General Measures

Samuel Lancaster Gerry (1813-1891) was an American painter.

We have sections

• Sigma-Field 
• Measurable Functions
• Measures
• Outer Measures
• Inner Measures

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

Sigma-Field.

The class of measurable functions plays a fundamental role in integration theory. We first introduce the concept of \(\sigma\)-field (or \(\sigma\)-algebra).

Definition. Let \(X\) be a universal set. A collection \(\mathfrak{M}\) of subsets of \(X\) is said to be a \(\sigma\)-field (or \(\sigma\)-algebra) when the following conditions are satisfied:

• \(X\in\mathfrak{M}\);
• if \(E\in\mathfrak{M}\), then \(E^{c}\in\mathfrak{M}\), where \(E^{c}\) is the complement of \(E\) relative to \(X\), i.e., \(E^{c}=X\setminus E\);
• for any sequence \(\{E_{k}\}_{k=1}^{\infty}\) in \(\mathfrak{M}\), we have \(\bigcup_{k=1}^{\infty}E_{k}\in\mathfrak{M}\).

Let \(\mathfrak{M}\) be a \(\sigma\)-field in \(X\). Then \((X,\mathfrak{M})\) is called a measurable space. The elements of \(\mathfrak{M}\) are called the measurable sets in \(X\). \(\sharp\)

For any sequence \(\{E_{k}\}_{k=1}^{\infty}\) in \(\mathfrak{M}\), we have \(\bigcap_{k=1}^{\infty}E_{k}\in\mathfrak{M}\), since

\[\bigcap_{k=1}^{\infty}E_{k}=\left (\bigcup_{k=1}^{\infty}E_{k}^{c}\right )^{c}.\]

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

Measurable Functions.

Let \(f\) be an extended real-valued functions defined on the measurable space \((X,\mathfrak{M})\). By referring to the proof of Theorem \ref{ma74}, the following statements are equivalent:

• \(\{x:f(x)<\alpha\}\in\mathfrak{M}\) for each \(\alpha\);
• \(\{x:f(x)\leq\alpha\}\in\mathfrak{M}\) for each \(\alpha\);
• \(\{x:f(x)>\alpha\}\in\mathfrak{M}\) for each \(\alpha\);
• $latex\{x:f(x)\geq\alpha\}\in\mathfrak{M}$ for each \(\alpha\).

We say that \(f\) is measurable with respect to \(\mathfrak{M}\) (or \(\mathfrak{M}\)-measurable) when any one of the above statements holds true.

Let \(E\) be a measurable set in \(X\). Then, the characteristic function
\[\chi_{E}(x)=\left\{\begin{array}{ll}
1 & \mbox{if \(x\in E\)}\\
0 & \mbox{if \(x\not\in E\)}
\end{array}\right .\]
is a measurable function.

Proposition. Let the extended real-valued functions \(f\) and \(g\) be measurable. Then, we have the following properties.

(i) For any constant \(c\), the functions \(f+c\) and \(cf\) are measurable.

(ii) The functions \(f+g\), \(fg\), \(\min\{f,g\}\) and \(\max\{f,g\}\) are measurable.

(iii) Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of measurable extended real-valued function. Then, the following functions

\[\sup_{k\geq 1}f_{k},\quad\inf_{k\geq}f_{k},\quad\limsup_{k\rightarrow\infty}f_{k}\mbox{ and }
\liminf_{k\rightarrow\infty}f_{k}\]

are measurable. \(\sharp\)

Let \(\phi :(X,\mathfrak{M})\rightarrow\bar{\mathbb{R}}\) be an extended real-valued function defined on a measurable space \((X,\mathfrak{M})\). We recall that \(\phi\) is a simple function when it can be expressed as the following form

\[\phi=\sum_{i=1}^{n}\alpha_{i}\chi_{A_{i}},\mbox{ where }A_{i}=\{x:\phi (x)=\alpha_{i}\}.\]

It is clear to see that the simple function \(\phi\) is measurable if and only if each of the sets \(A_{i}\) is measurable for \(i=1,\cdots,n\).

\begin{equation}{\label{rat90}}\tag{1}\mbox{}\end{equation}

Theorem \ref{rat90}. Suppose that the nonnegative and extended real-valued function \(f:(X,\mathfrak{M})\rightarrow [0,+\infty ]\) defined on a measurable space \((X,\mathfrak{M})\) is measurable. Then, there exists a sequence of simple functions \(\{s_{n}\}_{n=1}^{\infty}\) satisfying

\[\phi_{n+1}\geq\phi_{n}\mbox{ for all }n\mbox{ and }\lim_{n\rightarrow\infty}\phi_{n}(x)=f(x)\mbox{ for each }x\in X.\]

Next, we are going to consider the measurability of function whose range is a topological space.

Definition. Let \(X\) be a universal set. A collection \(\tau\) of subsets of \(X\) is said to be a topology in \(X\) when the following conditions are satisfied:

• \(\emptyset\in\tau\) and \(X\in\tau\);
• for any finite collection \(\{A_{1},\cdots ,A_{n}\}\) in \(\tau\), we have \(\bigcap_{i=1}^{n}A_{i}\in\tau\);
• for any arbitrary collection \(\{A_{\alpha}\}_{\alpha\in I}\) in \(\tau\), where \(I\) is the set of indices, we have \(\bigcup_{\alpha\in I}A_{\alpha}\in\tau\).

Let \(\tau\) be a topology in \(X\). Then \((X,\tau )\) is called a topological space. The elements of \(\tau\) are called the open sets in \(X\). \(\sharp\)

Given two universal sets \(X\) and \(Y\), we consider the function \(f:X\rightarrow Y\). The concepts of continuity and measurability of function \(f\) is based on the inverse image. Given any subset \(V\) of \(Y\), the inverse image $f^{-1}(V)$ of \(V\) is defined by

\[f^{-1}(V)=\left\{x\in X:f(x)\in V\right\}.\]

Let \((X,\tau_{X})\) and \((Y,\tau_{Y})\) be two topological spaces. The function \(f:(X,\tau_{X})\rightarrow (Y,\tau_{Y})\) is said to be continuous when, for any open set \(V\) in \(Y\), the inverse image \(f^{-1}(V)\) of \(V\) is an open set in \(X\); that is, \(V\in\tau_{Y}\) implies \(f^{-1}(V)\in\tau_{X}\).

