HW3

Exercise 2

19. Let $(X,\mathcal{A},\mu )$ be a finite measure space and $\left\{A_n\right\}\subset \mathcal{A}$ such that ${\sum }_n\mu (A_n-$ $A_{n+1})<\infty$. Prove:

• (a) $\mu \left(\mathrm{liminf}{A}_n\right)=\mu \left(\mathrm{limsup}{A}_n\right)=L$.
• (b) $\lim \mu \left(A_n\right)=L$.

\begin{proof} For any positive integer $N$, define $M_N={\cup }_{n=1}^N{\cap }_{m=n}^{\infty }{A}_n={\cap }_{m=N}^{\infty }{A}_n$, and define $K_N={\cap }_{n=1}^N{\cup }_{m=n}^{\infty }{A}_n={\cup }_{m=N}^{\infty }{A}_n$. Note that $M_N\uparrow \lim \inf A_n$ and $K_N\downarrow \lim \sup A_n$ as $N\to \infty$. Notice that for any $a\in K_N-M_N$, there exists $n_a⩾N$ such that $a\in A_{n_a}$ and $a\notin A_{n_a+1}$. Therefore, $\mu (K_N-M_N)⩽{\sum }_{n=N}^{\infty }\mu (A_n-A_{n+1})$. Since $\mu (K_1-M_1)⩽{\sum }_n\mu (A_n-A_{n+1})<\infty$ and $(X,\mathcal{A},\mu )$ is a finite measure space, there is

$$\underset{N\to \infty }{\lim }\mu (K_N-M_N)=0$$

and so ${\lim }_{n\to \infty }\mu (K_N)={\lim }_{n\to \infty }\mu (M_N)$, that is, $\mu \left(\mathrm{liminf}{A}_n\right)=\mu \left(\mathrm{limsup}{A}_n\right)$.

ii) Note that $\mu (M_n)⩽\mu (A_n)⩽\mu (K_n)$ and $\lim \mu (M_n)=\lim \mu (K_n)=L$. It follows that $\lim \mu \left(A_n\right)=L$. \end{proof}

21. Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space so that there exists a sequence $\left\{E_n\right\}\subset \mathcal{A}$ such that ${\cup }_{n=1}^{\infty }{E}_n=X$ and $\mu \left(E_n\right)<\infty$ for every $n\in \mathbb{N}$. Show that there exists a disjoint sequence $\left\{F_n\right\}\subset \mathcal{A}$ such that ${\cup }_{n=1}^{\infty }{F}_n=X$ and $\mu \left(F_n\right)<\infty$ for every $n\in \mathbb{N}$.

\begin{proof} Define $F_n=E_n\cap ({\cup }_{i=1}^{n-1}{E}_n{)}^c$, then it is easy to verify $\{F_n\}\subseteq \mathcal{A}$ is pairwise disjoint and ${\cup }_{n=1}^{\infty }{F}_n=X$. Since $\mu (F_n)⩽\mu (E_n)$, we have that $\mu \left(F_n\right)<\infty$ for every $n\in \mathbb{N}$. \end{proof}

Exercise 3

1. Find an example of a set $X$ and a monotone class $\mathcal{M}$ consisting of subsets of $X$ such that $\varnothing ,X\in \mathcal{M}$, but $\mathcal{M}$ is not a $\sigma$-algebra.

\begin{proof}

Let $X=\{1,2,3\}$ and let $\mathcal{M}=\{\varnothing ,\{1\},\{2\},\{3\},X\}$. Since $\{1,2\}\notin \mathcal{M}$, then $\mathcal{M}$ is not a $\sigma$-algebra. \end{proof}

2. Let X be a set and let ${\left(A_i\right)}_{i\in I}$ be a partition of $X$, i.e., $A_i$ is a nonempty subset of $X$ for each $i\in I,A_i\cap A_j=\varnothing$; for $i\ne j$, and $X={\cup }_{i=1}^{\infty }{A}_i$. Then

$$\mathcal{A}=\left\{A_J:={\cup }_{j\in J}{A}_j\mid J\subset I\right\}$$

is a $\sigma$-algebra.

\begin{proof} Note that $A_{\varnothing }=\varnothing$ and $A_I=X$ are contained in $\mathcal{A}$. For any $A_J\in \mathcal{A}$, there is $A_{J^c}\in \mathcal{A}$ and $A_{J^c}=A_J$. For any $\{A_{J_i}{\}}_{i=1}^{\infty }\subseteq \mathcal{A}$, define $J_{\infty }={\cup }_{i=1}^{\infty }{J}_i$. Since $J_{\infty }\subseteq I$, then $A_{J_{\infty }}={\cup }_{i=1}^{\infty }{A}_{J_i}\in \mathcal{A}$. Therefore, $\mathcal{A}$ is a $\sigma$-algebra, \end{proof}

