Measure Theory

1. Outer Measure

1.1. Definition and Example

Definition 1.1    Suppose \(E \subset \mathbb{R}^n\). The outer measure of \(E\) is \[m^*(E):=\inf\left\{\sum_{i=1}^\infty |I_i|: \bigcup_{i=1}^\infty I_i \supset E \right\},\] where \(I_i\) is open, and \(|I_i|\) is the volume of \(I_i\).

In definition 1.1, we need at most countable \(\{I_i\}\). For example, if \([0, 1] \cap \mathbb{Q}\) is covered by finite intervals, then those finite intervals and endpoints cover \([0, 1]\). Therefore, \(m^*([0, 1] \cap \mathbb{Q}) \geq 1\). Similarly, \(m^*([0, 1] \cap \mathbb{Q}^c) \geq 1\), and thus \(m^*([0, 1]) \geq 2\). In other cases, \(I_i\) can be an empty set and thus \(\{I_i\}\) can be finite.

Example 1.1    \(m^*(\varnothing)=0\).

Example 1.2    If \(p \in \mathbb{R}\), then \(m^*(\{p\})=0\).

Proof. \(\forall \varepsilon>0\), define \(\displaystyle I=\left(x-\frac{\varepsilon}{2}, x+\frac{\varepsilon}{2}\right)\). We have \(|I|=\varepsilon\) and \(\{p\} \subset I\). Therefore, \(m^*(\{p\}) \leq \varepsilon\), i.e., \(m^*(\{p\})=0\).

\(\square\)

Example 1.3    If \(p_1, \ldots, p_k \in \mathbb{R}\), then \(m^*(\{p_1, \ldots, p_k\})=0\).

Proof. \(\forall \varepsilon>0\), define \(\displaystyle I_i=\left(p_i-\frac{\varepsilon}{2k}, p_i+\frac{\varepsilon}{2k}\right)\), we have \(\displaystyle |I_i|=\frac{\varepsilon}{k}\). Hence, \(\displaystyle \{p_1, \ldots, p_k\} \subset \bigcup_{i=1}^k I_i\) and \(\displaystyle \sum_{i=1}^n |I_i|=\varepsilon\). Therefore, \(m^*(\{p_1, \ldots, p_n\}) \leq \varepsilon\), i.e., \(m^*(\{p_1, \ldots, p_n\})=0\).

\(\square\)

Example 1.4    If \(p_1, \ldots, p_n, \ldots \in \mathbb{R}\), then \(m^*(\{p_1, \ldots, p_n, \ldots\})=0\).

Hint of proof. Take \(\displaystyle I_i=\left(x_i-\frac{\varepsilon}{2^{i+1}}, x_i+\frac{\varepsilon}{2^{i+1}}\right)\), then \(\displaystyle |I_i|=\frac{\varepsilon}{2^i}\), and \(\displaystyle \sum_{i=1}^\infty \frac{\varepsilon}{2^i}=\varepsilon\).

\(\square\)

Example 1.5    \(m^*([0, 1] \cap \mathbb{Q})=0\).

1.2. Basic Property

Property 1.1 (Nonnegativity)    Suppose \(E \subset \mathbb{R}^n\), \(m^*(E) \geq 0\). Besides, \(m^*(\varnothing)=0\).

Property 1.2 (Countable Subadditivity)    Suppose \(A_i \subset \mathbb{R}^n\). \(\displaystyle m^*\left(\bigcup_{i=1}^\infty A_i\right) \leq \sum_{i=1}^\infty m^*(A_i)\).

Proof. Take an arbitrary \(\varepsilon>0\), there exists an open cover of \(A_i\), \(\displaystyle \bigcup_{m=1}^\infty I_{i, m}\) such that \[m^*(A_i) \geq \sum_{m=1}^\infty |I_{i, m}|-\frac{\varepsilon}{2^i}, \bigcup_{i=1}^\infty A_i \subset \bigcup_{i, m=1}^\infty I_{i, m}.\] Therefore, \[\begin{aligned} m^*\left(\bigcup_{i=1}^\infty A_i\right) &\leq \sum_{i, m=1}^\infty |I_{i, m}| \\ &\leq \sum_{i=1}^\infty m^*(A_i)+\sum_{i=1}^\infty \frac{\varepsilon}{2^i} \\ &=\sum_{i=1}^\infty m^*(A_i)+\varepsilon. \end{aligned}\] Hence, \(\displaystyle m^*\left(\bigcup_{i=1}^\infty A_i\right) \leq \sum_{i=1}^\infty m^*(A_i).\)

\(\square\)

Property 1.3 (Monotonicity)    Suppose \(A \subset B\), then \(m^*(A) \leq m^*(B)\).