Definition. Let \((X,\mathfrak{M})\) be a measurable space, and let \((Y,\tau_{Y})\) be a topological space. We say that the function \(f:(X,\mathfrak{M})\rightarrow (Y,\tau_{Y})\) is measurable when, for any open set \(V\) in \(Y\), the inverse image \(f^{-1}(V)\) of \(V\) is a measurable set in \(X\); that is, \(V\in\tau_{Y}\) implies \(f^{-1}(V)\in\mathfrak{M}\). \(\sharp\)

The definition of continuity given above is a global one. It is convenient to consider the continuity of local sense. Given a topological space \((X,\tau )\), a neighborhood of \(x\in X\) is an open set in \(X\) containing \(x\). The function \(f:(X,\tau_{X})\rightarrow (Y,\tau_{Y})\) is said to be continuous at the point \(x_{0}\in X\) when, for any neighborhood \(V\) of \(f(x_{0})\), there exists a neighborhood \(U\) of \(x_{0}\) satisfying \(f(U)\subseteq V\).

Proposition. The function \(f:(X,\tau_{X})\rightarrow (Y,\tau_{Y})\) is continuous if and only if \(f\) is continuous at each point of \(X\).

Proposition. Let \(g:(Y,\tau_{Y})\rightarrow (Z,\tau_{Z})\) be a continuous function from the topological space \((Y,\tau_{Y})\) into the topological space \((Z,\tau_{Z})\). We have the following properties.

(i) Let \((X,\tau_{X})\) be a topological space. Suppose that the function \(f:(X,\tau_{X})\rightarrow (Y,\tau_{Y})\) is continuous. Then, the composition \(h=g\circ f:(X,\tau_{X})\rightarrow (Z,\tau_{Z})\) is continuous.

(ii) Let \((X,\mathfrak{M}_{X})\) be a measurable space. Suppose that the function \(f:(X,\mathfrak{M}_{X})\rightarrow (Y,\tau_{Y})\) is measurable. Then, the composition \(h=g\circ f:(X,\tau_{X})\rightarrow (Z,\tau_{Z})\) is measurable. \(\sharp\)

We can also consider the measurable function without using topological space.

Definition. We consider a function \(f:(X,\mathfrak{M}_{X})\rightarrow (Y,\mathfrak{M}_{Y})\) from the measurable space \((X,\mathfrak{M}_{X})\) into another measurable space \((Y,\mathfrak{M}_{Y}))\). In this case, we say that \(f\) is measurable when \(V\in\mathfrak{M}_{Y}\) implies \(f^{-1}(Y)\in\mathfrak{M}_{X}\).

\begin{equation}{\label{rap1}}\tag{2}\mbox{}\end{equation}

Proposition \ref{rap1}. Let \(\mathfrak{F}\) be any collection of subsets of \(X\). Then, there exists a smallest \(\sigma\)-field \(\mathfrak{M}_{\mathfrak{F}}\) in \(X\) satisfying \(\mathfrak{F}\subseteq\mathfrak{M}_{\mathfrak{F}}\). \(\sharp\)

The smallest \(\sigma\)-field \(\mathfrak{M}_{\mathfrak{F}}\) described in Proposition \ref{rap1} is also called the \(\sigma\)-field generated by \(\mathfrak{F}\). Let \((X,\tau)\) be a topological space, and let \(\mathfrak{O}\) be the family of all open sets in \(X\). Proposition \ref{rap1} says that there exists a smallest \(\sigma\)-field \(\mathfrak{B}\) satisfying \(\mathfrak{O}\subset\mathfrak{B}\). In this case, the elements of \(\mathfrak{B}\) are called the Borel sets in \(X\), and \((X,\mathfrak{B})\) is called a Borel measurable space. Then, we have the following observations.

• The open sets are Borel sets.
• The closed sets are Borel sets, since it is the complement of some open set.
• The countable union of closed sets is a Borel set.
• The countable intersection of open sets is a Borel set.

Let \((X,\tau_{X})\) be a topological space. Then, we can induce a Borel measurable space \((X,\mathfrak{B}_{X})\) from \(\tau_{X}\). Now, we consider the function \(f:(X,{\cal B}_{X})\rightarrow (Y,\tau_{Y})\). Suppose that the function \(f\) is continuous. Then, we see that \(f\) is measurable with respect to the Borel measurable space \((X,\mathfrak{B}_{X})\), since \(f^{-1}(V)\in\mathfrak{B}\) for any open set \(V\) in \(Y\). In this case, the function is also called Borel measurable or simply, it is called a Borel function. In other words, every continuous function is a Borel function.

Proposition. We consider the function \(f:(X,\mathfrak{M}_{X})\rightarrow (Y,\tau_{Y})\) from the measurable space \((X,\mathfrak{M}_{X})\) into the topological space \((Y,\tau_{Y})\). Then, we have the following properties.

(i) Let \(\mathfrak{M}_{Y}\) be a collection of all sets \(E\subset Y\) satisfying \(f^{-1}(E)\in\mathfrak{M}_{X}\). Then \(\mathfrak{M}_{Y}\) is a \(\sigma\)-field in \(Y\).

(ii) Let \(f\) be a measurable function, and let \(E\) be a Borel set in \(Y\). Then \(f^{-1}(E)\in\mathfrak{M}_{X}\).

(iii) We take \(Y=\bar{\mathbb{R}}\). Suppose that \(f^{-1}((\alpha ,+\infty ])\in\mathfrak{M}_{X}\) for each \(\alpha\). Then \(f\) is a measurable function.

(iv) Given a topological space \((Z,\tau_{Z})\), suppose that \(f\) is measurable, and that \(g:(Y,\tau_{Y})\rightarrow (Z,\tau_{Z})\) is a Borel function. Then, the composition \(h=g\circ f:(X,\mathfrak{M}_{X})\rightarrow (Z,\tau_{Z})\) is measurable.

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

Measures.

Definition. The nonnegative and extended real-valued function \(\mu :(X,\mathfrak{M})\rightarrow [0,+\infty ]\) defined on the measurable space \((X,\mathfrak{M})\) is called a measure when, given any disjoint countable collection \(\{A_{i}\}_{i=1}^{\infty}\) of measurable sets, we have

\[\mu\left (\bigcup_{i=1}^{\infty}A_{i}\right )=\sum_{i=1}^{\infty}\mu (A_{i}).\]

The triple \((X,\mathfrak{M},\mu )\) is also called a measure space. \(\sharp\)

\begin{equation}{\label{rap92}}\tag{3}\mbox{}\end{equation}

Proposition \ref{rap92}. Let \(\mu\) be a nonnegative measure defined on a measurable space \((X,\mathfrak{M})\). Then, we have the following properties.