3. Prove that if $\mathcal{A}$ is an algebra of subsets of $X$ and $\mu ,\nu$ are finite measures on $\sigma (\mathcal{A})$ such that $\mu (A)=\nu (A)$ for all $A\in \mathcal{A}$, then $\mu =\nu$.

\begin{proof} By Corollary 4.6, since $\mathcal{A}$ is a $\pi$-system and $\mu ,\nu$ are finite on $\sigma (\mathcal{A})$, then $\mu (A)=\nu (A)$ for all $A\in \mathcal{A}$ yields that $\mu (A)=\nu (A)$ for all $A\in \sigma (\mathcal{A})$.

\end{proof}

4. Let $\mathcal{A}$ be an algebra of subsets of $X$ and $\mu ,\nu$ measures on $\sigma (\mathcal{A})$ that are $\sigma$-finite relative to $\mathcal{A}$, i.e., $X={\cup }_{n=1}^{\infty }{A}_n$ with the $A_n$ pairwise disjoint sets in $\mathcal{A}$ and $\mu \left(A_n\right)=\nu \left(A_n\right)<\infty$ for all $n$. Prove that if $\mu (A)=\nu (A)$ for all $A\in \mathcal{A}$, then $\mu =\nu$.

\begin{proof} By Corollary 4.7, since $\mathcal{A}$ is a $\pi$-system, $\mu ,\nu$ are $\sigma$-finite on $\mathcal{A}$ and $\mu {|}_{\mathcal{A}}=\nu {|}_{\mathcal{A}}$, then $\mu (A)=\nu (A)$ for all $A\in \sigma (\mathcal{A})$. \end{proof}

5. Let $\mathcal{F}$ be a $\pi$-system on $X$ such that $X\in \mathcal{F}$ and if $A\in \mathcal{F}$, then $A^c={\cup }_{k=1}^n{A}_k$ with $A_k\in \mathcal{F}$. Prove that $\mathcal{A}=\left\{A\subset X:A={\cup }_{k=1}^n{A}_k,A_k\in \mathcal{F}\right\}=\mathcal{A}(\mathcal{F})$, the algebra of subsets of $X$ generated by $\mathcal{F}$.

\begin{proof} First, we show that $\mathcal{F}$ is an algebra. Since $X\in \mathcal{F}$, then $X\in \mathcal{A}$. Furthermore, $\varnothing =X^c={\cup }_{n=1}^k{A}_k^{\prime }$ for some $A_k^{\prime }\in \mathcal{F}$ yields that $\varnothing \in \mathcal{F}$ and $\varnothing \in \mathcal{A}$. For any $A\in \mathcal{A}$, $A={\cup }_{k=1}^n{A}_k$ for some $A_k\in \mathcal{F}$, and

$$A^c=({\cup }_{k=1}^n{A}_k{)}^c={\cap }_{k=1}^n{A}_k^c.$$

Since $A_k^c={\cup }_{i=1}^{n_k}{A}_{k,i}$ for some $A_{k,i}\in \mathcal{F}$, then $A^c={\cup }_{(i_1,\cdots ,i_n)\in I}({\cap }_{k=1}^n{A}_{k,i_k})$ where $I=\{(i_1,\cdots ,i_n):1⩽i_j⩽n_j\}$. As $\mathcal{F}$ is a $\pi$-system, then ${\cap }_{k=1}^n{A}_{k,i_k}\in \mathcal{F}$ and so $A^c\in \mathcal{A}$. For any $A_1={\cup }_{k=1}^{m_1}{A}_k^{(1)}$ and $A_2={\cup }_{k=1}^{m_2}{A}_k^{(2)}$, $A_1\cup A_2$ is still a union of finite elements in $\mathcal{F}$, so $A_1\cup A_2\in \mathcal{A}$. Therefore, $\mathcal{A}$ is an algebra.