Proof. Take an arbitrary open cover of \(B\), denoted \(\{I_i\}_{i=1}^\infty\). Since \(A \subset B\), then \(\{I_i\}_{i=1}^\infty\) is an open cover of \(A\), and thus \(\displaystyle m^*(A) \leq \sum_{i=1}^\infty |I_i|\). By the definition of outer measure and infimum, we have \(m^*(B) \geq m^*(A)\).

\(\square\)

Theorem 1.4    Suppose \(I\) is an interval, then \(m^*(I)=|I|\).

Proof. Suppose \(I\) is a closed interval. Take an arbitrary \(\varepsilon>0\), we can find an open interval \(I' \supset I\) such that \(|I'| \leq |I|+\varepsilon\). Hence, \(m^*(I) \leq |I'| \leq |I|+\varepsilon\), i.e., \(m^*(I) \leq |I|\).

Besides, we can find an open cover of \(I\), denoted \(\{I_i\}_{i=1}^\infty\), such that \(\displaystyle m^*(I) \geq \sum_{i=1}^\infty |I_i|-\varepsilon\). Since \(I\) is bounded and closed, then there exists a finite open cover of \(I\), denoted \(\{I_i\}_{i=1}^m\), such that \(\displaystyle m^*(I) \geq \sum_{i=1}^m |I_i|-\varepsilon\). Since \(\displaystyle \sum_{i=1}^m |I_i| \geq |I|\), then \(m^*(I) \geq |I|-\varepsilon\), i.e., \(m^*(I) \geq |I|\).

Therefore, \(m^*(I)=|I|\).

We now suppose \(I\) is an arbitrary interval. We have \(\forall \varepsilon>0, \exists I_1, I_2\) s.t. \(I_1 \subset I \subset I_2\), where \(I_1, I_2\) are closed, \(|I_1|+\varepsilon>|I|\), and \(|I_2|-\varepsilon<|I|\). Therefore, \[|I|-\varepsilon<m^*(I_1) \leq m^*(I) \leq m^*(I_2)<|I|+\varepsilon.\] As a consequence, \(m^*(I)=|I|\).

\(\square\)

2. Measurable Set

2.1. Definition

Definition 2.1    Suppose \(E \subset \mathbb{R}^n\) is bounded, and \(I \supset E\) is open. The inner measure of \(E\) is \[m_*(E)=|I|-m^*(I-E).\]

显然，若\(E\)有界，且\(m^*(E)=m_*(E)\)，我们能得到可测的定义. 然而，我们希望能对任何一个\(E\)定义可测，而上述的定义则有所限制. 此外，这一定义需要同时使用内外测度的定义. 因此，Carathéodory提出了一个更简洁地等价定义.

Definition 2.2    Suppose \(E \subset \mathbb{R}^n\) is a set. If for any set \(T \subset \mathbb{R}^n\), we have \[m^*(T)=m^*(T \cap E)+m^*(T \cap E^c),\] then \(E\) is L-measurable or measurable. The L-measure, denoted \(m(E)\), equals to \(m^*(E)\). The collection of all measurable sets is called family of measurable sets, denoted as \(\mathscr{M}\).

Theorem 2.1    \(E\) is measurable if and only if for any \(A \subset E\) and \(B \subset E^c\), we have \[m^*(A \cup B)=m^*(A)+m^*(B).\]

Proof. \((\Rightarrow)\) Take \(T=A \cup B\) for any \(A \subset E\) and \(B \subset E^c\). Then \[m^*(A \cup B)+m^*((A \cup B) \cap E)+m^*((A \cup B) \cap E^c)=m^*(A)+m^*(B).\]

\((\Leftarrow)\) Take \(A=T \cap E\) and \(B=T \cap E^c\) for any \(T\). Then \[m^*(T \cap E)+m^*(T \cap E^c)=m^*(A)+m^*(B)=m^*(A \cup B)=m^*(T).\]

\(\square\)

2.2. Operation

Theorem 2.2    \(S\) is measurable if and only if \(S^c\) is measurable.

Proof. If \(S\) is measurable, then for any set \(T\), \[m^*(T)=m^*(T \cap S)+m^*(T \cap S^c)=m^*(T \cap (S^c)^c)+m^*(T \cap S^c),\] which is equivalent to say that \(S^c\) is measurable.

\(\square\)

Theorem 2.3    If \(S_1\) and \(S_2\) are measurable, then \(S_1 \cup S_2\) is measurable.