(i) We have \(\mu (\emptyset )=0\).

(ii) Let \(A_{1},\cdots ,A_{n}\) be pairwise disjoint measurable sets. Then, we have

\[\mu\left (\bigcup_{i=1}^{n}A_{i}\right )=\sum_{i=1}^{n}\mu (A_{i}).\]

(iii) Given any measurable sets \(A\) and \(B\) satisfying \(A\subseteq B\), we have \(\mu (A)\leq\mu (B)\).

(iv) Given any sequence \(\{A_{i}\}_{i=1}^{\infty}\) in \(\mathfrak{M}\) satisfying

\[A_{n}\subseteq A_{n+1}\mbox{ for all }n\mbox{ and }A=\bigcup_{n=1}^{\infty}A_{n}.\]

Then, we have

\[\lim_{n\rightarrow\infty}\mu (A_{n})=\mu (A).\]

(v) Given any sequence \(\{A_{i}\}_{i=1}^{\infty}\) in \(\mathfrak{M}\) satisfying

\[A_{n+1}\subseteq A_{n}\mbox{ for all }n,\quad\mu (A_{1})<\infty\mbox{ and }A=\bigcap_{n=1}^{\infty}A_{n}.\]

Then, we have

\[\lim_{n\rightarrow\infty}\mu (A_{n})=\mu (A).\]

(vi) Given any sequence \(\{A_{i}\}_{i=1}^{\infty}\) in \(\mathfrak{M}\), we have

\[\mu\left (\bigcup_{n=1}^{\infty}A_{n}\right )\leq\sum_{n=1}^{\infty}\mu (A_{n}).\]

Given a sequence of sets \(\{E_{k}\}_{k=1}^{\infty}\), we define

\[\limsup_{k\rightarrow\infty}E_{k}=\bigcap_{m=1}^{\infty}\bigcup_{k=m}^{\infty}E_{k}\mbox{ and }
\liminf_{k\rightarrow\infty}E_{k}=\bigcup_{m=1}^{\infty}\bigcap_{k=m}^{\infty}E_{k}.\]

We can show

\[\limsup_{k\rightarrow\infty}E_{k}\in\mathfrak{M}\mbox{ and }
\liminf_{k\rightarrow\infty}E_{k}\in\mathfrak{M}.\]

Proposition. Let \((X,\mathfrak{M},\mu )\) be a measure space. Given a sequence \(\{E_{k}\}_{k=1}^{\infty}\) of measurable sets, we have

\[\mu\left (\liminf_{k}E_{k}\right )\leq\liminf_{k\rightarrow\infty}\mu (E_{k}).\]

We further assume that \(\mu (\bigcup_{k=k_{0}}^{\infty}E_{k})<+\infty\) for some \(k_{0}\). Then, we have

\begin{align*} \mu\left (\liminf_{k\rightarrow\infty}E_{k}\right ) & \leq\liminf_{k\rightarrow\infty}\mu (E_{k})
\leq\limsup_{k\rightarrow\infty}\mu (E_{k})\\ &\leq\mu\left (\limsup_{k\rightarrow\infty}E_{k}\right ).\end{align*}

Let \(\mu\) be a nonnegative measure defined on a measurable space \((X,\mathfrak{M})\). For any \(E\in\mathfrak{M}\), the statement “Property P holds true almost everywhere on \(E\)” means that there exists an \(N\in\mathfrak{M}\) satisfying \(N\subseteq E\) with \(\mu (N)=0\) and the property P holds true at every point of \(E\setminus N\). The concept depends on the given measure \(\mu\). Therefore, we shall write “Property P holds a.e. \([\mu ]\) on \(E\)”. For example, given any two measurable extended real-valued functions \(f\) and \(g\), suppose that

\[\mu\left (\left\{x:f(x)\neq g(x)\right\}\right )=0.\]

Then, we say that \(f=g\) a.e. \([\mu ]\) on \(X\). Similarly, when we write \(f\leq g\) a.e. \([\mu ]\), it means

\[\mu\left (\left\{x:f(x)>g(x)\right\}\right )=0.\]

Definition. Given a measurable space \((X,\mathfrak{M})\), we also have many terminologies given below

• A measure \(\mu\) is called finite when \(\mu (X)<+\infty\). The Lebesgue measure on \([0,1]\) is an example of a finite measure.
• A measure \(\mu\) is called \(\sigma\)-finite when there exists a sequence \(\{E_{k}\}_{k=1}^{\infty}\) of measurable sets satisfying
  \[X=\bigcup_{k=1}^{\infty}E_{k}\mbox{ and }\mu (E_{k})<+\infty\mbox{ for all }k.\] The Lebesgue measure on \((-\infty ,+\infty )\) is an example of a \(\sigma\)-finite measure.
• A set \(E\) is said to be of finite measure when \(E\in\mathfrak{M}\) and \(\mu (E)<+\infty\).
• A measure \(\mu\) is said to be semifinite when each measurable set of infinite measure contains measurable sets of arbitrary large finite measure. Therefore, every \(\sigma\)-finite measure is semifinite.

A set \(E\) is said to be of \(\sigma\)-finite measure when \(E\) is the union of a countable collection of measurable sets of finite measure. Then, we have the following properties.

• Any measurable set contained in a set of \(\sigma\)-finite measure is itself of \(\sigma\)-finite measure.
• The union of a countable collection of sets of \(\sigma\)-finite measure is of \(\sigma\)-finite measure.
• Let \(\mu\) be a \(\sigma\)-finite measure. Then, every measurable set is of \(\sigma\)-finite measure.

Given a measure space \((X,\mathfrak{M},\mu)\), we say that \(N\) is a negligible set when \(N\in\mathfrak{M}\) and \(\mu (N)=0\). It is clear that every subset of a negligible set is negligible. But it may happen that some set \(N\in\mathfrak{M}\) with \(\mu (N)=0\) has a subset \(E\) that is not in \(\mathfrak{M}\). In this case, we can define \(\mu (E)=0\). The problem is whether this extension of \(\mu\) is still a measure or not. The answer will be presented in the sequel.

