)\right ],\end{align*}
which implies
\begin{equation}{\label{ma105}}\tag{42}
U({\cal P},f^{2},\alpha )-L({\cal P},f^{2},\alpha )\leq 2M\cdot\left [U({\cal P},|f|,\alpha )-L({\cal P},|f|,\alpha )\right ].
\end{equation}
Since $|f|\in R(\alpha )$ on $[a,b]$ by Proposition \ref{map414}, Theorem \ref{mat62} says that $|f|$ satisfies the Riemann’s condition. Therefore, given any $\epsilon>0$, there exists a partition ${\cal P}(\epsilon)$ such that
\[{\cal P}\supseteq {\cal P}(\epsilon)\mbox{ implies }U({\cal P},|f|,\alpha )-L({\cal P},|f|,\alpha )<\frac{\epsilon}{2M},\]
which also implies
\[U({\cal P},f^{2},\alpha )-L({\cal P},f^{2},\alpha )\leq 2M\cdot\frac{\epsilon}{2M}=\epsilon\]
by using (\ref{ma105}). This shows that $f^{2}$ also satisfies the Riemann’s condition. Therefore, Theorem \ref{mat62} says $f^{2}\in R(\alpha )$ on $[a,b]$. Now, we assume that $\alpha$ is of bounded variation on $[a,b]$. Since $f\in R(\alpha )$ on $[a,b]$, Proposition \ref{map75} says that $f\in R(\beta )$ on $[a,b]$ and $f\in R(\beta -\alpha )$ on $[a,b]$, where $\beta$ and $\beta -\alpha$ are increasing functions on $[a,b]$. It also follows $f^{2}\in R(\beta )$ on $[a,b]$ and $f^{2}\in R(\beta -\alpha )$ on $[a,b]$ according to the previous results. Using Proposition \ref{map40}, we have
\[f^{2}\in R(\beta-(\beta-\alpha))=R(\alpha)\mbox{ on }[a,b].\]
This completes the proof. $\blacjsquare$

\begin{equation}{\label{map82}}\tag{43}\mbox{}\end{equation}
Proposition \ref{map82}. Let $\alpha$ be of bounded variation on $[a,b]$. Suppose that $f,g$ are bounded and $f,g\in R(\alpha )$ on $[a,b]$. Then, the product $fg\in R(\alpha )$ on $[a,b]$.

Proof. Since
\[2f(x)g(x)=\left [f(x)+g(x)\right ]^{2}-f^{2}(x)-g^{2}(x),\]
the result follows immediately from Propositions \ref{maeq70} and \ref{map25}. $\blacksquare$

Next, we are going to prove the integration by parts which presents the connection between the integrand and integrator in the Riemann-Stieltjes integrals.

\begin{equation}{\label{mat39}}\tag{44}\mbox{}\end{equation}
Theorem \ref{mat39}. (Integration by Parts). Suppose that $f\in R(\alpha )$ on $[a,b]$. Then $\alpha\in R(f)$ on $[a,b]$ and
\[\int_{a}^{b}f(x)d\alpha (x)+\int_{a}^{b}\alpha (x)df(x)=f(b)\alpha (b)-f(a)\alpha (a).\]

Proof. Since $f\in R(\alpha )$ on $[a,b]$, given any $\epsilon >0$, there exists a partition ${\cal P}(\epsilon )$ of $[a,b]$ such that
\begin{equation}{\label{maeq20}}\tag{45}
{\cal P}’\supseteq {\cal P}(\epsilon )\mbox{ implies }\left |S({\cal P}’,f,\alpha )-\int_{a}^{b}fd\alpha\right |<\epsilon .
\end{equation}
Given a partition ${\cal P}=\{a=x_{0},x_{1},\cdots ,x_{n}=b\}$ with ${\cal P}\supseteq {\cal P}(\epsilon )$, we consider an arbitrary Riemann-Stieltjes sum
\begin{equation}{\label{maeq34}}\tag{46}
S({\cal P},\alpha ,f)=\sum_{k=1}^{n}\alpha (t_{k})\Delta f_{k}=\sum_{k=1}^{n}\alpha (t_{k})f(x_{k})-\sum_{k=1}^{n}\alpha (t_{k})f(x_{k-1}),
\end{equation}
where $t_{k}\in [x_{k-1},x_{k}]$. We define
\[A=f(b)\alpha (b)-f(a)\alpha (a)=f(x_{n})\alpha (x_{n})-f(x_{0})\alpha (x_{0}).\]
Then, we have
\begin{equation}{\label{maeq22}}\tag{47}
A=\sum_{k=1}^{n}f(x_{k})\alpha (x_{k})-\sum_{k=1}^{n}f(x_{k-1})\alpha (x_{k-1}).
\end{equation}
Let ${\cal P}’={\cal P}\cup\{t_{1},\cdots ,t_{n}\}$. From (\ref{maeq34}) and (\ref{maeq22}), we have
\begin{align*} A-S({\cal P},\alpha ,f) & =\sum_{k=1}^{n}f(x_{k})\left [\alpha (x_{k})-\alpha (t_{k})\right ]+\sum_{k=1}^{n}f(x_{k-1})\left [\alpha (t_{k})-\alpha (x_{k-1})\right ]\\ & =S({\cal P}’,f,\alpha ).\end{align*}
Since ${\cal P}’\supseteq {\cal P}$, it follows ${\cal P}’\supseteq {\cal P}(\epsilon )$. According to (\ref{maeq20}), we obtain
\begin{align*} \left |S({\cal P},\alpha ,f)-\left (A-\int_{a}^{b}fd\alpha\right )\right |
& =\left |A-S({\cal P},\alpha ,f)-\int_{a}^{b}fd\alpha\right |
\\ & =\left |S({\cal P}’,f,\alpha )-\int_{a}^{b}fd\alpha\right |<\epsilon\end{align*}
whenever ${\cal P}\supseteq {\cal P}(\epsilon )$. This shows that $\alpha\in R(f)$ and
\[\int_{a}^{b}\alpha(x)df(x)=A-\int_{a}^{b}f(x)d\alpha(x).\]
This completes the proof. $\blacksquare$

\begin{equation}{\label{ma104}}\tag{48}\mbox{}\end{equation}

Lemma \ref{ma104}. (Heine). Suppose that the function $f:(M_{1},d_{1})\rightarrow (M_{2},d_{2})$ is continuous on a compact subset $S$ of $M_{1}$. Then $f$ is uniformly continuous on $S$.

The above Lemma \ref{ma104} can refer to the page Continuity of Functions.

\begin{equation}{\label{mat18}}\tag{49}\mbox{}\end{equation}
Lemma \ref{mat18}. Let $f$ be a real-valued function defined on $[a,b]$. Then $f$ is of bounded variation on $[a,b]$ if and only if $f$ can be expressed as the difference of two increasing functions.

The above Lemma \ref{mat18} can refer to the page Functions of Bounded Variation.

\begin{equation}{\label{ma5}}\tag{50}\mbox{}\end{equation}
Proposition \ref{ma5}. Suppose that $f$ is continuous and $\alpha$ is of bounded variation on $[a,b]$. Then $f\in R(\alpha )$ on $[a,b]$.

Proof. We first assume that $\alpha$ is increasing on $[a,b]$. Suppose that $\alpha (a)=\alpha (b)$. Then $\alpha$ is constant on $[a,b]$ and the result follows obviously. Therefore, we assume $\alpha (a)\neq\alpha (b)$, i.e., $\alpha (a)<\alpha (b)$. Let $A=\alpha (b)-\alpha (a)$. Since $f$ is continuous on the closed and bounded interval, it follows that $f$ is also uniformly continuous on $[a,b]$ by using Lemma \ref{ma104}. Therefore, given any $\epsilon >0$, there exists $\delta>0$ (which depends only on $\epsilon$) such that
\begin{equation}{\label{maeq86}}\tag{51}
|x-y|<\delta\mbox{ implies }|f(x)-f(y)|<\frac{\epsilon}{A}.
\end{equation}
Let ${\cal P}(\epsilon )$ be a partition with norm $\parallel {\cal P}(\epsilon )\parallel<\delta$. Then, for any partition ${\cal P}=\{x_{0},x_{1},\cdots,x_{n}\}$ of $[a,b]$ with ${\cal P}\supseteq{\cal P}(\epsilon )$, we have $|x_{k}-x_{k-1}|<\delta$ for all $k$ since $\parallel {\cal P}(\epsilon )\parallel<\delta$. Using (\ref{maeq86}), we obtain
\begin{align}
M_{k}(f)-m_{k}(f) & =\sup_{x\in [x_{k-1},x_{k}]}f(x)-\inf_{y\in [x_{k-1},x_{k}]}f(y)
\nonumber\\ & =\sup_{x\in [x_{k-1},x_{k}]}f(x)+\sup_{y\in [x_{k-1},x_{k}]}(-f(y))\nonumber\\
& =\sup_{x,y\in [x_{k-1},x_{k}]}\left [f(x)-f(y)\right ]\nonumber\\ & \leq
\sup_{x,y\in [x_{k-1},x_{k}]}\left |f(x)-f(y)\right |<\frac{\epsilon}{A}.\label{maeq87}\tag{52}
\end{align}
Since $\Delta\alpha_{k}>0$ for all $k$, using (\ref{maeq87}), we also obtain
\begin{align*}
U({\cal P},f,\alpha )-L({\cal P},f,\alpha ) & =\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\cdot\Delta\alpha_{k}\\
& <\frac{\epsilon}{A}\cdot\sum_{k=1}^{n}\Delta\alpha_{k}\\
& =\frac{\epsilon}{A}\left [\alpha (b)-\alpha (a)\right ]=\epsilon ,
\end{align*}
which says that the Riemann’s condition is satisfied. Theorem \ref{mat62} says $f\in R(\alpha )$ on $[a,b]$. Now, we assume that $\alpha$ is of bounded variation on $[a,b]$. Lemma \ref{mat18} says $\alpha=\alpha_{1}-\alpha_{2}$ for some increasing functions $\alpha_{1}$ and $\alpha_{2}$ on $[a,b]$. It follows $f\in R(\alpha_{1})$ and $f\in R(\alpha_{2})$ on $[a,b]$. Proposition \ref{map40} says $f\in R(\alpha_{1}-\alpha_{2})=R(\alpha )$ on $[a,b]$. This completes the proof. $\blacksquare$

\begin{equation}{\label{ma161}}\tag{53}\mbox{}\end{equation}
Proposition \ref{ma161}. Let $\alpha$ be of bounded variation on $[a,b]$. Suppose that $f,g$ are bounded and $f,g\in R(\alpha )$ on $[a,b]$. Then $\max\{f,g\},\min\{f,g\}\in R(\alpha )$ on $[a,b]$.