Proof. Since \(S_1\) and \(S_2\) are measurable, then for any set \(T\), \[\begin{aligned} m^*(T)&=m^*(T \cap S_1)+m^*(T \cap S_1^c) \\ &=m^*((T \cap (S_1 \cup S_2)) \cap S_1)+[m^*((T \cap S_1^c) \cap S_2)+m^*((T \cap S_1^c) \cap S_2^c))] \\ &=m^*((T \cap (S_1 \cup S_2)) \cap S_1)+[m^*((T \cap (S_1 \cup S_2)) \cap S_1^c)+m^*(T \cap (S_1 \cup S_2)^c)] \\ &=m^*(T \cap (S_1 \cup S_2))+m^*(T \cap (S_1 \cup S_2)^c). \end{aligned}\] Therefore, \(S_1 \cup S_2\) is measurable.

\(\square\)

对于不熟悉集合运算的读者，我们建议利用韦恩图逆推.

Theorem 2.4    If \(S_1\) and \(S_2\) are measurable, and \(S_1 \cap S_2=\varnothing\), then for any set \(T\), \[m^*(T \cap (S_1 \cup S_2))=m^*(T \cap S_1)+m^*(T \cap S_2).\]

Proof. Since \(S_1 \cap S_2=\varnothing\), then for any set \(T\), we have \(T \cap S_1 \subset S_1\) and \(T \cap S^2 \subset S_1^c\). Since \(S_1\) is measurable, then by theorem 2.1, \[m^*(T \cap (S_1 \cup S_2))=m^*((T \cap S_1) \cup (T \cap S_2))=m^*(T \cap S_1)+m^*(T \cap S_2).\]

\(\square\)

Corollary 2.5    If \(S_1\) and \(S_2\) are measurable, and \(S_1 \cap S_2=\varnothing\), then \[m(S_1 \cup S_2)=m(S_1)+m(S_2).\]

Corollary 2.6    If \(S_1, \ldots, S_k\) are measurable, then \(\displaystyle \bigcup_{i=1}^k S_i\) and \(\displaystyle \bigcap_{i=1}^k S_i\) are measurable. Moreover, if \[S_1 \cap \cdots \cap S_k=\varnothing,\] then for any set \(T\), \[m^*(T \cap (S_1 \cup \cdots \cup S_k))=m^*(T \cap S_1)+\cdots+m^*(T \cap S_k).\] Specifically, \[m(S_1 \cup \cdots \cup S_k)=m(S_1)+\cdots+m(S_k),\] which is finite additivity.

Corollary 2.7    If \(S_1\) and \(S_2\) are measurable, then \(S_1-S_2\) is measurable.

Theorem 2.8    If \(S_1 \supset S_2\) and \(m(S_2)<\infty\), then \[m(S_1-S_2)=m(S_1)-m(S_2).\]

Proof. We know \(S_1=(S_1-S_2) \cup S_2\) and \((S_1-S_2) \cap S_2=\varnothing\). Then by corollary 2.5, \[m(S_1)=m(S_1-S_2)+m(S_2).\] Since \(m(S_2)<\infty\), then \(m(S_1-S_2)=m(S_1)-m(S_2)\).

\(\square\)

Theorem 2.9    If \(\{S_i\}_{i=1}^\infty\) is a collection of mutually disjoint measurable sets, then \(\displaystyle \bigcup_{i=1}^\infty S_i\) is measurable. Besides, \(\displaystyle m\left(\bigcup_{i=1}^\infty S_i\right)=\sum_{i=1}^\infty m(S_i)\), which is countable additivity.

Proof. We know \[\begin{aligned} m^*(T)&=m^*\left(T \cap \left(\bigcup_{i=1}^k S_i\right)\right)+m^*\left(T \cap \left(\bigcup_{i=1}^k S_i\right)^c\right) \\ &\geq m^*\left(T \cap \left(\bigcup_{i=1}^k S_i\right)\right)+m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)^c\right) & \text{(Monotonicity)} \\ &=\sum_{i=1}^k m^*(T \cap S_i)+m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)^c\right). \end{aligned}\] Take \(k \to \infty\), then \[\begin{aligned} m^*(T)&\geq \sum_{i=1}^\infty m^*(T \cap S_i)+m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)^c\right) \\ &\geq m^*\left(\bigcup_{i=1}^\infty (T \cap S_i)\right)+m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)^c\right) & \text{(Countable Subadditivity)} \\ &=m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)\right)+m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)^c\right). \end{aligned}\] Because of countable subadditivity, we know \[m^*(T) \leq m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)\right)+m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)^c\right),\] and thus \[m^*(T)=m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)\right)+m^*\left(T \cap \left(\bigcup_{i=1}^\infty S_i\right)^c\right),\] i.e., \(\displaystyle \bigcup_{i=1}^\infty S_i\) is measurable.