Definition. A measure space \((X,\mathfrak{M},\mu )\) is said to be complete when \(\mathfrak{M}\) contains all subsets of sets of measure zero. More precisely, if \(E\in\mathfrak{M}\) with \(\mu (E)=0\), then \(F\subset E\) implies \(F\in\mathfrak{M}\). \(\sharp\)

The Lebesgue measure is complete. However, if the Lebesgue measure that is restricted to the Borel \(\sigma\)-field is not complete. The following proposition shows that each measure space can be completed by the addition of subsets of measure zero.

\begin{equation}{\label{rap39}}\tag{4}\mbox{}\end{equation}

Proposition \ref{rap39}. Let \((X,\mathfrak{M},\mu )\) be a measure space. Then, we can find a complete measure space \((X,\mathfrak{M}_{0},\mu_{0})\) such that the following properties are satisfied.

• We have \(\mathfrak{M}\subseteq\mathfrak{M}_{0}\).
• \(E\in\mathfrak{M}\) implies \(\mu (E)=\mu_{0}(E)\).
• \(E\in\mathfrak{M}_{0}\) if and only if \(E=A\cup B\), where \(B\in\mathfrak{M}\) and \(A\subseteq N\) with \(N\in\mathfrak{M}\) and \(\mu (N)=0\). \(\sharp\)

\begin{equation}{\label{rat5}}\tag{5}\mbox{}\end{equation}

Theorem \ref{rat5}. Given a measure space \((X,\mathfrak{M},\mu )\), let \(\mathfrak{M}^{*}\) be the collection of all sets \(E\subseteq X\) such that there exist sets \(A,B\in\mathfrak{M}\) satisfying \(A\subseteq E\subseteq B\) and \(\mu (B\setminus A)=0\). In this situation, we define \(\mu (E)=\mu (A)\). Then \(\mathfrak{M}^{*}\) is a \(\sigma\)-field and \(\mu\) is a measure on \(\mathfrak{M}^{*}\). \(\sharp\)

The extended measure \(\mu\) in Theorem \ref{rat5} is complete, since all subsets of sets of measure zero are now measurable. The \(\sigma\)-field \(\mathfrak{M}^{*}\) is called the \(\mu\)-completion of \(\mathfrak{M}\). This theorem says that every measure can be completed. Therefore, for convenience, we may assume that any given measure is complete.

Theorem. Given a measure space \((X,\mathfrak{A},\mu )\), let \(\mathfrak{B}\) be a collection of subsets of \(X\) which is closed under countable unions and which has the property that, for each \(A\in\mathfrak{A}\) with \(A\subseteq B\in\mathfrak{B}\), we have \(\mu (A)=0\). Then, there is an extension \(\bar{\mu}\) of \(\mu\) to the smallest \(\sigma\)-field \(\mathfrak{M}\) containing \(\mathfrak{A}\) and \(\mathfrak{B}\) satisfying \(\bar{\mu}(B)=0\) for each \(B\in\mathfrak{B}\). \(\sharp\)

Proposition. Let \(f\) be a nonnegative and measurable extended real-valued function defined on \((X,\mathfrak{M})\). Then, there is a sequence \(\{\phi_{k}\}_{k=1}^{\infty}\) of simple functions satisfying \(\phi_{k+1}\geq\phi_{k}\) for all \(k\) and

\[\lim_{k\rightarrow\infty}\phi_{k}(x)=f(x)\mbox{ for each }x\in X.\]

If \(f\) is defined on a \(\sigma\)-finite measure space, then we may choose the functions \(\{\phi_{k}\}_{k=1}^{\infty}\) such that each vanishes outside a set of finite measure.

Proposition. Let \(\mu\) be a complete measure. Suppose that the extended real-valued function \(f\) defined on \((X,\mathfrak{M},\mu )\) is measurable. Then \(f=g\) a.e. on \(X\) implies \(g\) is measurable.

Theorem. (Egorov’s Theorem). Let \((X,\mathfrak{M},\mu )\) be a measure space, and let \(E\) be a measurable set with \(\mu (E)<+\infty\). Suppose that \(\{f_{k}\}_{k=1}^{\infty}\) is a sequence of measurable functions such that \(f_{k}<+\infty\) a.e. on \(E\) for all \(k\) and the sequence \(\{f_{k}\}_{k=1}^{\infty}\) converges a.e. on \(E\) to a finite limit function \(f\). Then, given any \(\epsilon >0\), there exists a measurable set \(F\subseteq E\) such that \(\mu (E\setminus F)<\epsilon\) and \(\{f_{k}\}_{k=1}^{\infty}\) converges uniformly to \(f\) on \(F\). \(\sharp\)

Signed Measures.

We are going to consider the measures that are allowed to take on both positive and negative values.

Definition. Let \((X,\mathfrak{M})\) be a measurable space, and let \(\lambda\) be an extended real-valued set function defined on \(\mathfrak{M}\). We say that \(\lambda\) is a signed measure when the following conditions are satisfied:

• \(\lambda\) assumes at most one of the values \(+\infty\) and \(-\infty\);
• \(\lambda (\emptyset )=0\);
•  for any sequence \(\{E_{k}\}_{k=1}^{\infty}\) of disjoint measurable sets, we have
  \[\lambda\left (\bigcup_{k=1}^{\infty}E_{k}\right )=\sum_{k=1}^{\infty}\lambda (E_{k}),\]
  where the equality means that if \(\lambda (\bigcup_{k=1}^{\infty}E_{k})<+\infty\), then the series on the right converges absolutely. Otherwise, it diverges. \(\sharp\)

It is clear to see that a measure is a special case of a signed measure. Sometimes the signed measure is called the \(\sigma\)-additive set function. We define some terminologies.

• We say that a set \(A\) is a positive set with respect to the signed measure \(\lambda\) when \(A\) is measurable and, for every measurable subset \(E\) of \(A\), we have \(\lambda (E)\geq 0\). Every measurable subset of a positive set is again positive. If we take the restriction of \(\lambda\) to positive sets, then we obtain a measure.
• A set \(B\) is called a negative set when \(B\) is measurable and, for every measurable subset \(F\) of \(B\), we have \(\lambda (F)\leq 0\).
• A set which is both positive and negative with respect to \(\lambda\) is called a null set. A measurable set \(N\) is a null set if and only if every measurable subset \(G\) of \(N\) has \(\lambda (G)=0\). We also note that every null set must have measure zero, and a set of measure zero may be a union of two sets whose measures are not zero but they are positive and negative of each other.