Proof. Since
\[\max\left\{f,g\right\}=\frac{1}{2}\left (f+g+|f-g|\right )\] and
\[\min\left\{f,g\right\}=\frac{1}{2}\left (f+g-|f-g|\right ),\]
the result follows immediately from Propositions \ref{map25} and \ref{map414}. $\blacksquare$

The formula of change of variable in the Riemann-Stieltjes integral can also be established below.

Theorem. (Change of Variable). Let $f\in R(\alpha )$ on $[a,b]$, and let $g$ be a strictly monotonic continuous function defined on an interval $[c,d]$. Assume that $a=g(c)$ and $b=g(d)$. We define the composite functions $h(x)=f(g(x))=(f\circ g)(x)$ and $\beta (x)=\alpha (g(x))=(\alpha\circ g)(x)$ for $x\in [c,d]$. Then $h\in R(\beta )$ on $[c,d]$ and
\[\int_{a}^{b}f(x)d\alpha(x)=\int_{c}^{d}h(x)d\beta(x);\]
that is,
\[\int_{g(c)}^{g(d)}f(t)d\alpha (t)=\int_{c}^{d}(f\circ g)(x)d(\alpha\circ g)(x)=\int_{c}^{d}f(g(x))d\alpha (g(x)).\]

Proof. Without loss of generality, we may assume that $g$ is strictly increasing on $[c,d]$. Therefore, the function $g$ is one-to-one, and has a strictly increasing and continuous inverse $g^{-1}$ defined on $[a,b]$. For every partition ${\cal P}=\{y_{0},y_{1},\cdots ,y_{n}\}$ of $[c,d]$, there is a corresponding one and only one partition ${\cal P}’=\{x_{0},x_{1},\cdots ,x_{n}\}$ of $[a,b]$ satisfying $x_{k}=g(y_{k})$. In this case, we write ${\cal P}’=g({\cal P})$ and ${\cal P}=g^{-1}({\cal P}’)$. Since $f\in R(\alpha )$, given any $\epsilon >0$, there exists a partition ${\cal P}'(\epsilon )$ of $[a,b]$ such that
\begin{equation}{\label{maeq23}}\tag{54}
{\cal P}’\supseteq{\cal P}'(\epsilon )\mbox{ implies }\left |S({\cal P}’,f,\alpha )-\int_{a}^{b}fd\alpha\right |<\epsilon .
\end{equation}
Let ${\cal P}(\epsilon )=g^{-1}({\cal P}'(\epsilon ))$ be the corresponding partition of $[c,d]$, and let ${\cal P}=\{y_{0},y_{1},\cdots ,y_{n}\}$ be a partition of $[c,d]$ finer than ${\cal P}(\epsilon )$. Then, the Riemann-Stieltjes sum is given by
\begin{align*} S({\cal P},h,\beta ) & =\sum_{k=1}^{n}h(u_{k})\Delta\beta_{k}\\ & =\sum_{k=1}^{n}h(u_{k})\left [\beta (y_{k})-\beta (y_{k-1})\right ],\end{align*}
where $u_{k}\in [y_{k-1},y_{k}]$. Now, we take $t_{k}=g(u_{k})$ and $x_{k}=g(y_{k})$. Then ${\cal P}’=\{x_{0},x_{1},\cdots ,x_{n}\}$ is a partition of $[a,b]$ finer than ${\cal P}'(\epsilon )$. Moreover, since $t_{k}\in [x_{k-1},x_{k}]$, we have
\begin{align*}
S({\cal P},h,\beta ) & =\sum_{k=1}^{n}f(g(u_{k}))\left [\alpha (g(y_{k}))-\alpha (g(y_{k-1}))\right ]\\
& =\sum_{k=1}^{n}f(t_{k})\left [\alpha (x_{k})-\alpha (x_{k-1})\right ]=S({\cal P}’,f,\alpha ).
\end{align*}
Therefore, from (\ref{maeq23}), we obtain
\[\left |S({\cal P},h,\beta )-\int_{a}^{b}fd\alpha\right |<\epsilon .\]
This completes the proof. $\blacksquare$

\begin{equation}{\label{mat8}}\tag{55}\mbox{}\end{equation}

Theorem \ref{mat8}. (Intermediate-Value Theorem). Let $f$ be a real-valued and continuous function defined on a closed interval $I$ in $\mathbb{R}$. Suppose that $\alpha <\beta$ in $I$ satisfying $f(\alpha )\neq f(\beta )$. Then $f$ takes every value between $f(\alpha )$ and $f(\beta )$ in the interval $(\alpha ,\beta )$.

The above Theorem \ref{mat8} can refer to page Continuity of Functions.

\begin{equation}{\label{mat91}}\tag{56}\mbox{}\end{equation}
Theorem \ref{mat91}. (First Mean-Value Theorem for Riemann-Stieltjes Integral). Let $\alpha$ be increasing on $[a,b]$, and let
\[M=\sup_{x\in [a,b]}f(x)\mbox{ and }m=\inf_{x\in [a,b]}f(x).\]
Suppose that $f\in R(\alpha )$ on $[a,b]$. Then, there exists a real number $c\in [m,M]$ satisfying
\begin{equation}{\label{maeq90}}\tag{57}
\int_{a}^{b}f(x)d\alpha (x)=c\int_{a}^{b}d\alpha (x)=c\left [\alpha (b)-\alpha (a)\right ].
\end{equation}
In particular, if $f$ is continuous on $[a,b]$, then $c=f(x_{0})$ for some $x_{0}\in [a,b]$.

Proof. Suppose that $\alpha (a)=\alpha (b)$. Since $\alpha$ is assumed to be increasing, it says that $\alpha$ is a constant function. Therefore, the result holds true obviously, where both sides are zero. Therefore, we assume $\alpha (a)\neq\alpha (b)$, i.e., $\alpha (a)<\alpha (b)$. We first have
\[m=\inf_{x\in [a,b]}f(x)\leq\inf_{x\in [x_{k-1},x_{k}]}f(x)=m_{k}\]
and
\[M=\sup_{x\in [a,b]}f(x)\geq\sup_{x\in [x_{k-1},x_{k}]}f(x)=M_{k}\]
for all $x$. Since $\Delta\alpha_{k}\geq 0$ for all $k$, we obtain
\begin{align*}
m\left [\alpha (b)-\alpha (a)\right ] & =\sum_{k=1}^{n}m\cdot\Delta\alpha_{k}\leq
\sum_{k=1}^{n}m_{k}\cdot\Delta\alpha_{k}=L({\cal P},f,\alpha )\\
& \leq U({\cal P},f,\alpha )=\sum_{k=1}^{n}M_{k}\cdot\Delta\alpha_{k}\leq
\sum_{k=1}^{n}M\cdot \Delta\alpha_{k}\\ & =M\left [\alpha (b)-\alpha (a)\right ],
\end{align*}
which implies
\begin{align*} m\left [\alpha (b)-\alpha (a)\right ] & \leq\sup_{{\cal P}\in\mathfrak{P}[a,b]}L({\cal P},f,\alpha )
\\ & =\underline{\int}_{a}^{b}fd\alpha\end{align*}
and
\begin{align*} \overline{\int}_{a}^{b}fd\alpha & =\inf_{{\cal P}\in\mathfrak{P}[a,b]}U({\cal P},f,\alpha )
\\ & \leq M\left [\alpha (b)-\alpha (a)\right ]\end{align*}
Since $f\in R(\alpha )$ on $[a,b]$, i.e.,
\[\underline{\int}_{a}^{b}fd\alpha=\overline{\int}_{a}^{b}fd\alpha,\]
we obtain
\begin{align*} m\left [\alpha (b)-\alpha (a)\right ] & \leq\int_{a}^{b}f(x)d\alpha (x)\\ & \leq M\left [\alpha (b)-\alpha (a)\right ],\end{align*}
which implies
\[m\leq\frac{1}{\alpha (b)-\alpha (a)}\cdot\int_{a}^{b}f(x)d\alpha (x)\equiv c\leq M.\]
This proves (\ref{maeq90}). If $f$ is continuous on $[a,b]$, the intermediate-value Theorem \ref{mat8} says $c=f(x_{0})$ for some $x_{0}\in [a,b]$. This completes the proof. $\blacksquare$

\begin{equation}{\label{mat113}}\tag{58}\mbox{}\end{equation}
Theorem \ref{mat113}. (Second Mean-Value Theorem for Riemann-Stieltjes integral). Suppose that $\alpha$ is continuous on $[a,b]$, and that $f$ is increasing on $[a,b]$. Then, there exists $x_{0}\in [a,b]$ satisfying
\begin{align*}
\int_{a}^{b}f(x)d\alpha (x) & =f(a)\int_{a}^{x_{0}}d\alpha (x)+f(b)\int_{x_{0}}^{b}d\alpha (x)\\
& =f(a)\left [\alpha (x_{0})-\alpha (a)\right ]+f(b)\left [\alpha (b)-\alpha (x_{0})\right ].
\end{align*}

Proof. We first have $\alpha\in R(f)$ by Proposition \ref{ma5}. Then, using Theorems \ref{mat39} and \ref{mat91} (in order) we obtain
\begin{align*}
& \int_{a}^{b}f(x)d\alpha (x)=f(b)\alpha (b)-f(a)\alpha (a)-\int_{a}^{b}\alpha (x)df(x)\\
& \quad =f(b)\alpha (b)-f(a)\alpha (a)-\alpha (x_{0})\left [f(b)-f(a)\right ]\mbox{ (where $x_{0}\in [a,b]$)}\\
& \quad =f(a)\left [\alpha (x_{0})-\alpha (a)\right ]+f(b)\left [\alpha (b)-\alpha (x_{0})\right ].
\end{align*}
This completes the proof. $\blacksquare$

\begin{equation}{\label{mat280}}\tag{59}\mbox{}\end{equation}
Theorem \ref{mat280}. (Cauchy-Schwartz Inequality). Let $\alpha$ be increasing on $[a,b]$. Suppose that $f\in R(\alpha )$, $f^{2}\in R(\alpha )$, $g\in R(\alpha )$ and $g^{2}\in R(\alpha )$ on $[a,b]$. Then, we have
\[\left (\int_{a}^{b}f(x)g(x)d\alpha (x)\right )^{2}\leq\left (\int_{a}^{b}f^{2}(x)d\alpha (x)\right )\left (\int_{a}^{b}g^{2}(x)d\alpha (x)\right ).\]

Suppose that the integrator $\alpha$ has a continuous derivative $\alpha^{\prime}$. Then, the Riemann-Stieltjes is reduced to the Riemann integral, which will be presented below.

\begin{equation}{\label{mat41}}\tag{60}\mbox{}\end{equation}
Theorem \ref{mat41}. (Reduction to the Riemann Integral).
Let $f\in R(\alpha )$ on $[a,b]$. Suppose that $\alpha$ has a continuous derivative $\alpha^{\prime}$ on $[a,b]$. Then, the product $f\cdot\alpha^{\prime}$ is Riemann-integrable on $[a,b]$ and we have
\[\int_{a}^{b}f(x)d\alpha (x)=\int_{a}^{b}f(x)\alpha^{\prime}(x)dx.\]