Note that the algebra generated by $\mathcal{F}$ contains elements ${\cup }_{k=1}^n{A}_k$ for any $A_k\in \mathcal{F}$, so $\mathcal{A}(\mathcal{F})\supseteq \mathcal{A}$. Therefore, $\mathcal{A}=\mathcal{A}(\mathcal{F})$ by $\mathcal{A}$ being an algebra. \end{proof}

6. Let $\mathcal{C}$ be a collection of subsets of $X$ and $\mathcal{M}$ is a monotone class of a collection of subsets of $X$. Show that $\mathcal{N}=\{A\in \mathcal{M}:A\cap C\in \mathcal{M}$ for all $C\in \mathcal{C}\}$ is a monotone class. Show that given a collection $\mathcal{C}$ of subsets of $X$, there is a smallest monotone class of subsets of $X$ that contains $\mathcal{C}$. If a monotone class $\mathcal{M}$ is an algebra of subsets of $X$, then $\mathcal{M}$ is a $\sigma$-algebra.

\begin{proof} i) Assume that $A_i\uparrow A$ and $A_i\in \mathcal{N}$, then $(A_i\cap C)\uparrow (A\cap C)$ for all $C\in \mathcal{C}$. Since $\mathcal{M}$ is a monotone class and $A_i\cap C\in \mathcal{M}$, then $A\cap C\in \mathcal{M}$ for all $C\in \mathcal{C}$. Therefore, $A\in \mathcal{N}$. Similarly, we can prove that if $A_i\downarrow A$ and $A_i\in \mathcal{N}$, then $A\in \mathcal{N}$. Therefore, $\mathcal{N}$ is a monotone class.

ii) Let $\mathcal{S}$ be the set of monotone classes which contains $\mathcal{C}$. We claim that $\mathcal{L}:={\cap }_{S\in \mathcal{S}}S$ is the smallest monotone class that contains $\mathcal{C}$. Since $S\supseteq \mathcal{C}$ for all $S\in \mathcal{S}$, then $\mathcal{L}\supseteq \mathcal{C}$. For any $A_i\in \mathcal{L}$ such that $A_i\uparrow A$, there is $A_i\in S$ and $A\in S$ for all $S\in \mathcal{S}$, which yields that $A\in \mathcal{L}$. Similarly, if $A_i\in \mathcal{L}$ and $A_i\downarrow A$, then $A\in \mathcal{L}$. Therefore, $\mathcal{L}$ is a monotone class that contains $\mathcal{C}$. Furthermore, by definition, $\mathcal{L}$ is the smallest one in $\mathcal{S}$.

iii) Assume that $\mathcal{M}$ is an algebra, then to show $\mathcal{M}$ is a $\sigma$-algebra, it suffices to show for $\{A_i{\}}_{i=1}^n\subseteq \mathcal{M}$, there is ${\cup }_{i=1}^{\infty }{A}_i\in \mathcal{M}$. Define $B_k={\cup }_{i=1}^k{A}_i$. Since $\mathcal{M}$ is an algebra, then $B_k\in \mathcal{M}$. Since $B_k\uparrow {\cup }_{i=1}^{\infty }{A}_i$ and $\mathcal{M}$ is a monotone class, then ${\cup }_{i=1}^{\infty }{A}_i\in \mathcal{M}$. Therefore, $\mathcal{M}$ is a $\sigma$-algebra. \end{proof}

7. Let $(X,\mathcal{A},\mu )$ be a probability measure space and let $\mathcal{F},{\mathcal{F}}_1\subset \mathcal{A}$ be $\pi$ systems in $X$. Prove that $\mu (B\cap C)=\mu (B)\mu (C)$ for all $B\in \sigma (\mathcal{F}),C\in \sigma \left({\mathcal{F}}_1\right)$ iff $\mu (B\cap C)=\mu (B)\mu (C)$ for all $B\in \mathcal{F},C\in {\mathcal{F}}_1$.

\begin{proof} Since $\sigma (\mathcal{F})\supseteq \mathcal{F}$ and $\sigma ({\mathcal{F}}_1)\supseteq {\mathcal{F}}_1$, one direction is easy.

Now assume that $\mu (B\cap C)=\mu (B)\mu (C)$ for all $B\in \mathcal{F}$ and $C\in {\mathcal{F}}_1$.

Define $\mathcal{L}=\{B\in \sigma (\mathcal{F}):\mu (B\cap C)=\mu (B)\mu (C),\forall C\in \mathcal{F}\}$, then $\mathcal{L}\supseteq \mathcal{F}$. Since $X\in \sigma (\mathcal{F})$ and $\mu (X\cap C)=\mu (X)\mu (C)$ for all $C\in \mathcal{F}$, then $X\in \mathcal{L}$. For $B_1,B_2\in \mathcal{L}$ with $B_2\subseteq B_1$, there is

$$\begin{array}{rl}\mu ((B_1-B_2)\cap C)& =\mu ((B_1\cap C)-(B_2\cap C))=\mu (B_1\cap C)-\mu (B_2\cap C)\\ & =\mu (B_1)\mu (C)-\mu (B_2)\mu (C)=\mu (B_1-B_2)\mu (C)\end{array}$$