Proposition. We have the following properties.

(i) Every measurable subset of a positive set is also positive.

(ii) The union of a countable collection of positive sets is also positive.

(iii) Let \(E\) be a measurable set satisfying \(0<\lambda (E)<+\infty\). Then, there is a positive set \(A\subseteq E\) satisfying \(\lambda (A)>0\). \(\sharp\)

Definition. Let \(\lambda\) be a signed measure on the measurable space \((X,\mathfrak{M})\). A decomposition of \(X\) into two disjoint sets \(A\) and \(B\) such that \(A\) is positive and \(B\) is negative for \(\lambda\) is called a Hahn decomposition for \(\lambda\). \(\sharp\)

\begin{equation}{\label{rat40}}\tag{6}\mbox{}\end{equation}

Theorem \ref{rat40}. (Hahn Decomposition Theorem). Let \(\lambda\) be a signed measure defined on \((X,\mathfrak{M})\). Then, there exist a positive set \(A\) and a negative set \(B\) satisfying \(X=A\cup B\) and \(A\cap B=\emptyset\). \(\sharp\)

Theorem \ref{rat40} states the existence of a Hahn decomposition for a signed measure. However, the Hahn decomposition is not unique.

Definition. Two measures \(\mu_{1}\) and \(\mu_{2}\) on \((X,\mathfrak{M})\) are said to be mutually singular when there exist two disjoint measurable sets \(A\) and \(B\) satisfying \(X=A\cup B\) and \(\mu_{1}(A)=\mu_{2}(B)=0\). \(\sharp\)

Suppose that \(\{A,B\}\) is a Hahn decomposition for \(\lambda\). Then, we may define two measures \(\lambda^{+}\) and \(\lambda^{-}\) with \(\lambda =\lambda^{+}-\lambda^{-}\) by setting

\[\lambda^{+}(E)=\lambda (E\cap A)\mbox{ and }\lambda^{-}(E)=-\lambda (E\cap B).\]

Thus, the measures \(\lambda^{+}\) and \(\lambda^{-}\) defined are mutually singular.

\begin{equation}{\label{rat41}}\tag{7}\mbox{}\end{equation}

Theorem \ref{rat41}. (Jordan Decomposition). Let \(\lambda\) be a signed measure on the measurable space \((X,\mathfrak{M})\). Then, there exist two mutually singular measures \(\lambda^{+}\) and \(\lambda^{-}\) on \((X,\mathfrak{M})\) satisfying \(\lambda =\lambda^{+}-\lambda^{-}\), where \(\lambda^{+}\) and \(\lambda^{-}\) are the only pair of mutually singular measures. \(\sharp\)

The decomposition of \(\lambda\) given in Theorem \ref{rat41} is called the Jordan decomposition of \(\lambda\). The measures of \(\lambda^{+}\) and \(\lambda^{-}\) are called the positive and negative parts of \(\lambda\). Since \(\lambda\) assumes at most one of the values \(+\infty\) and \(-\infty\), either \(\lambda^{+}\) or \(\lambda^{-}\) must be finite. When they are both finite, we call \(\lambda\) a finite signed measure.

Definition. The measure \(|\lambda |\) defined by

\[|\lambda |(E)=\lambda^{+}(E)+\lambda^{-}(E)\]

is called the total variation (or absolute value) of \(\lambda\). \(\sharp\)

We have the following observations.

• A set \(E\) is positive for \(\lambda\) if \(\lambda^{-}(E)=0\).
• A set \(E\) is a null set if \(|\lambda |(E)=0\).

Definition. Let \(\mu_{1}\) and \(\mu_{2}\) be two measures on \((X,\mathfrak{M})\). The measure \(\mu_{2}\) is said to be absolutely continuous with respect to \(\mu_{1}\) when \(\mu_{1}(E)=0\) implies \(\mu_{2}(E)=0\). In this case, we write \(\mu_{2}\ll\mu_{1}\). \(\sharp\)

Suppose that \(\mu_{2}\ll\mu_{1}\). If a property holds true a.e. \([\mu_{1}]\), then it also holds true a.e. \([\mu_{2}]\).

Theorem. (Lebesgue Decomposition). Let \((X,\mathfrak{M},\mu_{1})\) be a \(\sigma\)-finite measure space. Given any \(\sigma\)-finite measure \(\mu_{2}\) defined on \((X,\mathfrak{M})\), there exist a measure \(\mu_{s}\) which is mutually singular with respect to \(\mu_{1}\) and a measure \(\mu_{a}\) which is absolutely continuous with respect to \(\mu_{1}\) satisfying \(\mu_{2}=\mu_{s}+\mu_{a}\). The measures \(\mu_{s}\) and \(\mu_{a}\) are unique.

Proposition. Let \(\lambda\) be a finite signed measure defined on a measurable space \((X,\mathfrak{M})\). Then, we have the following properties.

(i) Given any sequence \(\{A_{n}\}_{n=1}^{\infty}\) in \(\mathfrak{M}\), suppose that

\[A_{n}\subseteq A_{n+1}\mbox{ for all }n\mbox{ and }A=\bigcup_{n=1}^{\infty}A_{n}.\]

Then, we have

\[\lim_{n\rightarrow\infty}\lambda (A_{n})=\lambda (A).\]

(ii) Given any sequence \(\{A_{n}\}_{n=1}^{\infty}\) in \(\mathfrak{M}\), suppose that

\[A_{n+1}\subseteq A_{n}\mbox{ for all }n\mbox{ with }\lambda (A_{1})<\infty\mbox{ and }A=\bigcap_{n=1}^{\infty}A_{n}.\]

Then, we have

\[\lim_{n\rightarrow\infty}\lambda (A_{n})=\lambda (A).\]

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

Outer Measures.

Definition. Let \(\mu^{*}\) be a set function defined on \(X\). We say that \(\mu^{*}\) is an outer measure when the following conditions are satisfied:

• \(\mu^{*}(\emptyset )=0\);
• \(A\subseteq B\) implies \(\mu^{*}(A)\subseteq\mu^{*}(B)\);
• for any countable collection of sets \(\{E_{k}\}_{k=1}^{\infty}\), we have
  \[\mu^{*}\left (\bigcup_{k=1}^{\infty}E_{k}\right )\leq\sum_{k=1}^{\infty}\mu^{*}\left (E_{k}\right ).\]

The outer measure \(\mu^{*}\) is said to be finite when \(\mu^{*}(X)<+\infty\). \(\sharp\)

The Lebesgue outer measure defined in (\ref{raeq51}) is an outer measure on \(\bar{\mathbb{R}}^{n}\). Inspired by Proposition \ref{rat45}, the concept of measurable set is defined below.