Proof. Let $g(x)=f(x)\alpha^{\prime}(x)$. We consider the Riemann sum
\[R({\cal P},g)=\sum_{k=1}^{n}g(t_{k})\Delta x_{k}=\sum_{k=1}^{n}f(t_{k})\alpha^{\prime}(t_{k})\Delta x_{k}.\]
The same partition ${\cal P}$ and the same choice of the $t_{k}$ can be used to form the Riemann-Stieltjes sum
\[S({\cal P},f,\alpha )=\sum_{k=1}^{n}f(t_{k})\Delta\alpha_{k}.\]
Applying the mean-value theorem for differentiation, we have
\[\Delta\alpha_{k}=\alpha (x_{k})-\alpha (x_{k-1})=\alpha^{\prime}(v_{k})\Delta x_{k}\]
for some $v_{k}\in (x_{k-1},x_{k})$. Therefore, we obtain
\begin{equation}{\label{maeq37}}\tag{61}
S({\cal P},f,\alpha )-R({\cal P},g)=\sum_{k=1}^{n}f(t_{k})\left [\alpha^{\prime}(v_{k})-\alpha^{\prime}(t_{k})\right ]\Delta x_{k}.
\end{equation}
Since $f$ is bounded on $[a,b]$, we have $|f(x)|\leq M$ for all $x\in [a,b]$ and for some $M>0$. Since $\alpha^{\prime}$ is continuous on the compact interval $[a,b]$, it follows that $\alpha^{\prime}$ is also uniform continuous on $[a,b]$ by referring to Theorem \ref{ma104}. This says that, given any $\epsilon >0$, there exists $\delta >0$ (which depends only on $\epsilon$) such that
\[0\leq |x-y|<\delta\mbox{ implies }\left |\alpha^{\prime}(x)-\alpha^{\prime}(y)\right |<\frac{\epsilon}{2M(b-a)}.\]
Now, we take a partition ${\cal P}'(\epsilon )$ with norm $\parallel {\cal P}'(\epsilon )\parallel<\delta$. Then, for any partition ${\cal P}=\{x_{0},x_{1},\cdots,x_{n}\}$ finer than ${\cal P}'(\epsilon )$, we have $|x_{k}-x_{k-1}|<\delta$ for all $k$. Therefore, we obtain
\[\left |\alpha^{\prime}(v_{k})-\alpha^{\prime}(t_{k})\right |<\frac{\epsilon}{2M(b-a)},\]
where $t_{k},v_{k}\in [x_{k},x_{k-1}]$, i.e.,
\[|v_{k}-t_{k}|\leq |x_{k}-x_{k-1}|<\delta.\]
For this partition ${\cal P}$, from (\ref{maeq37}), we obtain
\begin{equation}{\label{maeq35}}\tag{62}
\left |S({\cal P},f,\alpha )-R({\cal P},g)\right |<M\cdot\frac{\epsilon}{2M(b-a)}\sum_{k=1}^{n}\Delta x_{k}=\frac{\epsilon}{2}.
\end{equation}

On the other hand, since $f\in R(\alpha )$ on $[a,b]$, there exists a partition ${\cal P}”(\epsilon )$ such that
\begin{equation}{\label{maeq36}}\tag{63}
{\cal P}\supseteq {\cal P}”(\epsilon )\mbox{ implies }\left |S({\cal P},f,\alpha )-\int_{a}^{b}fd\alpha\right |<\frac{\epsilon}{2}.
\end{equation}
We take ${\cal P}(\epsilon )={\cal P}'(\epsilon )\cup {\cal P}”(\epsilon )$. For ${\cal P}\supseteq {\cal P}(\epsilon )$, by combining (\ref{maeq35}) and (\ref{maeq36}), we obtain
\begin{align*}
\left |R\left ({\cal P},f\alpha^{\prime}\right )-\int_{a}^{b}fd\alpha\right | & =\left |R({\cal P},g)-\int_{a}^{b}fd\alpha\right |\\
& \leq\left |R({\cal P},g)-S({\cal P},f,\alpha )\right |+\left |S({\cal P},f,\alpha )-\int_{a}^{b}fd\alpha\right |\\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon ,
\end{align*}
which says that $f\alpha^{\prime}$ is Riemann-integrable and
\[\int_{a}^{b}f(x)d\alpha (x)=\int_{a}^{b}f(x)\alpha^{\prime}(x)dx.\]
This completes the proof. $\blacksquare$

\begin{equation}{\label{mat101}}\tag{64}\mbox{}\end{equation}
Theorem \ref{mat101}. Let $\alpha$ be of bounded variation on $[a,b]$. Suppose that $f,g$ are bounded and $f,g\in R(\alpha )$ on $[a,b]$. For $x\in [a,b]$, we define
\[F(x)=\int_{a}^{x}f(t)d\alpha (t)\mbox{ and }G(x)=\int_{a}^{x}g(t)d\alpha (t).\]
Then $f\in R(G), g\in R(F)$ and $fg\in R(\alpha )$ on $[a,b]$. Moreover, we have
\[\int_{a}^{b}f(x)g(x)d\alpha (x)=\int_{a}^{b}f(x)dG(x)=\int_{a}^{b}g(x)dF(x).\]

Proof. We first assume that $\alpha$ is increasing on $[a,b]$. Then, Proposition \ref{map82} says $fg\in R(\alpha )$ on $[a,b]$. For every partition ${\cal P}$ of $[a,b]$, we have
\begin{align}
S({\cal P},f,G) & =\sum_{k=1}^{n}f(t_{k})\Delta G_{k}=\sum_{k=1}^{n}f(t_{k})\left [G(x_{k})-G(x_{k-1})\right ]\nonumber\\
& =\sum_{k=1}^{n}\left [f(t_{k})\int_{x_{k-1}}^{x_{k}}g(t)d\alpha (t)\right ]=\sum_{k=1}^{n}\int_{x_{k-1}}^{x_{k}}f(t_{k})g(t)d\alpha (t)\label{maeq83}\tag{65}
\end{align}
and
\begin{equation}{\label{maeq84}}\tag{66}
\int_{a}^{b}f(x)g(x)d\alpha (x)=\sum_{k=1}^{n}\int_{x_{k-1}}^{x_{k}}f(t)g(t)d\alpha (t).
\end{equation}
Let
\[M=\sup_{x\in [a,b]}|g(x)|.\]
From (\ref{maeq83}) and (\ref{maeq84}), we have
\begin{align}
& \left |S({\cal P},f,G)-\int_{a}^{b}f(x)g(x)d\alpha (x)\right |=\left |\sum_{k=1}^{n}
\int_{x_{k-1}}^{x_{k}}\left [f(t_{k})-f(t)\right ]g(t)d\alpha (t)\right |\nonumber\\
& \quad\leq M\cdot\sum_{k=1}^{n}\int_{x_{k-1}}^{x_{k}}\left |f(t_{k})-f(t)\right |d\alpha (t)
\leq M\cdot\sum_{k=1}^{n}\int_{x_{k-1}}^{x_{k}}\left [M_{k}(f)-m_{k}(f)\right ]d\alpha (t)\nonumber\\
& \quad =M\cdot\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\int_{x_{k-1}}^{x_{k}}d\alpha (t)
=M\cdot\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta\alpha_{k}\nonumber\\
& \quad =M\cdot\left [U({\cal P},f,\alpha )-L({\cal P},f,\alpha )\right ].\label{maeq85}\tag{67}
\end{align}
Since $f\in R(\alpha )$ on $[a,b]$, using Theorem \ref{mat62} by considering the Riemann’s condition, for every $\epsilon >0$, there exits a partition ${\cal P}(\epsilon )$ such that
\[{\cal P}\supseteq {\cal P}(\epsilon )\mbox{ implies }0\leq U({\cal P},f,\alpha )-L({\cal P},f,\alpha )\leq\frac{\epsilon}{M}.\]
Therefore, from (\ref{maeq85}), we also obtain
\[\left |S({\cal P},f,G)-\int_{a}^{b}f(x)g(x)d\alpha (x)\right |<\epsilon ,\]
which says that $f\in R(G)$ on $[a,b]$ and
\[\int_{a}^{b}f(x)g(x)d\alpha (x)=\int_{a}^{b}f(x)dG(x).\]
Similarly, we can show that $g\in R(F)$ on $[a,b]$ and
\[\int_{a}^{b}f(x)g(x)d\alpha (x)=\int_{a}^{b}g(x)dF(x).\]

Next, we assume that $\alpha$ is of bounded variation on $[a,b]$. Since $f\in R(\alpha )$, from Proposition \ref{map75}, we have $f\in R(\beta )$ and $f\in R(\beta-\alpha )$. Let $\alpha_{1}=\beta$ and $\alpha_{2}=\beta-\alpha$. Then $\alpha=\alpha_{1}-\alpha_{2}$, where $\alpha_{1}$ and $\alpha_{2}$ are increasing on $[a,b]$. Since $g\in R(\alpha )$, we also have $g\in R(\alpha_{1})$ and $g\in R(\alpha _{2})$. Then, using Proposition \ref{map40}, we obtain
\begin{align*} F(x) & =\int_{a}^{x}f(t)d\alpha (t)\\ & =\int_{a}^{x}f(t)d\alpha_{1}(t)-\int_{a}^{x}f(t)d\alpha_{2}(t)\\ & \equiv F_{1}(x)-F_{2}(x)\end{align*}
and
\begin{align*} G(x) & =\int_{a}^{x}g(t)d\alpha (t)\\ & =\int_{a}^{x}g(t)d\alpha_{1}(t)-\int_{a}^{x}g(t)d\alpha_{2}(t)\\ & \equiv G_{1}(x)-G_{2}(x).\end{align*}
Then, we have
\[\int_{a}^{b}f(x)dG(x)=\int_{a}^{b}f(x)dG_{1}(x)-\int_{a}^{b}f(x)dG_{2}(x).\]
Since $\alpha_{1}$ and $\alpha_{2}$ are increasing on $[a,b]$, according to the previous results, we have $f\in R(G_{1})$ and $f\in R(G_{2})$ satisfying
\[\int_{a}^{b}f(x)g(x)d\alpha_{1}(x)=\int_{a}^{b}f(x)dG_{1}(x)\] and
\[\int_{a}^{b}f(x)g(x)d\alpha_{2}(x)=\int_{a}^{b}f(x)dG_{2}(x).\]
Since $fg\in R(\alpha_{1})$ and $fg\in R(\alpha_{2})$ on $[a,b]$ by using Proposition \ref{map75} again, it follows
\begin{align*}
\int_{a}^{b}f(x)g(x)d\alpha (x) & =\int_{a}^{b}f(x)g(x)d\alpha_{1}(x)-\int_{a}^{b}f(x)g(x)d\alpha_{2}(x)\\
& =\int_{a}^{b}f(x)dG_{1}(x)-\int_{a}^{b}f(x)dG_{2}(x)=\int_{a}^{b}f(x)dG(x)
\end{align*}
We can similarly obtain
\[\int_{a}^{b}f(x)g(x)d\alpha (x)=\int_{a}^{b}g(x)dF(x).\]
This completes the proof. $\blacksquare$

\begin{equation}{\label{map99}}\tag{68}\mbox{}\end{equation}
Proposition \ref{map99}. Let $\alpha$ be of bounded variation on $[a,b]$. Suppose that $f\in R(\alpha )$ on $[a,b]$. We define
\[F(x)=\int_{a}^{x}f(x)d\alpha (x)\mbox{ for }x\in [a,b].\]
Then, we have the following properties.