Let \(\displaystyle T=\bigcup_{i=1}^\infty S_i\), then \(\displaystyle m\left(\bigcup_{i=1}^\infty S_i\right) \geq \sum_{i=1}^\infty m(S_i)\). Because of countable subadditivity, we have \(\displaystyle m\left(\bigcup_{i=1}^\infty S_i\right) \leq \sum_{i=1}^\infty m(S_i)\). Hence, \(\displaystyle m\left(\bigcup_{i=1}^\infty S_i\right)=\sum_{i=1}^\infty m(S_i)\).

\(\square\)

Theorem 2.10    If \(\{S_i\}_{i=1}^\infty\) is a collection of measurable sets, then \(\displaystyle \bigcup_{i=1}^\infty S_i\) is measurable.

Proof. Let \(S_0=\varnothing\), and \(\displaystyle \widetilde{S}_i=S_i-\bigcup_{k=0}^{i-1} S_k\), where \(i \geq 1\). Then \(\displaystyle \bigcup_{i=1}^\infty S_i=\bigcup_{i=1}^\infty \widetilde{S}_i\), and \(\{\widetilde{S}_i\}_{i=1}^\infty\) is a collection of mutually disjoint measurable sets. Hence, \(\displaystyle \bigcup_{i=1}^\infty \widetilde{S}_i\) is measurable, i.e., \(\displaystyle \bigcup_{i=1}^\infty S_i\) is measurable.

\(\square\)

Corollary 2.11    If \(\{S_i\}_{i=1}^\infty\) is a collection of measurable sets, then \(\displaystyle \bigcap_{i=1}^\infty S_i\) is measurable.

Theorem 2.12    If \(\{S_i\}_{i=1}^\infty\) is a collection of increasing measurable sets, then \(\displaystyle \lim_{i \to \infty} S_i\) is measurable, and \(\displaystyle m\left(\lim_{i \to \infty} S_i\right)=\lim_{i \to \infty} m(S_i)\).

Proof. Since \(\displaystyle \lim_{i \to \infty} S_i=\bigcup_{i=1}^\infty S_i\), then \(\displaystyle \lim_{i \to \infty} S_i\) is measurable.

Let \(S_0=\varnothing\), and \(\widetilde{S}_i=S_i-S_{i-1}\), where \(i \geq 1\). Then \(\displaystyle \lim_{i \to \infty} S_i=\bigcup_{i=1}^\infty S_i=\bigcup_{i=1}^\infty \widetilde{S}_i\), and \(\{\widetilde{S}_i\}_{i=1}^\infty\) is a collection of mutually disjoint measurable sets. Hence, \[\begin{aligned} m\left(\lim_{i \to \infty} S_i\right)&=\sum_{i=1}^\infty m(\widetilde{S}_i) \\ &=\sum_{i=1}^\infty m(S_i-S_{i-1}) \\ &=\lim_{k \to \infty}\sum_{i=1}^k m(S_i-S_{i-1}) \\ &=\lim_{k \to \infty}m\left(\bigcup_{i=1}^k (S_i-S_{i-1})\right) \\ &=\lim_{k \to \infty}m(S_k)=\lim_{i \to \infty}m(S_i). \end{aligned}\]

\(\square\)

Theorem 2.13    If \(\{S_i\}_{i=1}^\infty\) is a collection of decreasing measurable sets, then \(\displaystyle \lim_{i \to \infty} S_i\) is measurable. Moreover, when \(m(S_1)<\infty\), \(\displaystyle m\left(\lim_{i \to \infty} S_i\right)=\lim_{i \to \infty} m(S_i)\).

Proof. Since \(\displaystyle \lim_{i \to \infty} S_i=\bigcap_{i=1}^\infty S_i\), then \(\displaystyle \lim_{i \to \infty} S_i\) is measurable.