Definition. Let \(E\) be a subset of \(X\). We say that \(E\) is measurable with respect to \(\mu^{*}\) when, for any subset \(A\) of \(X\), we have

\[\mu^{*}(A)=\mu^{*}(A\cap E)+\mu^{*}(A\cap E^{c})=\mu^{*}(A\cap E)+\mu^{*}(A\setminus E).\]

In order to prove the measurability of set \(E\), since \(\mu^{*}\) is subadditive by the third condition, it is only necessary to show

\begin{equation}{\label{ma75}}\tag{8}
\mu^{*}(A)\geq\mu^{*}(A\cap E)+\mu^{*}(A\cap E^{c})\mbox{ for each }A.
\end{equation}

This inequality is trivially true if \(\mu^{*}(A)=+\infty\). Therefore, we need only to consider sets \(A\) with \(\mu^{*}(A)<+\infty\).

Proposition. Suppose that \(\mu^{*}(N)=0\). Then \(N\) is \(\mu^{*}\)-measurable.

Proof. For any \(A\subseteq X\), the second condition of \(\mu^{*}\) gives

\[\mu^{*}(A\cap N)+\mu^{*}(A\setminus N)\leq\mu^{*}(N)+\mu^{*}(A)=\mu^{*}(A).\]

This says that \(N\) is \(\mu^{*}\)-measurable by referring to (\ref{ma75}). \(\blacksquare\)

Proposition. Suppose that \(E_{1}\) and \(E_{2}\) are \(\mu^{*}\)-measurable. Then \(E_{1}\setminus E_{2}\) is also \(\mu^{*}\)-measurable.

\begin{equation}{\label{rap46}}\tag{9}\mbox{}\end{equation}

Proposition \ref{rap46}. Let \(\mathfrak{M}^{*}\) be a family of all \(\mu^{*}\)-measurable sets. Then \(\mathfrak{M}^{*}\) is a \(\sigma\)-field. Suppose that \(\bar{\mu}\) is \(\mu^{*}\) restricted to \(\mathfrak{M}^{*}\). Then \(\bar{\mu}\) is a complete measure on \(\mathfrak{M}^{*}\).

Proposition. Let \(\mu^{*}\) be an outer measure on \(X\). Then, we have the following properties.

(i) Let \(\{E_{k}\}_{k=1}^{\infty}\) be a countable collection of disjoint measurable sets. Then,we have

\[\mu^{*}\left (\bigcup_{k=1}^{\infty}E_{k}\right )=\sum_{k=1}^{\infty}\mu^{*}\left (E_{k}\right ).\]

In other words, if the outer measure \(\mu^{*}\) is defined on a \(\sigma\)-field \((X,\mathfrak{M})\), then it becomes a measure on \((X,\mathfrak{M})\).

(ii) For any \(A\subseteq X\), we have

\[\mu^{*}\left (A\bigcap\left (\bigcup_{k=1}^{\infty}E_{k}\right )\right )=\sum_{k=1}^{\infty}\mu^{*}\left (A\cap E_{k}\right )\]

and

\[\mu^{*}(A)=\mu^{*}\left (A\setminus\left (\bigcup_{k=1}^{\infty}E_{k}\right )\right )+\sum_{k=1}^{\infty}\mu^{*}\left (A\cap E_{k}\right ).\]

Proposition. Let \(\mu^{*}\) be an outer measure on \(X\), and let \(\{E_{k}\}_{k=1}^{\infty}\) be a sequence of \(\mu^{*}\)-measurable sets. Given any subset \(A\) of \(X\), we have the following properties.

(i) Suppose that \(E_{k}\subseteq E_{k+1}\) for all \(k\) or \(E_{k+1}\subseteq E_{k}\) for all \(k\) with \(\mu^{*}(A\cap E_{k_{0}})<+\infty\) for some \(k_{0}\). Then, we have

\begin{equation}{\label{raeq50}}\tag{10}
\mu^{*}\left (A\cap\lim_{k\rightarrow\infty}E_{k}\right )=\lim_{k\rightarrow\infty}\mu^{*}\left (A\cap E_{k}\right ).
\end{equation}

(ii) We have

\[\mu^{*}\left (A\cap\liminf_{k\rightarrow\infty}E_{k}\right )\leq\liminf_{k\rightarrow\infty}\mu^{*}\left (A\cap E_{k}\right ).\]

(iii) Suppose that

\[\mu^{*}\left (A\cap\left (\bigcup_{k=k_{0}}^{\infty}E_{k}\right )\right )<+\infty\mbox{ for some }k_{0}.\]

Then, we have

\[\mu^{*}\left (A\cap\limsup_{k\rightarrow\infty}E_{k}\right )\geq\limsup_{k\rightarrow\infty}\mu^{*}\left (A\cap E_{k}\right ).\]

Definition. Let \(\mathfrak{A}\) be a family of subsets of \(X\). We say that \(\mathfrak{A}\) is an algebra when the following conditions are satisfied:

• \(A_{1}\cup A_{2}\in\mathfrak{A}\) for any \(A_{1},A_{2}\in\mathfrak{A}\);
• \(A^{c}\in\mathfrak{A}\) for any \(A\in\mathfrak{A}\). \(\sharp\)

It is clear to see that any \(\sigma\)-field is an algebra.