(i) $F$ is of bounded variation on $[a,b]$.

(ii) If $\alpha$ is continuous at $x_{0}\in[a,b]$, then $F$ is continuous at $x_{0}\in[a,b]$.

Proof. We first assume that $\alpha$ is increasing on $[a,b]$. For $x,y\in [a,b]$ with $x<y$, Proposition \ref{mat95} says that $f\in R(\alpha )$ on $[x,y]$. Using Proposition \ref{map414}, we also have $|f|\in R(\alpha )$ on $[x,y]$. Define
\begin{equation}{\label{maeq98}}\tag{69}
M(x,y)=\sup_{t\in [x,y]}|f(t)|\mbox{ and }m(x,y)=\inf_{t\in [x,y]}|f(t)|
\end{equation}
and
\[M=\sup_{t\in [a,b]}|f(t)|.\]
Since $\alpha$ is increasing, using the mean-value Theorem \ref{mat91}, we obtain
\begin{align}
\left |F(y)-F(x)\right | & =\left |\int_{x}^{y}f(t)d\alpha (t)\right |\leq\int_{x}^{y}\left |f(t)\right |d\alpha (t)=c[\alpha (y)-\alpha (x)]\nonumber\\
& \leq M(x,y)\cdot\left [\alpha (y)-\alpha (x)\right ]\leq M\cdot\left [\alpha (y)-\alpha (x)\right ],\label{maeq96}\tag{70}
\end{align}
where $c\in [m(x,y),M(x,y)]$. Given any partition ${\cal P}=\{a=x_{0},x_{1},\cdots ,x_{n}=b\}$ of $[a,b]$, using (\ref{maeq96}), since $\alpha$ is increasing, we have
\begin{align*} \sum_{k=1}^{n}|\Delta F_{k}| & =\sum_{k=1}^{n}\left |F(x_{k})-F(x_{k-1})\right |\\ & \leq M\cdot\sum_{k=1}^{n}\left [\alpha (x_{k})-\alpha (x_{k-1})\right ]\\ & =M\cdot [\alpha (b)-\alpha (a)],\end{align*}
which proves that $F(x)$ is of bounded variation. Using (\ref{maeq96}) again, we also see that if $\alpha$ is continuous at $x_{0}\in[a,b]$, then $F$ is continuous at $x_{0}\in[a,b]$. Now, we assume that $\alpha$ is of bounded variation on $[a,b]$. Since $f\in R(\alpha )$ on $[a,b]$, Proposition \ref{map75} says that $f\in R(\beta )$ on $[a,b]$ and $f\in R(\beta -\alpha )$ on $[a,b]$, where $\beta$ and $\beta -\alpha$ are increasing functions on $[a,b]$. Therefore, we have
\begin{align*}
F(x) & =\int_{a}^{x}f(x)d\alpha (x)=\int_{a}^{x}f(x)d(\beta (x)-(\beta (x)-\alpha (x)))\\
& =\int_{a}^{x}f(x)d\beta (x)-\int_{a}^{x}f(x)d(\beta (x)-\alpha (x))\equiv F_{1}(x)-F_{2}(x),
\end{align*}
where the functions
\[F_{1}(x)=\int_{a}^{x}f(x)d\beta (x)\mbox{ and }F_{2}(x)=\int_{a}^{x}f(x)d(\beta (x)-\alpha (x))\]
are of bounded variation on $[a,b]$. Proposition \ref{map17} says that $F=F_{1}-F_{2}$ is of bounded variation on $[a,b]$. Suppose that $\alpha$ is continuous at $x_{0}$. Proposition \ref{map20} says that $\beta$ is continuous at $x_{0}$, which also says that $\beta-\alpha$ is continuous at $x_{0}$. Therefore $F_{1}$ and $F_{2}$ are also continuous at $x_{0}$. It follows that $F$ is continuous at $x_{0}$, and the proof is complete. $\blacksquare$

\begin{equation}{\label{map109}}\tag{71}\mbox{}\end{equation}
Proposition \ref{map109}. Suppose that $f$ and $g$ are Riemann-integrable on $[a,b]$. We define
\[F(x)=\int_{a}^{x}f(t)dt\mbox{ and }G(x)=\int_{a}^{x}g(t)dt\mbox{ for }x\in [a,b].\]
Then $F$ and $G$ are continuous functions of bounded variation on $[a,b]$. Moreover, we also have $f\in R(G)$ on $[a,b]$, $g\in R(F)$ on $[a,b]$ and
\begin{equation}{\label{maeq100}}\tag{72}
\int_{a}^{b}f(x)g(x)dx=\int_{a}^{b}f(x)dG(x)=\int_{a}^{b}g(x)dF(x).
\end{equation}

Proof. Parts (i) and (ii) of Proposition \ref{map99} shows that $F$ and $G$ are continuous functions of bounded variation on $[a,b]$. The existence of the integrals and the formula (\ref{maeq100}) follows immediately from Theorem \ref{mat101}. $\blacksquare$

\begin{equation}{\label{mat110}}\tag{73}\mbox{}\end{equation}
Theorem \ref{mat110}. (First Fundamental Theorem of Calculus). Let $\alpha$ be increasing on $[a,b]$. Suppose that $f$ is continuous on $[a,b]$, and that $\alpha^{\prime}(x)$ exists at each point $x\in (a,b)$. We define
\[F(x)=\int_{a}^{x}f(t)d\alpha (t)\mbox{ for }x\in [a,b].\]
Then, the derivative $F'(x)$ exists at each point $x\in (a,b)$ and we have
\[F'(x)=f(x)\alpha^{\prime}(x).\]

Proof. For $x,y\in [a,b]$ with $x<y$, suppose that $x$ is fixed. We consider $y\rightarrow x$. Using Proposition \ref{mat95}, we have $f\in R(\alpha )$ on $[x,y]$. We define
\begin{equation}{\label{maeq*98}}\tag{74}
M(x,y)=\sup_{t\in [x,y]}f(t)\mbox{ and }m(x,y)=\inf_{t\in [x,y]}f(t).
\end{equation}
Then, using mean-value Theorem \ref{mat91}, we obtain
\begin{equation}{\label{maeq*96}}\tag{75}
F(y)-F(x)=\int_{x}^{y}f(t)d\alpha (t)=c[\alpha (y)-\alpha (x)],
\end{equation}
where $c\in [m(x,y),M(x,y)]$. Since we consider $y\rightarrow x$, it follows that the number $c$ in (\ref{maeq*96}) is a function of $y$. Therefore, we can write $c(y)$. From (\ref{maeq*98}), since $f$ is continuous, we have
\[\lim_{y\rightarrow x}c(y)=f(x).\]
Finally, using (\ref{maeq*96}), we obtain
\begin{align*} F'(x) & =\lim_{y\rightarrow x}\frac{F(y)-F(x)}{y-x}\\ & =\lim_{y\rightarrow x}
c(y)\cdot\frac{\alpha (y)-\alpha (x)}{y-x}\\ & =f(x)\alpha^{\prime}(x).\end{align*}
This completes the proof. $\blacksquare$

\begin{equation}{\label{ma8}}\tag{76}\mbox{}\end{equation}
Theorem \ref{ma8}. (Generalized Mean-Value Theorem). Suppose that the real-valued functions $f,g:[a,b]\rightarrow\mathbb{R}$ are defined on the
closed interval $[a,b]$ such that the following conditions are satisfied.

The limits $f(a+)$, $g(a-)$, $f(b+)$ and $g(b-)$ exist as finite values.
$f$ and $g$ are differentiable on $(a,b)$ in which each derivative can be finite or infinite.
There is no $c\in (a,b)$ at which $f'(c)$ and $g'(c)$ are infinite.

Then, there exists some point $c\in (a,b)$ satisfying
\[f'(c)\cdot [g(b-)-g(a+)]=g'(c)\cdot [f(b-)-f(a+)].\]

The above Theorem \ref{ma8} can refer to the page Differentiation.

\begin{equation}{\label{mat107}}\tag{77}\mbox{}\end{equation}
Theorem \ref{mat107}. (Second Fundamental Theorem of Calculus). Let $f$ be Riemann-integrable on $[a,b]$, and let $g$ be a function defined on $[a,b]$ such that the derivative $g'(x)$ exists at each $x\in (a,b)$ and $g'(x)=f(x)$. Suppose that $g(a+)$ and $g(b-)$ exists satisfying
\begin{equation}{\label{maeq101}}\tag{78}
g(a)-g(a+)=g(b)-g(b-).
\end{equation}
Then, we have
\begin{equation}{\label{maeq106}}\tag{79}
\int_{a}^{b}f(x)dx=g(b-)-g(a+)=g(b)-g(a)=\int_{a}^{b}g'(x)dx.
\end{equation}