Let \(\widetilde{S}_i=S_1-S_i\), then \(\{\widetilde{S}_i=S_1-S_i\}_{i=1}^\infty\) is a collection of increasing sets. Therefore, \[m\left(\lim_{i \to \infty} (S_1-S_i)\right)=\lim_{i \to \infty} m(S_1-S_i).\] Since \(m(S_1)<\infty\), then \(m(S_i)<\infty\) and \(\displaystyle \lim_{i \to \infty} S_i<\infty\). Hence, \[m\left(\lim_{i \to \infty} (S_1-S_i)\right)=m\left(S_1-\lim_{i \to \infty} S_i\right)=m(S_1)-m\left(\lim_{i \to \infty} S_i\right),\] and \[\lim_{i \to \infty} m(S_1-S_i)=\lim_{i \to \infty} (m(S_1)-m(S_i))=m(S_1)-\lim_{i \to \infty} m(S_i).\] Therefore, \(\displaystyle m\left(\lim_{i \to \infty} S_i\right)=\lim_{i \to \infty} m(S_i)\).

\(\square\)

The condition \(m(S_1)<\infty\) is important. For example, \(S_n=(n, \infty)\). It is obvious that \(\{S_n\}\) is decreasing. Therefore, \(\displaystyle \lim_{n \to \infty} S_n=\bigcap_{n=1}^\infty S_n=\varnothing\), and \(\displaystyle m\left(\lim_{n \to \infty} S_n\right)=0\). However, \(m(S_n)=\infty\), and \(\displaystyle \lim_{n \to \infty} m(S_n)=\infty\).

Theorem 2.14    Suppose \(\{S_i\}_{i=1}^\infty\) is a collection of measurable sets. If the upper limit and lower limit of \(\{S_i\}_{i=1}^\infty\) and measure exist, then \(\displaystyle m\left(\varliminf_{i \to \infty} S_i \right) \leq \varliminf_{i \to \infty} m(S_i) \leq \varlimsup_{i \to \infty} m(S_i) \leq m\left(\varlimsup_{i \to \infty} S_i\right)\). Note that we need \(\displaystyle m\left(\bigcup_{i=1}^\infty S_i\right)<\infty\) to prove \(\displaystyle \varlimsup_{i \to \infty} m(S_i) \leq m\left(\varlimsup_{i \to \infty} S_i\right)\).

3. \(\sigma\)-Algebra

Definition 3.1    Suppose \(\{F_i\}_{i=1}^\infty\) is a collection of closed sets, and \(\{G_i\}_{i=1}^\infty\) is a collection of open sets. If \(\displaystyle F=\bigcup_{i=1}^\infty F_i\), then \(F\) is an \(F_\sigma\) set. If \(\displaystyle G=\bigcap_{i=1}^\infty G_i\), then \(G\) is a \(G_\delta\) set.

\(F\)可能来自于法语中的fermé，意为闭. \(\sigma\)可能来自于德语中的Summe，意为总和. 因此\(F_\sigma\)型集表示闭集的可数并. \(G\)可能来自于德语中的Gebiet，意为区域，可以理解为连通的开集. \(\delta\)可能来自于德语中的Durchschnitt，意为横断. 因此\(G_\delta\)型集表示开集的可数交.

Definition 3.2    Suppose \(\Omega\) is the family of sets including some subsets of \(X\). \(\Omega\) is called a \(\sigma\)-algebra if it has the following properties:

    (1) \(X \in \Omega\);

    (2) \(\Omega\) is closed under countable union;

    (3) \(\Omega\) is closed under complement.

显然，可测集类\(\mathscr{M}\)是\(\mathbb{R}^n\)上的\(\sigma\)-代数.

Definition 3.3    Suppose \(\Sigma\) is the family of sets including some subsets of \(X\). \(\sigma\)-algebra generated by \(\Sigma\) is the intersection of all \(\sigma\)-algebras containing \(\Sigma\) on \(X\).

Definition 3.4    Borel \(\sigma\)-algebra is a \(\sigma\)-algebra generated by all open sets on \(\mathbb{R}^n\), denoted as \(\mathscr{B}\). The element in \(\mathscr{B}\) is called Borel set.

A Borel set is any set that can be formed from open sets through the operations of countable union, countable intersection, and complement.

Example 3.1    Intervals, open sets, closed sets, \(F_\sigma\) sets, and \(G_\delta\) sets are Borel sets.

Theorem 3.1    Suppose \(f: E \subset \mathbb{R} \to \mathbb{R}\) is a function, and \(E \in \Sigma\), where \(\Sigma\) is a \(\sigma\)-algebra on \(\mathbb{R}\). Let \[\mathscr{A}=\{X \subset \mathbb{R}: f^{-1}(X) \in \Sigma\},\] then \(\mathscr{A}\) is a \(\sigma\)-algebra.

Proof. We check three properties of \(\sigma\)-algebra:

(1) Since \(f^{-1}(\mathbb{R})=E \in \Sigma\), then \(\mathbb{R} \in \mathscr{A}\).