Definition. Let \(\hat{\mu}\) be a nonnegative extended real-valued set function defined on an algebra \(\mathfrak{A}\). We say that \(\hat{\mu}\) is a measure on an algebra \(\mathfrak{A}\) when the following conditions are satisfied:

• \(\hat{\mu}(\emptyset )=0\);
• for any disjoint countable collection \(\{A_{i}\}_{i=1}^{\infty}\) in \(\mathfrak{A}\) satisfying \(\bigcup_{i=1}^{\infty}A_{i}\in\mathfrak{A}\), we have
  \[\hat{\mu}\left (\bigcup_{i=1}^{\infty}A_{i}\right )=\sum_{i=1}^{\infty}\hat{\mu}(A_{i}).\]

Let \(\hat{\mu}\) be a measure on an algebra \(\mathfrak{A}\). We are going to extend it to be a measure defined on a \(\sigma\)-field \(\mathfrak{M}^{*}\) containing \(\mathfrak{A}\). We can do this by using \(\hat{\mu}\) to construct an outer measure \(\mu^{*}\) and show that the measure \(\bar{\mu}\)
induced by \(\mu^{*}\) as shown in Proposition \ref{rap46} is the desired extension of \(\hat{\mu}\). Now, we define

\begin{equation}{\label{ma76}}\tag{11}
\mu^{*}(E)=\inf\sum_{k=1}^{\infty}\hat{\mu}(A_{k}),
\end{equation}

where the infimum is taken over all sequences \(\{A_{k}\}_{k=1}^{\infty}\) of sets in \(\mathfrak{A}\) satisfying \(E\subseteq\bigcup_{k=1}^{\infty}A_{k}\).

Proposition. Let \(\hat{\mu}\) be a measure on an algebra \(\mathfrak{A}\). The set function \(\mu^{*}\) is defined in \((\ref{ma76})\). Then, we have the following properties.

(i) Let \(A\in\mathfrak{A}\) and \(\{A_{k}\}_{k=1}^{\infty}\) be any sequence of sets in \(\mathfrak{A}\) satisfying \(A\subseteq\bigcup_{k=1}^{\infty}A_{k}\). Then, we have

\[\hat{\mu}(A)\leq\sum_{k=1}^{\infty}\hat{\mu}(A_{k}).\]

(ii) Suppose that \(A\in\mathfrak{A}\). Then \(\mu^{*}(A)=\hat{\mu}(A)\).

(iii) The set function \(\mu^{*}\) is an outer measure.

(iv) Suppose that \(A\in\mathfrak{A}\). Then \(A\) is measurable with respect to \(\mu^{*}\). \(\sharp\)

Given an algebra \(\mathfrak{A}\), we use \(\mathfrak{A}_{\sigma}\) to denote the family of sets which are countable union of sets of \(\mathfrak{A}\). We also use \(\mathfrak{A}_{\sigma\delta}\) to denote the family of sets which are countable intersection of sets of \(\mathfrak{A}_{\sigma}\).

Proposition. Let \(\hat{\mu}\) be a measure on an algebra \(\mathfrak{A}\), and let \(\mu^{*}\) be an outer measures induced by \(\hat{\mu}\). Given any set \(E\) and \(\epsilon >0\), there exists a set \(A\in\mathfrak{A}_{\sigma}\) satisfying

\[E\subseteq A\mbox{ and }\mu^{*}(A)\leq\mu^{*}(E)+\epsilon .\]

There also exists a set \(B\in\mathfrak{A}_{\sigma\delta}\) satisfying

\[E\subseteq B\mbox{ and }\mu^{*}(E)=\mu^{*}(B).\]

Proposition. Let \(\hat{\mu}\) be a \(\sigma\)-finite measure on an algebra \(\mathfrak{A}\), and let \(\mu^{*}\) be an outer measures induced by \(\hat{\mu}\). A set \(E\) is \(\mu^{*}\)-measurable if and only if \(E\) is the proper difference \(A\setminus B\) of a set \(A\in\mathfrak{A}_{\sigma\delta}\) and a set \(B\) with \(\mu^{*}(B)=0\). Each set \(B\) with \(\mu^{*}(B)=0\) is contained in a set \(C\in\mathfrak{A}_{\sigma\delta}\) with \(\mu^{*}(C)=0\).

\begin{equation}{\label{rat49}}\tag{12}\mbox{}\end{equation}

Theorem \ref{rat49}. (Caratheodory). Let \(\hat{\mu}\) be a measure on an algebra \(\mathfrak{A}\), and let \(\mu^{*}\) be an outer measures induced by \(\hat{\mu}\). Then, the restriction \(\bar{\mu}\) of \(\mu^{*}\) to the \(\mu^{*}\)-measurable sets is an extension of \(\hat{\mu}\) to a \(\sigma\)-field containing \(\mathfrak{A}\). We also have the following properties.

(i) Suppose that \(\hat{\mu}\) is finite \((\)resp. \(\sigma\)-finite$)$. Then \(\bar{\mu}\) is also finite \((\)resp. \(\sigma\)-finite$)$.

(ii) Suppose that \(\hat{\mu}\) is \(\sigma\)-finite. Then \(\bar{\mu}\) is the only measure on the smallest \(\sigma\)-field containing \(\mathfrak{A}\). \(\sharp\)

It is convenient to start with a set function on a collection \(\mathfrak{C}\) of sets having less structure than an algebra of sets.

Definition. We say that a collection \(\mathfrak{C}\) of subsets of \(X\) is a  semialgebra when the following conditions are satisfied.

• The intersection of any two sets in \(\mathfrak{C}\) is again in \(\mathfrak{C}\).
• The complement of any set in \(\mathfrak{C}\) is a finite disjoint union of sets in \(\mathfrak{C}\). \(\sharp\)

Let \(\mathfrak{C}\) be a semialgebra. Then the collection \(\mathfrak{A}\) consisting of the empty set and all finite disjoint unions of sets in \(\mathfrak{C}\) is an algebra, which is called the algebra generated by \(\mathfrak{C}\).

\begin{equation}{\label{rap47}}\tag{13}\mbox{}\end{equation}

Proposition \ref{rap47}. Let \(\mathfrak{C}\) be a semialgebra, and let \(\tilde{\mu}\) be a nonnegative set function defined on \(\mathfrak{C}\) satisfying \(\tilde{\mu}(\emptyset )=0\) if \(\emptyset\in\mathfrak{C}\). Then \(\tilde{\mu}\) has a unique extension to a measure \(\hat{\mu}\) on the algebra \(\mathfrak{A}\) generated by \(\mathfrak{C}\) when the following conditions are satisfied.