Proof. Let ${\cal P}=\{a=x_{0},x_{1},\cdots ,x_{n}=b\}$ be a partition of $[a,b]$. Since $g’$ exists on $(a,b)$, it says that $g$ is continuous on $(a,b)$. Therefore, using the mean-value theorem, for $k=2,\cdots ,n-1$, we have
\begin{equation}{\label{maeq102}}\tag{80}
g(x_{k})-g(x_{k-1})=g'(t_{k})\Delta x_{k}
\end{equation}
for some $t_{k}\in (x_{k-1},x_{k})$. For $k=1$, using the generalized mean-value Theorem \ref{ma8}, we have
\begin{equation}{\label{maeq103}}\tag{81}
g(x_{1})-g(x_{0})=\left [g(x_{1})-g(x_{0}+)\right ]+\left [g(x_{0}+)-g(x_{0})\right ]=g'(t_{1})\Delta x_{1}+g(a+)-g(a)
\end{equation}
for some $t_{1}\in (x_{0},x_{1})=(a,x_{1})$. For $k=n$, using the generalized mean-value Theorem \ref{ma8} again, we have
\begin{equation}{\label{maeq104}}\tag{82}
g(x_{n})-g(x_{n-1})=\left [g(x_{n})-g(x_{n}-)\right ]+\left [g(x_{n}-)-g(x_{n-1})\right ]=g'(t_{n})\Delta x_{n}+g(b)-g(b-)
\end{equation}
for some $t_{n}\in (x_{n-1},x_{n})=(x_{n-1},b)$. Combining (\ref{maeq101}), (\ref{maeq102}), (\ref{maeq103}) and (\ref{maeq104}), we have
\begin{equation}{\label{maeq105}}\tag{83}
g(b)-g(a)=\sum_{k=1}^{n}\left [g(x_{k})-g(x_{k-1})\right ]=\sum_{k=1}^{n}g'(t_{k})\Delta x_{k}=\sum_{k=1}^{n}f(t_{k})\Delta x_{k}.
\end{equation}
Since $f$ is Riemann-integrable on $[a,b]$, given any $\epsilon >0$, there is a partition ${\cal P}(\epsilon )$ such that
\[{\cal P}\supseteq {\cal P}(\epsilon )\mbox{ implies }
\left |\sum_{k=1}^{n}f(t_{k})\Delta x_{k}-\int_{a}^{b}f(x)dx\right |<\epsilon .\]
Using (\ref{maeq105}), we have
\begin{align*} \left |g(b)-g(a)-\int_{a}^{b}f(x)dx\right |
& =\left |\sum_{k=1}^{n}g'(t_{k})\Delta x_{k}-\int_{a}^{b}f(x)dx\right |
\\ & =\left |\sum_{k=1}^{n}f(t_{k})\Delta x_{k}-\int_{a}^{b}f(x)dx\right |<\epsilon .\end{align*}
which shows that $g’$ is Riemann-integrable on $[a,b]$ and
\[\int_{a}^{b}g'(x)dx=\int_{a}^{b}f(x)dx.\]
Since $\epsilon>0$ is any positive number, it follows
\[\int_{a}^{b}f(x)dx=g(b)-g(a)=g(b-)-g(a+).\]
This completes the proof. $\blacksquare$

Proposition. Let $f$ be Riemann-integrable on $[a,b]$, and let $\alpha$ be continuous at $a$ and $b$. Suppose that $\alpha$ is Riemann-integrable on $[a,b]$, and that $\alpha^{\prime}$ exists on $(a,b)$. Then, we have
\[\int_{a}^{b}f(x)d\alpha (x)=\int_{a}^{b}f(x)\alpha^{\prime}(x)dx.\]

Proof. Using Theorem \ref{mat107}, for each $x\in [a,b]$, we have
\[\alpha (x)-\alpha (a)=\int_{a}^{x}\alpha^{\prime}(t)dt\equiv G(x).\]
Using Proposition \ref{map109}, we have $f\in R(G)$ on $[a,b]$. Since $\alpha (x)=\alpha (a)+G(x)$, using Proposition \ref{ma7}, we also have $f\in R(\alpha)$ on $[a,b]$ and
\[\int_{a}^{b}f(x)d\alpha (x)=\int_{a}^{b}f(x)dG(x).\]
Using Proposition \ref{map109} by considering $g=\alpha^{\prime}$, we have
\begin{align*} \int_{a}^{b}f(x)d\alpha(x) & =\int_{a}^{b}f(x)dG(x)\\ & =\int_{a}^{b}f(x)g(x)dx\\ & =\int_{a}^{b}f(x)\alpha^{\prime}(x)dx.\end{align*}
This completes the proof. $\blacksquare$

A related result is also given below.

Proposition. Let $f$ be defined and bounded on $[a,b]$, and let $\alpha$ be continuous at $a$ and $b$ such that the derivative $\alpha^{\prime}(x)$ is bounded at each $x\in (a,b)$. Suppose that the following integrals
\[\int_{a}^{b}f(x)d\alpha (x)\mbox{ and }\int_{a}^{b}f(x)\alpha^{\prime}(x)dx\]
exist. Then, we have
\[\int_{a}^{b}f(x)d\alpha (x)=\int_{a}^{b}f(x)\alpha^{\prime}(x)dx.\]

\begin{equation}{\label{mat112}}\tag{84}\mbox{}\end{equation}
Theorem \ref{mat112}. (Change of Variable in Riemann Integral). Suppose that $g$ has a continuous derivative $g’$ on the closed interval $[a,b]$, and that $f$ is continuous on $g([a,b])$. Then, for each $x\in [a,b]$, the function $f(g(t))g'(t)$ is Riemann-integrable on $[a,x]$ and
\[\int_{a}^{x}f(g(t))g'(t)dt=\int_{g(a)}^{g(x)}f(t)dt.\]
In particular, we have
\[\int_{a}^{b}f(g(t))g'(t)dt=\int_{g(a)}^{g(b)}f(x)dx.\]

Proof. Since $g’$ and the composition $f\circ g$ are continuous on $[a,b]$, we see that the function $f(g(t))g'(t)$ is also continuous on $[a,b]$, which says that $f(g(t))g'(t)$ is Riemann-integrable on $[a,b]$. We define a function $F$ on $g([a,b])$ by
\[F(x)=\int_{g(a)}^{x}f(t)dt\mbox{ for }x\in g([a,b]),\]
and a function $G$ on $[a,b]$ by
\[G(x)=\int_{a}^{x}f(g(t))g'(t)dt.\]
Using Theorem \ref{mat110}, we have $F'(x)=f(x)$ and $G'(x)=f(g(x))g'(x)$. Therefore, we obtain
\[(F(g(x)))’=F'(g(x))g'(x)=f(g(x))g'(x)=G'(x),\]
which says that $G(x)-F(g(x))$ is constant on $[a,b]$. Since $G(a)=0$ and $F(g(a))=0$, this constant must be zero. This shows that $G(x)=F(g(x))$ for all $x\in [a,b]$. In particular, if $x=b$, then we have $G(b)=F(g(b))$. This completes the proof. $\blacksquare$

Theorem. (Mean-Value Theorem for Riemann Integrals)
Let $g$ be continuous and $f$ be increasing on $[a,b]$. Suppose that $A$ and $B$ are two real numbers satisfying $A\leq f(a+)$ and $B\geq f(b-)$. Then, there exists a point $x_{0}\in [a,b]$, which depends on $A$ and $B$, satisfying
\[\int_{a}^{b}f(x)g(x)dx=A\int_{a}^{x_{0}}g(x)dx+B\int_{x_{0}}^{b}g(x)dx.\]
In particular, suppose that $f(x)\geq 0$ for all $x\in [a,b]$. Then, by taking $A=0$, we have
\[\int_{a}^{b}f(x)g(x)dx=B\int_{x_{0}}^{b}g(x)dx\]
for some $x_{0}\in [a,b]$.

Proof. We define a new function $\hat{f}$ as follows:
\[\hat{f}(x)=\left\{\begin{array}{ll}
A & \mbox{if $x=a$}\\
f(x) & \mbox{if $a<x<b$}\\
B & \mbox{if $x=b$}.
\end{array}\right .\]
Then, it is clear that
\[\int_{a}^{b}f(x)g(x)dx=\int_{a}^{b}\hat{f}(x)g(x)dx.\]
Since $A\leq f(a+)$ and $B\geq f(b-)$, we see that $\hat{f}$ is still increasing on $[a,b]$. Let
\[\alpha (x)=\int_{a}^{x}g(t)dt.\]
Then $\alpha$ is continuous by Proposition \ref{map99}. Now, using Proposition \ref{map109}, Theorem \ref{mat113} and Proposition \ref{map109} (in order) we obtain
\begin{align}
\int_{a}^{b}f(x)g(x)dx & =\int_{a}^{b}\hat{f}(x)g(x)dx=\int_{a}^{b}\hat{f}(x)d\alpha (x)\nonumber\\
& =\hat{f}(a)\int_{a}^{x_{0}}d\alpha (x)+\hat{f}(b)\int_{x_{0}}^{b}d\alpha (x)\nonumber\\
& = \hat{f}(a)\int_{a}^{x_{0}}g(x)dx+\hat{f}(b)\int_{x_{0}}^{b}g(x)dx\nonumber\\
& =A\int_{a}^{x_{0}}g(x)dx+B\int_{x_{0}}^{b}g(x)dx.\label{maeq115}\tag{85}
\end{align}
This completes the proof. $\blacksquare$

We are going to provide the sufficient and necessary conditions for the existence of Riemann-Stieltjes integrals.

\begin{equation}{\label{mat92}}\tag{86}\mbox{}\end{equation}
Theorem \ref{mat92}. (Sufficient Conditions). We have the following results.

(i) Suppose that $f$ is continuous on $[a,b]$, and that $\alpha$ is of bounded variation on $[a,b]$. Then $f\in R(\alpha )$ on $[a,b]$.

(ii) Suppose that $\alpha$ is continuous on $[a,b]$, and that $f$ is of bounded variation on $[a,b]$. Then $f\in R(\alpha )$ on $[a,b]$.

Proof. Part (i) follows from Proposition \ref{ma5}. To prove part (ii), we immediately have $\alpha\in R(f)$ by part (i). Finally, by Theorem \ref{mat39}, we obtain $f\in R(\alpha )$ on $[a,b]$. This completes the proof. $\blacksquare$

Corollary. Suppose that any one of the following conditions is satisfied.

$f$ is continuous on $[a,b]$.
$f$ is of bounded variation on $[a,b]$.

Then $f$ is Riemann-integrable on $[a,b]$.

Proof. The result follows from Theorem \ref{mat92} immediately by taking $\alpha (x)=x$. $\blacksquare$

Theorem. (Necessary Conditions). Let $\alpha$ be increasing on $[a,b]$. Suppose that $\alpha$ and $f$ are discontinuous from the right or from the left at $c$. Then $f\not\in R(\alpha )$ on $[a,b]$.

Proof. It suffices to prove the case that both $\alpha$ and $f$ are discontinuous from the right at $x=c$. In this case, there exists $\epsilon >0$ such that, for every $\delta >0$, there exist $x,y\in (c,c+\delta )$ satisfying
\begin{equation}{\label{maeq88}}\tag{87}
\left |\alpha (y)-\alpha (c)\right |\geq\epsilon
\end{equation}
and
\begin{equation}{\label{ma111}}\tag{88}
\left |f(x)-f(c)\right |\geq\epsilon.
\end{equation}
Let ${\cal P}$ be a partition of $[a,b]$ containing $c$ as a point of subdivision. We consider the difference
\[U({\cal P},f,\alpha )-L({\cal P},f,\alpha )=\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta\alpha_{k}.\]
Since each term is non-negative, if the $i$th sub-interval has $c$ as its left endpoint, we have
\begin{equation}{\label{maeq89}}\tag{8}
U({\cal P},f,\alpha )-L({\cal P},f,\alpha )\geq\left [M_{i}(f)-m_{i}(f)\right ]\cdot\left [\alpha (x_{i})-\alpha (c)\right ].
\end{equation}
Since $c$ is a common discontinuity from the right, using (\ref{maeq88}), we can choose $x_{i}$ satisfying $\alpha (x_{i})-\alpha (c)\geq\epsilon$. Using (\ref{ma111}), we also have
\[M_{i}(f)-m_{i}(f)=\sup_{x\in [c,x_{i}]}f(x)-\inf_{x\in [c,x_{i}]}f(x)\geq\epsilon .\]
Therefore, using (\ref{maeq89}), we obtain
\[U({\cal P},f,\alpha )-L({\cal P},f,\alpha )\geq\epsilon^{2},\]
which says that the Riemann’s condition cannot be satisfied. This completes the proof. $\blacksquare$

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Step Functions as Integrators.