(2) Since \(\displaystyle f^{-1}\left(\bigcup_{i=1}^\infty X_i\right)=\bigcup_{i=1}^\infty f^{-1}(X_i)\), then if \(X_i \in \mathscr{A}\), we have \(\displaystyle \bigcup_{i=1}^\infty X_i \in \mathscr{A}\).

(3) Since \(f^{-1}(X^c)=(f^{-1}(X))^c\), then if \(X \in \mathscr{A}\), we have \(X^c \in \mathscr{A}\).

Therefore, \(\mathscr{A}\) is a \(\sigma\)-algebra.

\(\square\)

Theorem 3.2    Suppose \(f: \mathbb{R} \to \mathbb{R}\) is continuous, then the preimage of a Borel set is a Borel set.

Proof. Suppose \(\mathscr{B}\) is the Borel \(\sigma\)-algebra on \(\mathbb{R}\), and \(\mathscr{A}=\{X \subset \mathbb{R}: f^{-1}(X) \in \mathscr{B}\}\). We know \(\mathscr{A}\) is a \(\sigma\)-algebra. Since the preimage of an open set is open, then all open sets belong to \(\mathscr{A}\). Since \(\mathscr{B}\) is generated by all open sets, then \(\mathscr{B} \subset \mathscr{A}\). Therefore, the preimage of a Borel set is a Borel set.

\(\square\)

4. Family of Measurable Sets

Theorem 4.1    Any interval \(I\) in \(\mathbb{R}^n\) is measurable.

Theorem 4.2    Any Borel set is measurable.

Theorem 4.3    Any zero measure set \(A\) is measurable.

Proof. For any set \(T\), we have \(0 \leq m^*(T \cap A) \leq m^*(A)=0\), and thus \(m^*(T \cap A)=0\). Besides, \(m^*(T \cap A^c) \leq m^*(T)\). Therefore, \(m^*(T) \geq m^*(T \cap A)+m^*(T \cap A^c)\).

Because of countable subadditivity, we have \(m^*(T)=m^*(T \cap A)+m^*(T \cap A^c)\), i.e., \(A\) is measurable.

\(\square\)

Corollary 4.4    Suppose \(A\) has measure zero. Any subset of \(A\) has measure zero.

Proof. Take an arbitrary subset \(A'\) of \(A\), then \(A' \subset A\). Hence, \(0 \leq m^*(A') \leq m^*(A)=0\), i.e., \(m^*(A')=0\).

\(\square\)

Corollary 4.5     The at most countable union of zero measure sets has measure zero.

Proof. Suppose \(\{A_i\}_{i=1}^\infty\) is a collection of zero measure sets. Because of countable subadditivity, we have \[0 \leq m^*\left(\bigcup_{i=1}^\infty A_i\right) \leq \sum_{i=1}^\infty m^*(A_i)=0,\] i.e., \(\displaystyle m^*\left(\bigcup_{i=1}^\infty A_i\right)=0.\)

\(\square\)

Theorem 4.6    If \(E\) is measurable, \(G\) is open, and \(K\) is compact, then \[m(E)=\inf\{m(G): E \subset G\}=\sup\{m(K): E \supset K\}.\] We define \(\inf\{m(G): E \subset G\}\) as external regularity, and \(\sup\{m(K): E \supset K\}\) as inner regularity. On the contrary, if \(E\) is bounded, and \(\inf\{m(G): E \subset G\}=\sup\{m(K): E \supset K\}\), where \(G\) is open and \(K\) is compact, then \(E\) is measurable.

Lemma 4.6.1    Suppose \(E\) is measurable. \(\forall \varepsilon>0\), \(\exists G \supset E\), s.t. \(m(G-E)<\varepsilon\), where \(G\) is open.

Proof. Take an arbitrary \(\varepsilon>0\), there exists an open cover of \(E\), \(\{I_i\}_{i=1}^\infty\), such that \[\sum_{i=1}^\infty |I_i|<m(E)+\varepsilon.\] Let \(\displaystyle G=\bigcup_{i=1}^\infty I_i\), where \(G\) is open, and \(\displaystyle m(G) \leq \sum_{i=1}^\infty m(I_i)<m(E)+\varepsilon\), i.e., \(m(G)-m(E)<\varepsilon\).

When \(m(E)<\infty\), \(m(G)-m(E)=m(G-E)<\varepsilon\).