• If a set \(C\in\mathfrak{C}\) is the union of a finite disjoint collection \(\{C_{i}\}_{i=1}^{n}\) of sets in \(\mathfrak{C}\), then
  \[\tilde{\mu}(C)=\sum_{i=1}^{n}\tilde{\mu}(C_{i}).\]
• If a set \(C\in\mathfrak{C}\) is the union of a countable disjoint collection \(\{C_{k}\}_{k=1}^{\infty}\) of sets in \(C\), then
  \[\tilde{\mu}(C)\leq\sum_{k=1}^{\infty}\tilde{\mu}(C_{k}).\]

Definition. Two sets \(A\) and \(B\) are said to be separated by the function \(\phi\) when there exist real numbers \(a\) and \(b\) with \(a>b\) such that \(\phi(x)\geq a\) for all \(x\in A\) and \(\phi (x)\leq b\) for all \(x\in B\). \(\sharp\)

Definition. Let \(\Gamma\) be a family of real-valued functions defined on \(X\). An outer measure \(\mu^{*}\) is called a Caratheodory outer measure with respect to \(\Gamma\) when, given any two sets \(A\) and \(B\) that are separated by some function in \(\Gamma\), we have

\[\mu^{*}(A\cup B)=\mu^{*}(A)+\mu^{*}(B).\]

Theorem. Suppose that \(\mu^{*}\) is Carath\'{e}odory outer measure with respect to \(\Gamma\). Then, every function in \(\Gamma\) is \(\mu^{*}\)-measurable. \(\sharp\)

We consider the metric space \((X,d)\). The distance between two sets \(A\) and \(B\) is defined by

\[d(A,B)=\inf\left\{d(a,b):a\in A\mbox{ and }b\in B\right\}.\]

Definition. An outer measure \(\mu^{*}\) on \(X\) is called a metric outer measure or an outer measure in the sense of Caratheodory when

\[\mu^{*}(A\cup B)=\mu^{*}(A)+\mu^{*}(B)\mbox{ whenever }d(A,B)>0.\]

Let \((X,d)\) be a metric space, and let \(\tau\) be a topology induced by the metric \(d\). Each \(O\in\tau\) is said to be open when, for every \(x\in O\), there exists \(\delta>0\) such that the open ball \(\{y\in X:d(x,y)<\delta\}\) is contained in \(O\). Recall that the Borel \(\sigma\)-field is the smallest \(\sigma\)-field containing all the open sets.

Theorem. Let \(\mu^{*}\) be a metric outer measure on a metric space \((X,d)\). Then, every Borel set is \(\mu^{*}\)-measurable.

Proposition. Let \(\mu^{*}\) be a metric outer measure on a metric space \((X,d)\). Then, every lower and upper semicontinuous function is \(\mu^{*}\)-measurable.

\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}

Inner Measures.

Let \(\hat{\mu}\) be a measure on an algebra and let \(\mu^{*}\) be the outer measure induced by \(\hat{\mu}\). Then \(\mu^{*}(E)\) may be thought of as the largest possible measure for \(E\) compatible with \(\hat{\mu}\). We can also define an inner measure \(\mu_{*}\) which assigns to a given set \(E\) the smallest measure compatible with \(\hat{\mu}\).

Definition. Let \(\hat{\mu}\) be a measure on an algebra \(\mathfrak{A}\), and let \(\mu^{*}\) be the outer measure induced by \(\hat{\mu}\). We define the inner measure $\mu_{*}$ by

\[\mu_{*}(E)=\sup\left [\hat{\mu}(A)-\mu^{*}(A\setminus E)\right ],\]

where the supremum is taken over all sets \(A\in\mathfrak{A}\) satisfying \(\mu^{*}(A\setminus E)<+\infty\). \(\sharp\)

Proposition. We have the following properties.

(i) We have \(\mu_{*}(E)\leq\mu^{*}(E)\). If \(E\in\mathfrak{A}\), then \(\mu_{*}(E)=\hat{\mu}(E)\).

(ii) If \(E\subseteq F\), then \(\mu_{*}(E)\leq\mu_{*}(F)\).

(iii) Let \(A,B\in\mathfrak{A}\) satisfying \(\mu^{*}(A\setminus E)<+\infty\) and \(\mu^{*}(B\setminus E)<+\infty\). If \(A\subseteq B\), then

\[\hat{\mu}(A)-\mu^{*}(A\setminus E)\leq\hat{\mu}(B)-\mu^{*}(B\setminus E).\]

If we further assume \(E\subseteq A\), then

\[\mu_{*}(E)=\hat{\mu}(A)-\mu^{*}(A\setminus E).\]

(iv)}If \(A\in\mathfrak{A}\), then

\[\hat{\mu}(A)=\mu_{*}(A\cap E)+\mu^{*}(A\cap E^{c}).\]

(v) Let \(\{A_{k}\}_{k=1}^{\infty}\) be a disjoint sequence of sets in \(\mathfrak{A}\). Then, we have

\[\mu_{*}\left (E\bigcap\left (\bigcup_{k=1}^{\infty}A_{k}\right )\right )=\sum_{k=1}^{\infty}\mu_{*}(E\cap A_{k}).\]

Theorem. Let \(\hat{\mu}\) be a measure on an algebra \(\mathfrak{A}\), and let \(E\) be any subset of \(X\). Let \(\mathfrak{B}\) be an algebra generated by \(\mathfrak{A}\) and \(E\). Suppose that \(\bar{\mu}\) is any extension of \(\hat{\mu}\) to \(\mathfrak{B}\). Then, we have

\[\mu_{*}(E)\leq\bar{\mu}(E)\leq\mu^{*}(E).\]

Moreover, there exist extensions \(\bar{\mu}\) and \(\underline{\mu}\) of \(\hat{\mu}\) to \(\mathfrak{B}\) \((\)and hence also to the \(\sigma\)-field generated by \(\mathfrak{B})\) satisfying

\[\bar{\mu}(E)=\mu^{*}(E)\mbox{ and }\underline{\mu}(E)=\mu_{*}(E).\]

Proposition. Let \(E\) be a set satisfying \(\mu_{*}(E)<+\infty\). Then, there exists a set \(H\in\mathfrak{A}_{\delta\sigma}\) satisfying \(H\subseteq E\) and \(\bar{\mu}(E)=\mu_{*}(E)\).

Theorem. Suppose that \(\mu^{*}(E)<+\infty\). Then \(E\) is measurable if and only if \(\mu_{*}(E)=\mu^{*}(E)\).

Theorem. Let \(E\) and \(F\) be disjoint sets. Then, we have

\begin{align*} \mu_{*}(E)+\mu_{*}(F) & \leq\mu_{*}(E\cup F)\leq\mu_{*}(E)+\mu^{*}(F)\\ & \leq\mu^{*}(E\cup F)\leq\mu^{*}(E)+\mu^{*}(F).\end{align*}