Given a function $\alpha$, the right limit $\alpha(x+)$ and left limit $\alpah(x-)$ can refer to the page Discontinuity and Semi-continuity of Functions. We are going to investigate the Riemann-Stieltjes integrals when the integrators are taken to be the step functions. A function $\alpha$ defined on $[a,b]$ is called a step function when there is a partition
\[a=x_{1}<x_{2}<\cdots <x_{n}=b\]
such that $\alpha$ is constant on each open sub-interval $(x_{k-1},x_{k})$. The number $\alpha (x_{k}+)-\alpha (x_{k}-)$ is called the jump at $x_{k}$ for $1<k<n$. The jump at $x_{1}$ is given by $\alpha (x_{1}+)-\alpha (x_{1})$ and the jump at $x_{n}$ is given by $\alpha (x_{n})-\alpha (x_{n}-)$.

\begin{equation}{\label{map30}}\tag{90}\mbox{}\end{equation}
Proposition \ref{map30}. Given $a<c<b$, we define $\alpha$ on $[a,b]$ as follows
\[\alpha (x)=\left\{\begin{array}{ll}
\alpha (a) & \mbox{if $a\leq x<c$}\\
\alpha (c) & \mbox{if $x=c$}\\
\alpha (b) & \mbox{if $c<x\leq b$,}
\end{array}\right .\]
where $\alpha (a)$, $\alpha (b)$ and $\alpha (c)$ are arbitrary values. Let $f$ be defined on $[a,b]$ in such a way that not both $f$ and $\alpha$ are discontinuous from the right and from the left at $c$. Then $f\in R(\alpha )$ on $[a,b]$ and we have
\begin{align*} \int_{a}^{b}f(x)d\alpha(x) & =f(c)\left [\alpha (c+)-\alpha (c-)\right ]\\ & =f(c)\left [\alpha (b)-\alpha (a)\right ].\end{align*}

Proof. Let ${\cal P}$ be any partition of $[a,b]$. For $c\in {\cal P}$, every term in the Reimann-Stieltjes sum $S({\cal P},f,\alpha )$ is zero except two terms arising from the sub-intervals separated by $c$. Therefore, we have
\[S({\cal P},f,\alpha )=f(t_{k-1})\left [\alpha (c)-\alpha (c-)\right ]+f(t_{k})\left [\alpha (c+)-\alpha (c)\right ]\]
where $t_{k-1}\leq c\leq t_{k}$, which implies
\begin{align}
& \left |S({\cal P},f,\alpha )-f(c)\left [\alpha (c+)-\alpha (c-)\right ]\right |
=\left |S({\cal P},f,\alpha )-f(c)\left [\left (\alpha (c+)-\alpha(c)\right )+
\left (\alpha(c)-\alpha (c-)\right )\right ]\right |\nonumber\\
& \quad\leq\left |f(t_{k-1})-f(c)\right |\cdot\left |\alpha (c)-\alpha (c-)\right |
+\left |f(t_{k})-f(c)\right |\cdot\left |\alpha (c+)-\alpha (c)\right |.\label{maeq29}\tag{91}
\end{align}
Now, we consider the following cases.

Suppose that $f$ is continuous at $c$. Then, given $\epsilon >0$, there exists $\delta>0$ such that
\[0\leq c-t_{k-1}<\delta\mbox{ implies }\left |f(t_{k-1})-f(c)\right |<\frac{\epsilon}{2\cdot\left |\alpha (c)-\alpha (c-)\right |}\]
and
\[0\leq t_{k}-c<\delta\mbox{ implies }\left |f(t_{k})-f(c)\right |<\frac{\epsilon}{2\cdot\left |\alpha (c+)-\alpha (c)\right |}.\]
We take a partition ${\cal P}_{\epsilon}$ satisfying $\parallel {\cal P}_{\epsilon}\parallel <\delta$. Then, for any partition ${\cal P}$ finer than ${\cal P}_{\epsilon}$, from (\ref{maeq29}), we obtain the inequality
\[\left |S({\cal P},f,\alpha )-f(c)\left [\alpha (c+)-\alpha (c-)\right ]\right |<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\]
Suppose that $f$ is discontinuous both from the right and from the left at $c$. Then, by the hypotheses, $\alpha$ is continuous at $c$, i.e,
\[\alpha (c)=\alpha (c-)=\alpha (c+).\]
In this case, from (\ref{maeq29}), we have
\[\left |S({\cal P},f,\alpha )-f(c)\left [\alpha (c+)-\alpha (c-)\right ]\right |\leq 0.\]
Suppose that $f$ is continuous from the left and discontinuous from the right at $c$. Then, by the hypotheses, $\alpha$ is continuous from the right, i.e., $\alpha (c)=\alpha (c+)$. The left-continuity of $f$ at $c$ says that, given $\epsilon >0$, there exists $\delta>0$ such that
\[0\leq c-t_{k-1}<\delta\mbox{ implies }\left |f(t_{k-1})-f(c)\right |<\frac{\epsilon}{\left |\alpha (c)-\alpha (c-)\right |}.\]
Using (\ref{maeq29}), we have
\[\left |S({\cal P},f,\alpha )-f(c)\left [\alpha (c+)-\alpha (c-)\right ]\right |<\epsilon.\]
Suppose that $f$ is continuous from the right and discontinuous from the left at $c$. Then, by the hypotheses, $\alpha$ is continuous from the left, i.e., $\alpha (c)=\alpha (c-)$. The right-continuity of $f$ at $c$ says that, given $\epsilon >0$, there exists $\delta>0$ such that
\[0\leq t_{k}-c<\delta\mbox{ implies }\left |f(t_{k})-f(c)\right |<\frac{\epsilon}{2\cdot\left |\alpha (c+)-\alpha (c)\right |}.\]
Using (\ref{maeq29}), we have
\[\left |S({\cal P},f,\alpha )-f(c)\left [\alpha (c+)-\alpha (c-)\right ]\right |<\epsilon.\]

This shows that $f\in R(\alpha)$ and
\[\int_{a}^{b}f(x)d\alpha (x)=f(c)\left [\alpha (c+)-\alpha(c-)\right ].\]
This completes the proof. $\blacksquare$

Proposition \ref{map30} says that the Riemann-Stieltjes integral is equal to $f(c)$ multiplying the jumps $\alpha (c+)-\alpha (c-)$ of $\alpha$ at $c$. The general case of step function is presented below.

\begin{equation}{\label{mat32}}\tag{92}\mbox{}\end{equation}
Theorem \ref{mat32}. Let $\alpha$ be a step function defined on $[a,b]$ with jump $\alpha_{k}$ at $x_{k}$. Let $f$ be defined on $[a,b]$ in such a way that not both $f$ and $\alpha$ are discontinuous from the right and from the left at each $x_{k}$. Then $f\in R(\alpha )$ on $[a,b]$ and
\[\int_{a}^{b}f(x)d\alpha (x)=\sum_{k=1}^{n}f(x_{k})\alpha_{k}.\]

Proof. The result follows from Proposition \ref{map30} immediately. $\blacksquare$

One of the simplest step function is the greatest-integer function. Its value at $x$ is the greatest integer which is less than or equal to $x$ and is denoted by $[x]$. Therefore, $[x]$ is the unique integer satisfying the inequalities $[x]\leq x\leq [x]+1$.

\begin{equation}{\label{mat*32}}\tag{93}\mbox{}\end{equation}
Corollary \ref{mat*32}. Every finite sum can be written as a Riemann-Stieltjes integral. As a matter of fact, given a finite sum $\sum_{k=1}^{n}a_{k}$, we define a function $f$ on $[0,n]$ as follows
\[f(x)=\left\{\begin{array}{ll}
a_{k} & \mbox{if $k-1<x\leq k$ for $k=1,2,\cdots ,n$}\\
0 & \mbox{if $x=0$.}
\end{array}\right .\]
Then, we have
\[\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}f(k)=\int_{0}^{n}f(x)d[x].\]

Proof. It is obvious that $[x]$ is a step function, continuous from the right and having jump $1$ at each integer. The function $f$ is continuous from the left at the integers $1,2,\cdots ,n$. The result follows immediately from Theorem \ref{mat32}. $\blacksquare$

\begin{equation}{\label{mat**32}}\tag{94}\mbox{}\end{equation}
Corollary \ref{mat**32}. Let $f$ be defined on $[a,b]$ such that it is continuous from the left at $a,[a]+1,[a]+2,\cdots, [b]$ and from the right at $b$. Then, we have
\[\int_{a}^{b}f(x)d[x]=\sum_{k=[a]+1}^{[b]}f(k).\]

Proof. Since $[x]$ is a step function, continuous from the right at $a,[a]+1,[a]+2,\cdots,[b]$ and from the left at $b$, and having jump $1$ at $[a]+1 [a]+2,\cdots,[b]$ and jump $0$ at $a$ and $b$, the result follows immediately from Theorem \ref{mat32}. $\blacksquare$

Theorem. (Euler’s Summation Formula). Suppose that $f$ has a continuous derivative $f’$ on $[a,b]$. Then
\begin{equation}{\label{maeq42}}\tag{95}
\sum_{n=[a]+1}^{[b]}f(n)=\int_{a}^{b}f(x)dx+\int_{a}^{b}f'(x)(x-[x])dx+f(a)(a-[a])-f(b)(b-[b]).
\end{equation}
When $a$ and $b$ are integers, we have
\begin{equation}{\label{maeq46}}\tag{96}
\sum_{n=a}^{b}f(n)=\int_{a}^{b}f(x)dx+\int_{a}^{b}f'(x)\left (x-[x]-\frac{1}{2}\right )dx+\frac{f(a)+f(b)}{2}.
\end{equation}

Proof. Using the integration by parts in Theorem \ref{mat39}, we have
\begin{equation}{\label{maeq43}}\tag{97}
\int_{a}^{b}f(x)d(x-[x])+\int_{a}^{b}(x-[x])df(x)=f(b)(b-[b])-f(a)(a-[a]).
\end{equation}
Since $[x]$ has jump $1$ at the integers $[a]+1,[a]+2,\cdots ,[b]$, by Corollary \ref{mat**32}, we have
\begin{equation}{\label{maeq44}}\tag{98}
\int_{a}^{b}f(x)d[x]=\sum_{n=[a]+1}^{[b]}f(n).
\end{equation}
Using Proposition \ref{map40} and Theorem \ref{mat41}, we also have
\begin{equation}{\label{maeq45}}\tag{99}
\int_{a}^{b}f(x)d(x-[x])=\int_{a}^{b}f(x)dx-\int_{a}^{b}f(x)d[x]
\end{equation}
and
\begin{equation}{\label{maeq*45}}\tag{100}
\int_{a}^{b}(x-[x])df(x)=\int_{a}^{b}(x-[x])f'(x)dx.
\end{equation}
The equality (\ref{maeq42}) follows from (\ref{maeq43}), (\ref{maeq44}), (\ref{maeq45}) and (\ref{maeq*45}). When $a$ and $b$ are integers, since
\[\sum_{n=a}^{b}f(n)=f(a)+\sum_{n=a+1}^{b}f(n),\]
the equality (\ref{maeq46}) can be obtained from (\ref{maeq42}). This completes the proof. $\blacksquare$

\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}

Riemann-Stieltjes Integrals Depending on a Parameter.