When \(m(E)=\infty\), let \(E_n=E \cap B_n(0)\), then \(\displaystyle E=\bigcup_{i=1}^\infty E_n\), and \(m(E_n)<\infty\). Hence, \(\forall \varepsilon>0\), \(\exists G_n \supset E_n\) s.t. \[m(G_n-E_n)<\sum_{i=1}^\infty \frac{\varepsilon}{2^n},\] where \(G_n\) is open. We also know \(\displaystyle G=\bigcup_{n=1}^\infty G_n\) is open, and \(G \supset E\). Since \[G-E=\bigcup_{n=1}^\infty G_n-\bigcup_{n=1}^\infty E_n \subset \bigcup_{n=1}^\infty (G_n-E_n),\] then \[m(G-E) \leq \sum_{n=1}^\infty m(G_n-E_n)<\sum_{n=1}^\infty \frac{\varepsilon}{2^n}=\varepsilon.\]

\(\square\)

Lemma 4.6.2    Suppose \(E\) is measurable. Then \[\forall \varepsilon>0, \exists F \subset E\ \text{s.t.}\ m(E-F)<\varepsilon,\] where \(F\) is closed.

Proof. Take an arbitrary \(\varepsilon>0\), there exists an open set \(G \supset E^c\) such that \(m(G-E^c)<\varepsilon\). Let \(F=G^c \subset E\), where \(F\) is closed. Therefore, \(m(G-E^c)=m(E-F)<\varepsilon\).

\(\square\)

Proof. External Regularity. Take an arbitrary \(\varepsilon>0\), there exists an open set \(G \supset E\) such that \(m(G-E)<\varepsilon\). When \(m(E)<\infty\), \(m(G)=m(E)+m(G-E)<m(E)+\varepsilon\), i.e., \(m(E)>m(G)-\varepsilon\). Since \(G \supset E\), then \(m(E) \leq m(G)\). Hence, \(m(E)=\inf\{m(G): E \subset G\}\).

When \(m(E)=\infty\), it is obvious that \(m(G)=\infty\), and thus \(m(E)=\inf\{m(G): E \subset G\}\).

Inner Regularity. When \(E\) is bounded, take an arbitrary \(\varepsilon>0\), there exists a closed set \(K \subset E\) such that \(m(E-K)<\varepsilon\). Since \(E\) is bounded, then \(K\) is bounded and closed, i.e., \(K\) is compact. Besides, we have \(m(E)<\infty\), and therefore \(m(E)=m(K)+m(E-K)<m(K)+\varepsilon\). Since \(K \subset E\), then \(m(K) \leq m(E)\). Hence, \(m(E)=\sup\{m(K): E \supset K\}\).

When \(E\) is an arbitrary set, let \(E_n=E \cap B_n(0)\), then \(E_n\) is bounded, and \(\displaystyle E=\bigcup_{n=1}^\infty E_n\). Since \(\{E_n\}_{n=1}^\infty\) is a collection of increasing measurable sets, then \(\displaystyle m(E)=\lim_{n \to \infty} m(E_n)\). We can take a compact set \(K_n \subset E_n\) such that \(\displaystyle m(K_n) \geq m(E_n)-\frac{1}{n}\) and \(m(K_n) \leq m(E_n)\). Hence, \(\displaystyle \lim_{n \to \infty} m(K_n)=\lim_{n \to \infty} m(E_n)=m(E)\), and thus \(m(E)=\sup\{m(K): E \supset K\}\).

Measurability. Let \(\alpha=\inf\{m(G): E \subset G\}=\sup\{m(K): E \supset K\}\). Take an arbitrary \(n \in \mathbb{N}\), there exists a open set \(G_n \supset E\) and compact set \(K_n \subset E\) such that \(\displaystyle m(G_n)<\alpha+\frac{1}{2n}\) and \(\displaystyle m(K_n)>\alpha-\frac{1}{2n}\). Hence, \[0 \leq m(G_n)-m(K_n)<\frac{1}{n}.\] Take \(\displaystyle A=\bigcap_{n=1}^\infty G_n\) and \(\displaystyle B=\bigcup_{n=1}^\infty K_n\), then \(A\) and \(B\) are measurable. Since \(m(A) \leq m(G_n)\), \(m(B) \geq m(K_n)\), and \(m(A) \geq m(B)\), then \[0 \leq m(A)-m(B) \leq m(G_n)-m(K_n)<\frac{1}{n},\] i.e., \(m(A)=m(B)\). Since \(E \supset B\) is bounded, then \(m(B)<\infty\), and \(m(A-B)=m(A)-m(B)=0\). Since \(E-B \subset A-B\), then \(m(E-B)=0\). Therefore, \(E=B \cup (E-B)\) is measurable.