We are going to consider the differentiation under the integral sign and interchanging the order of integration.

\begin{equation}{\label{ma20}}\tag{101}\mbox{}\end{equation}

Lemma \ref{ma20}. Suppose that two real-valued functions $f_{1},f_{2}:(M,d)\rightarrow\mathbb{R}$ are uniformly continuous on a subset $S$ of $M$. Then $f_{1}-f_{2}$ is also uniformly continuous on $S$.

Proof. Given any $\epsilon>0$, there exist $\delta_{1},\delta_{2}>0$ such that $d(x_{1},x_{2})<\delta_{1}$ implies $|f_{1}(x_{1})-f_{1}(x_{2})|<\frac{\epsilon}{2}$, and that $d(x_{1},x_{2})<\delta_{2}$ implies $|f_{2}(x_{1})-f_{2}(x_{2})|<\frac{\epsilon}{2}$ for $x_{1},x_{2}\in S$. Let $\delta=\min{\delta_{1},\delta_{2}}$, and let $f=f_{1}-f_{2}$. Then, for $d(x_{1},x_{2})<\delta$, we have

\begin{align*} & \left |f(x_{1})-f(x_{2})\right |\\ & quad =\left |f_{1}(x_{1})-f_{2}(x_{1})-f_{1}(x_{2})+f_{2}(x_{2})\right |\\ & \quad\leq\left |f_{1}(x_{1})-f_{1}(x_{2})\right |+\left |f_{2}(x_{2})-f_{2}(x_{1})\right |\\ & \quad<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \end{align*}

This completes the proof. $\blacksquare$

\begin{equation}{\label{map117}}\tag{102}\mbox{}\end{equation}
Proposition \ref{map117}. Let $f$ be continuous on the rectangle $[a,b]\times [c,d]$, and let $\alpha$ be of bounded variation on $[a,b]$. We define a function $F$ on $[c,d]$ by
\[F(y)=\int_{a}^{b}f(x,y)d\alpha (x).\]
Then $F$ is uniformly continuous on $[c,d]$. For any $y_{0}\in [c,d]$, we have
\begin{align*} \lim_{y\rightarrow y_{0}}\int_{a}^{b}f(x,y)d\alpha (x) & =\int_{a}^{b}\left [
\lim_{y\rightarrow y_{0}}f(x,y)\right ]d\alpha (x)\\ & =\int_{a}^{b}f(x,y_{0})d\alpha (x).\end{align*}

Proof. We first assume that $\alpha$ is increasing on $[a,b]$. Since the rectangle $Q\equiv [a,b]\times [c,d]$ is a compact set, the function $f$ is uniformly continuous on $Q$ by referring to Theorem \ref{ma104}. Therefore, given any $\epsilon >0$, there exists $\delta >0$ (which depends only on $\epsilon$) such that, for every pair of points $(x_{1},y_{1}),(x_{2},y_{2})\in Q$,
\[\parallel (x_{1},y_{1})-(x_{2},y_{2})\parallel <\delta\mbox{ implies }
\left |f(x_{1},y_{1})-f(x_{2},y_{2})\right |<\frac{\epsilon}{2[\alpha (b)-\alpha (a)]},\]
where $\alpha (b)-\alpha (a)\geq 0$ by the increasing assumption. For $y_{1},y_{2}\in [c,d]$ with $|y_{1}-y_{2}|<\delta$, given any fixed $x$, we have
$\parallel (x,y_{1})-(x,y_{2})\parallel <\delta$, which implies, by using Proposition \ref{map414},
\begin{align*}
\left |F(y_{1})-F(y_{2})\right | & =\left |\int_{a}^{b}\left [f(x,y_{1})-f(x,y_{2})\right ]d\alpha (x)\right |\\
& \leq\int_{a}^{b}\left |f(x,y_{1})-f(x,y_{2})\right |d\alpha (x)\\
& \quad\mbox{(since $\alpha$ is increasing on $[a,b]$)}\\
& \leq\int_{a}^{b}\frac{\epsilon}{2[\alpha (b)-\alpha (a)]}d\alpha (x)\\ & =\frac{\epsilon}{2[\alpha (b)-\alpha (a)]}\cdot [\alpha (b)-\alpha (a)]=\frac{\epsilon}{2}<\epsilon,
\end{align*}
which shows the uniform continuity of $F$ on $[c,d]$. Now, we assume that $\alpha$ is of bounded variation on $[a,b]$. Given any fixed $y$, the continuity of $f$ says $f(\cdot ,y)\in R(\alpha )$ on $[a,b]$. From Proposition \ref{map75}, it follows $f(\cdot ,y)\in R(\beta )$ on $[a,b]$ and $f(\cdot ,y)\in R(\beta -\alpha )$ on $[a,b]$, where $\beta$ and $\beta -\alpha$ are increasing functions on $[a,b]$. Therefore, we have
\begin{align*}
F(y) & =\int_{a}^{b}f(x,y)d\alpha (x)=\int_{a}^{b}f(x,y)d(\beta (x)-(\beta (x)-\alpha (x)))\\
& =\int_{a}^{b}f(x,y)d\beta (x)-\int_{a}^{b}f(x,y)d(\beta (x)-\alpha (x))\\ & \equiv F_{1}(y)-F_{2}(y),
\end{align*}
where the functions
\[F_{1}(y)=\int_{a}^{b}f(x,y)d\beta (x)\mbox{ and }F_{2}(y)=\int_{a}^{b}f(x,y)d(\beta (x)-\alpha (x))\]
are uniformly continuous on $[a,b]$. Using Lemma \ref{ma20}, this completes the proof. $\blacksquare$

Suppose that $\alpha (x)=x$. Then, Proposition \ref{map117} becomes a continuity theorem for Riemann integrals involving a parameter. However, we can derive a much more general result for Riemann integral below.

Proposition. Let $f$ be continuous on the rectangle $[a,b]\times [c,d]$. Suppose that the function $g$ is Riemann-integrable on $[a,b]$. Define a function $F$ on $[c,d]$ by
\begin{equation}{\label{maeq130}}\tag{103}
F(y)=\int_{a}^{b}g(x)f(x,y)dx.
\end{equation}
Then $F$ is uniformly continuous on $[c,d]$. For any $y_{0}\in [c,d]$, we also have
\begin{align*} \lim_{y\rightarrow y_{0}}\int_{a}^{b}g(x)f(x,y)dx & =\int_{a}^{b}\left [
\lim_{y\rightarrow y_{0}}g(x)f(x,y)\right ]dx\\ & =\int_{a}^{b}g(x)f(x,y_{0})dx.\end{align*}

Proof. Define
\[\alpha (x)=\int_{a}^{x}g(t)dt.\]
Using Theorem \ref{mat101}, we see that the function $F$ in (\ref{maeq130}) can be written as
\[F(y)=\int_{a}^{b}f(x,y)d\alpha (x)=\int_{a}^{b}g(x)f(x,y)dx.\]
The desired results follow immediately from Proposition \ref{map117}. $\blacksquare$

\begin{equation}{\label{mat367}}\tag{104}\mbox{}\end{equation}
Theorem \ref{mat367}. (Differentiation under the Integral Sign).
Let $Q=[a,b]\times [c,d]$, and let $\alpha$ be of bounded variation on $[a,b]$. For each fixed $y\in [c,d]$, suppose that the Riemann-Stieltjes integral
\[F(y)=\int_{a}^{b}f(x,y)d\alpha (x)\]
exists. We also assume that the partial derivative $\partial f/\partial y$ is continuous on $Q$. Then, the function $F$ is differentiable at each $y\in (c,d)$ and is given by
\[F'(y)=\int_{a}^{b}\left [\frac{\partial f}{\partial y}(x,y)\right ]d\alpha (x).\]
In particular, suppose that $g$ is Riemann-integrable on $[a,b]$ and define
\[\alpha (x)=\int_{a}^{x}g(t)dt.\]
Then, we have
\[F(y)=\int_{a}^{b}g(x)f(x,y)dx\] and \[F'(y)=\int_{a}^{b}\left [g(x)\cdot\frac{\partial f}{\partial y}(x,y)\right ]dx.\]

Proof. Given $y,y_{0}\in (c,d)$, we have
\begin{align*}
\frac{F(y)-F(y_{0})}{y-y_{0}} & =\int_{a}^{b}\frac{f(x,y)-f(x,y_{0})}{y-y_{0}}d\alpha (x)\\
& =\int_{a}^{b}\frac{\partial f}{\partial y}(x,\bar{y})d\alpha (x)\mbox{ (by the mean-value theorem for differentiation)}
\end{align*}
for some $\bar{y}$ between $y$ and $y_{0}$. Since $\partial f/\partial y$ is continuous on $Q$, using Proposition \ref{map117}, we have
\begin{align*}
F'(y_{0}) & =\lim_{y\rightarrow y_{0}}\frac{F(y)-F(y_{0})}{y-y_{0}}\\ & =\lim_{y\rightarrow y_{0}}\int_{a}^{b}\frac{\partial f}{\partial y}(x,\bar{y})d\alpha (x)\\
& =\int_{a}^{b}\left [\lim_{y\rightarrow y_{0}}\frac{\partial f}{\partial y}(x,\bar{y})\right ]d\alpha (x)\\
& =\int_{a}^{b}\left [\lim_{\bar{y}\rightarrow y_{0}}\frac{\partial f}{\partial y}(x,\bar{y})\right ]d\alpha (x)\mbox{ (since $\bar{y}$ is between $y$ and $y_{0}$)}\\
& =\int_{a}^{b}\frac{\partial f}{\partial y}(x,y_{0})d\alpha (x).
\end{align*}
Let
\[\alpha(x)=\int_{a}^{x}g(t)dt.\]
Then $\alpha$ is of bounded variation. Using Proposition \ref{mat112}, the proof is complete. $\blacksqyare$

\begin{equation}{\label{mat142}}\tag{105}\mbox{}\end{equation}
Theorem \ref{mat142}. (Interchanging the Order of Integration). Let $Q=[a,b]\times [c,d]$, and let $\alpha$ be of bounded variation on $[a,b]$
and $\beta$ be of bounded variation on $[c,d]$. For any $(x,y)\in Q$, we define
\[F(y)=\int_{a}^{b}f(x,y)d\alpha (x)\mbox{ and }G(x)=\int_{c}^{d}f(x,y)d\beta (y).\]
Suppose that $f$ is continuous on $Q$. Then $F\in R(\beta )$ on $[c,d]$ and $G\in R(\alpha )$ on $[a.b]$. Moreover, we have
\begin{equation}{\label{maeq140}}\tag{106}
\int_{c}^{d}F(y)d\beta (y)=\int_{a}^{b}G(x)d\alpha (x).
\end{equation}
In other words, we can interchange the order of integration as follows
\[\int_{a}^{b}\left [\int_{c}^{d}f(x,y)d\beta (y)\right ]d\alpha (x)=\int_{c}^{d}\left [\int_{a}^{b}f(x,y)d\alpha (x)\right ]d\beta (y).\]

Proof. Using Proposition \ref{map117}, we see that $F$ and $G$ are continuous on $[c,d]$ and $[a,b]$, respectively. Therefore $F\in R(\beta )$ on $[c,d]$ and $G\in R(\alpha )$ on $[a,b]$. For proving the equality (\ref{maeq140}), we first assume that $\alpha$ and $\beta$ are increasing on $[a,b]$ and $[c,d]$, respectively. Since the function $f$ is also uniformly continuous on the compact set $Q$, given any $\epsilon >0$, there exists $\delta >0$ (which depends only on $\epsilon$) such that, for every ${\bf z}_{1}=(x_{1},y_{1}),{\bf z}_{2}=(x_{2},y_{2})\in Q$,
\begin{equation}{\label{ma112}}\tag{107}
\parallel {\bf z}_{1}-{\bf z}_{2}\parallel <\delta\mbox{ implies }
|f(x_{1},y_{1})-f(x_{2},y_{2})|<\frac{\epsilon}{[\beta(d)-\beta(c)][\alpha(b)-\alpha(a)]}.
\end{equation}
We subdivide $Q$ into $n^{2}$ equal rectangles by subdividing $[a,b]$ and $[c,d]$ simultaneously into $n$ equal sub-intervals, where $n$ is chosen to satisfy
\begin{equation}{\label{maeq141}}\tag{108}
\frac{b-a}{n}<\frac{\delta}{\sqrt{2}}\mbox{ and }\frac{d-c}{n}<\frac{\delta}{\sqrt{2}}.
\end{equation}
We take
\[x_{k}=a+\frac{k(b-a)}{n}\mbox{ and }y_{k}=c+\frac{k(d-c)}{n}\]
for $k=0,1,2,\cdots ,n$. Using Theorem \ref{mat91}, we have
\begin{align*}
\int_{a}^{b}G(x)d\alpha (x) & =\int_{a}^{b}\left [\int_{c}^{d}f(x,y)d\beta (y)\right ]d\alpha (x)\\
& =\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\int_{x_{k}}^{x_{k+1}}\left [\int_{y_{j}}^{y_{j+1}}f(x,y)d\beta (y)\right ]d\alpha (x)\\
& =\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\int_{x_{k}}^{x_{k+1}}f(x,\hat{y}_{j})\cdot\left [\beta (y_{j+1})-\beta (y_{j})\right ]d\alpha (x)\\
& =\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}f(\hat{x}_{k},\hat{y}_{j})\left [\beta (y_{j+1})-\beta (y_{j})\right ]\cdot\left [\alpha (x_{k+1})-\alpha (x_{k})\right ],
\end{align*}
where $(\hat{x}_{k},\hat{y}_{j})$ is in the rectangle $Q_{kj}$ having $(x_{k},y_{j})$ and $(x_{k+1},y_{j+1})$ as the opposite vertices. Similarly, we have
\begin{align*}
\int_{c}^{d}F(y)d\beta (y) & =\int_{c}^{d}\left [\int_{a}^{b}f(x,y)d\alpha (x)\right ]d\beta (y)\\
& =\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}f(\bar{x}_{k},\bar{y}_{j})\left [\beta (y_{j+1})-\beta (y_{j})\right ]\cdot\left [\alpha (x_{k+1})-\alpha (x_{k})\right ],
\end{align*}
where $(\bar{x}_{k},\bar{y}_{j})$ is in the rectangle $Q_{kj}$. Now, by the assumption (\ref{maeq141}), we have
\begin{align*} \parallel (\hat{x}_{k},\hat{y}_{j})-(\bar{x}_{k},\bar{y}_{j})\parallel
& =\sqrt{(\hat{x}_{k}-\bar{x}_{k})^{2}+(\hat{y}_{k}-\bar{y}_{k})^{2}}
\\ & <\sqrt{\frac{\delta^{2}}{2}+\frac{\delta^{2}}{2}}=\delta.\end{align*}
Since $\alpha$ and $\beta$ are assumed to be increasing, by referring to (\ref{ma112}), the uniform continuity of $f$ says that
\[\left |f(\hat{x}_{k},\hat{y}_{j})-f(\bar{x}_{k},\bar{y}_{j})\right |<\frac{\epsilon}
{\left [\beta (d)-\beta (c)\right ]\cdot\left [\alpha (b)-\alpha (a)\right ]}.\]
Therefore, we obtain
\begin{align*}
& \left |\int_{a}^{b}G(x)d\alpha (x)-\int_{c}^{d}F(y)d\beta (y)\right |\\
& \quad <\frac{\epsilon}{\left [\beta (d)-\beta (c)\right ]\cdot\left [\alpha (b)-\alpha (a)\right ]}
\cdot\left (\sum_{j=1}^{n-1}\left [\beta (y_{j+1})-\beta (y_{j})\right ]\right )\cdot
\left (\sum_{k=0}^{n-1}\left [\alpha (x_{k+1})-\alpha (x_{k})\right ]\right )\\
& \quad =\frac{\epsilon}{\left [\beta (d)-\beta (c)\right ]\cdot\left [\alpha (b)-\alpha (a)\right ]}
\cdot\left [\beta (d)-\beta (c)\right ]\cdot\left [\alpha (b)-\alpha (a)\right ]=\epsilon .
\end{align*}
Since $\epsilon$ can be arbitrary positive number, this implies the equality (\ref{maeq140}). Now, we assume that $\alpha$ and $\beta$ are of bounded variation. Using Proposition \ref{map75}, there exist increasing functions $\alpha_{1}$ and $\alpha_{2}$ on $[a,b]$ satisfying $\alpha=\alpha_{1}-\alpha_{2}$ and
\[F(y)=\int_{a}^{b}f(x,y)d\alpha (x)=\int_{a}^{b}f(x,y)d\alpha_{1}(x)-\int_{a}^{b}f(x,y)d\alpha_{2}(x)\equiv F_{1}(y)-F_{2}(y).\]
Also, there exist increasing functions $\beta_{1}$ and $\beta_{2}$ on $[c,d]$ satisfying $\beta=\beta_{1}-\beta_{2}$ and
\[G(x)=\int_{c}^{d}f(x,y)d\beta (y)=\int_{a}^{b}f(x,y)d\beta_{1}(y)-\int_{c}^{d}f(x,y)d\beta_{2}(y)\equiv G_{1}(x)-G_{2}(x).\]
According to the previous result for the increasing case, we have
\[\int_{c}^{d}F_{1}(y)d\beta_{1}(y)=\int_{a}^{b}G_{1}(x)d\alpha_{1}(x)\mbox{ and }
\int_{c}^{d}F_{1}(y)d\beta_{2}(y)=\int_{a}^{b}G_{1}(x)d\alpha_{2}(x)\]
and
\[\int_{c}^{d}F_{2}(y)d\beta_{1}(y)=\int_{a}^{b}G_{2}(x)d\alpha_{1}(x)\mbox{ and }
\int_{c}^{d}F_{2}(y)d\beta_{2}(y)=\int_{a}^{b}G_{2}(x)d\alpha_{2}(x).\]
Finally, we obtain
\begin{align*}
\int_{c}^{d}F(y)d\beta (y) & =\int_{c}^{d}\left (F_{1}(y)-F_{2}(y)\right )d\left (\beta_{1}(y)-\beta_{2}(y)\right )\\
& =\int_{c}^{d}F_{1}(y)d\beta_{1}(y)-\int_{c}^{d}F_{1}(y)d\beta_{2}(y)-\int_{c}^{d}F_{2}(y)d\beta_{1}(y)+\int_{c}^{d}F_{2}(y)d\beta_{2}(y)\\
& =\int_{a}^{b}G_{1}(x)d\alpha_{1}(x)-\int_{a}^{b}G_{1}(x)d\alpha_{2}(x)-\int_{a}^{b}G_{2}(x)d\alpha_{1}(x)+\int_{a}^{b}G_{2}(x)d\alpha_{2}(x)\\
& =\int_{a}^{b}\left (G_{1}(x)-G_{2}(x)\right )d\left (\alpha_{1}(x)-\alpha_{2}(x)\right )\\
& =\int_{a}^{b}G(x)d\alpha (x).
\end{align*}
This completes the proof. $\blacksquare$

\begin{equation}{\label{mac349}}\tag{109}\mbox{}\end{equation}
Corollary \ref{mac349}. Let $f$ be continuous on the rectangle $[a,b]\times [c,d]$. Suppose that $g$ is Riemann-integrable on $[a,b]$, and that $h$ is Riemann-integrable on $[c,d]$. Then, we have
\[\int_{a}^{b}\left [\int_{c}^{d}g(x)h(y)f(x,y)dy\right ]dx=\int_{c}^{d}\left [\int_{a}^{b}g(x)h(y)f(x,y)dx\right ]dy.\]

Proof. We define
\[\alpha (x)=\int_{a}^{x}g(u)du\mbox{ and }\beta (y)=\int_{c}^{y}h(v)dv.\]
Then, the desired results follow from immediately Theorems \ref{mat101} and \ref{mat142}. $\blacksquare$

Riemann-Stieltjes sum.

Upper and Lower Riemann-Stieltjes Integrals.

Step Functions as Integrators.

Riemann-Stieltjes Integrals Depending on a Parameter.

Hsien-Chung Wu

發佈留言取消回覆

Riemann-Stieltjes sum.

Upper and Lower Riemann-Stieltjes Integrals.

Step Functions as Integrators.

Riemann-Stieltjes Integrals Depending on a Parameter.

Hsien-Chung Wu

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