\(\square\)

Theorem 4.7    If \(E\) is measurable, then there exists a \(G_\delta\) set \(G\) such that \(G \supset E\) and \(m(G-E)=0\).

Proof. Take an arbitrary \(n \in \mathbb{N}\). By lemma 4.6.1, there exists an open set \(G \supset E\) such that \(\displaystyle m(G-E)<\frac{1}{n}\). Take \(\displaystyle G=\bigcap_{n=1}^\infty G_n\), which is a \(G_\delta\) set. Since \(G-E \subset G_n-E\), then \[m(G-E) \leq m(G_n-E)<\frac{1}{n},\] i.e., \(m(G-E)=0\).

\(\square\)

Corollary 4.8    If \(E\) is measurable, then there exists an \(F_\sigma\) set \(F\) such that \(F \subset E\) and \(m(E-F)=0\).

根据定理4.7与其推论，如果我们有了所有的\(G_\delta\)型集（或\(F_\sigma\)型集）与零测集，那么我们就知道了所有的可测集. 可测集一定可以表示为\[E=G_\delta-M_0=F_\sigma \cup M_0.\]

5. Non-Measurable Set in \(\mathbb{R}\)

此前，我们一直在回答什么是可测集，并发现可测集是存在的——我们构造集合通常都是从区间出发，经过一系列并、交、差等运算来获得，而这样的集合都是Borel集，总是可测的. 我们还需要回答不可测集是否存在的问题——我们可以利用Lebesgue测度的平移不变性，构造一个不可测集. 当然，在一般的数学实践中，遇到不可测集的机会是极少的，它通常只是被用来构成各种特例.

Property 5.1 (Translation Invariance)    Suppose \(E\) is measurable and \(\tau_\alpha E=\{x+\alpha: x \in E\}\), then \(\tau_\alpha E\) is measurable and \(m^*(E)=m^*(\tau_\alpha E)\).

Property 5.2 (Reflection Invariance)    Suppose \(E\) is measurable and \(\tau E=\{-x: x \in E\}\), then \(\tau E\) is measurable and \(m^*(E)=m^*(\tau E)\).

Example 5.1    We define a relation on \([0, 1]\): \(x \sim y \Leftrightarrow x-y \in \mathbb{Q}\), which is an equivalence relation. From this equivalence relation, we can find equivalence classes. We take a representative element from each equivalence class (Axiom of choice). Let \(Z=\{\text{All representative elements}\}\). Take an arbitrary \(\xi \in Z\). All elements in the class including \(\xi\) have the form of \(\xi+r \in [0, 1]\), where \(r \in \mathbb{Q} \cap [-\xi, 1-\xi]\).

Therefore, \(\displaystyle \bigcup_{r \in \mathbb{Q} \cap [-\xi, 1-\xi]} \{\tau_r \xi\}\) represents all elements in the class including \(\xi\), and \(\displaystyle \bigcup_{r \in \mathbb{Q} \cap [-1, 1]} \tau_r Z \supset [0, 1]\).

If \(\xi \in \tau_{r_1} Z \cap \tau_{r_2} Z\), where \(r_1 \neq r_2\), then \(\xi-r_1, \xi-r_2 \in Z\). Since \(\xi-r_1 \sim \xi-r_2\), then we have two representative elements for one equivalence class, which is a contradiction. Therefore, \(\tau_{r_1} Z \cap \tau_{r_2} Z=\varnothing\).

We know order \(\mathbb{Q} \cap [-1, 1]\), denoted \(r_1, r_2, \ldots\). Let \(Z_n=\tau_{r_n} Z\). We know \(Z_n \subset [-1, 2]\), is bounded and mutually disjoint from \(Z_{n'}\), where \(n \neq n'\). Hence, \(\displaystyle [0, 1] \subset \bigcup_{n=1}^\infty Z_n\). If \(Z\) is measurable, then \(Z_n\) is measurable, and \[m^*\left(\bigcup_{n=1}^\infty Z_n\right)=\sum_{n=1}^\infty m^*(Z_n)=\sum_{n=1}^\infty m^*(Z)<\infty.\] Therefore, \[m^*(Z)=0=m^*\left(\bigcup_{n=1}^\infty Z_n\right),\] which is a contradiction. As a consequence, \(Z\) is a non-measurable set, a.k.a. Vitali set.

There will be non-measurable subsets in any set with measure greater than zero, and we can also construct non-measurable sets in \(\mathbb{R}^n\).