and so $B_1-B_2\in \mathcal{L}$. For any $B_i\uparrow B$ with $B_i\in \mathcal{L}$, since $\mu$ is finite, we have that $\mu (B_i)\to \mu (B)$, $\mu (B_i\cap C)=\mu (B_i)\mu (C)\to \mu (B)\mu (C)$ and $B_i\cap C\to B\cap C$ for any $C\in \mathcal{F}$. Hence, $\mathcal{L}$ is a $d$-system. By Theorem 4.5, $\mathcal{L}$ contains the $d$-system generated by $\mathcal{F}$, which coincides with $\sigma (\mathcal{F})$. Therefore, $\mathcal{L}\supseteq \sigma (\mathcal{F})$.

Define ${\mathcal{L}}_1=\{C\in \sigma ({\mathcal{F}}_1):\mu (B\cap C)=\mu (B)\mu (C),\forall B\in \sigma (\mathcal{F})\}$. Similarly, we can prove that $X\in {\mathcal{L}}_1$, $C_1-C_2\in {\mathcal{L}}_1$ for any $C_1,C_2\in {\mathcal{L}}_1$ with $C_2\subseteq C_1$, and $C\in {\mathcal{L}}_1$ for any $C_i\uparrow C$ with $C_i\in {\mathcal{L}}_1$. Therefore, ${\mathcal{L}}_1$ is a $d$-system and by Theorem 4.5 ${\mathcal{L}}_1\supseteq \sigma ({\mathcal{F}}_1)$. Therefore, for all $B\in \sigma (\mathcal{F})$ and $C\in \sigma ({\mathcal{F}}_1)$, $\mu (B\cap C)=\mu (B)\mu (C)$. \end{proof}

8. Show that a $d$-system is a $\sigma$-algebra if, and only if, it is closed with respect to finite intersections.

\begin{proof} Let $\mathcal{D}$ be a $d$-system. Assume that $\mathcal{D}$ is a $\sigma$-algebra. For any $\{D_i{\}}_{i=1}^n\in \mathcal{D}$, $D_i^c\in \mathcal{C}$ and so $({\cup }_{i=1}^n{D}_i^c{)}^c={\cap }_{i=1}^n{D}_i\in \mathcal{D}$. Thus $\mathcal{D}$ is closed with respect to finite intersections.

Now assume that $\mathcal{D}$ is closed with respect to finite intersections. Since $X\in \mathcal{D}$, for any $A\in \mathcal{D}$, $A^c=X-A\in \mathcal{D}$ and so $\varnothing \in \mathcal{D}$. For $\{A_i{\}}_{i=1}^{\infty }\subseteq \mathcal{D}$, we aim to show ${\cup }_{i=1}^{\infty }{A}_i\in \mathcal{D}$. Define $B_k={\cup }_{i=1}^k{A}_i$. Since $B_k=({\cap }_{i=1}^k{A}_i^c{)}^c$ and $\mathcal{D}$ is closed with respect to finite intersections, then $B_k\in \mathcal{D}$. As $B_k\uparrow {\cup }_{i=1}^{\infty }{A}_i$ and $\mathcal{D}$ is a $d$-system, then ${\cup }_{i=1}^{\infty }{A}_i\in \mathcal{D}$. Therefore, $\mathcal{D}$ is a $\sigma$-algebra. \end{proof}

Exercise 4

In exercises $1-5$, we assume that ${\mu }^{\ast }$ is an outer measure on some set $X$.

1. Let $E$ be a measurable subset of $X$. Show that for every subset $A$ of $X$ the following equality holds:

$${\mu }^{\ast }(E\cup A)+{\mu }^{\ast }(E\cap A)={\mu }^{\ast }(E)+{\mu }^{\ast }(A)$$

\begin{proof} Since $E$ is measurable, then ${\mu }^{\ast }(A)={\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c)$. Then it suffices to show ${\mu }^{\ast }(E\cup A)={\mu }^{\ast }(E)+{\mu }^{\ast }(A\cap E^c)$. Define $B=E\cup A$, then

$${\mu }^{\ast }(B)={\mu }^{\ast }(B\cap E)+{\mu }^{\ast }(B\cap E^c)={\mu }^{\ast }(E)+{\mu }^{\ast }(A\cap E^c).$$

Now we finish the proof. \end{proof}

2. If $A$ is a non-measurable subset of $X$ and $E$ is a measurable set such that $A\subset E$, then show that ${\mu }^{\ast }(E-A)>0$.

\begin{proof} Assume that ${\mu }^{\ast }(E-A)=0$. For any $M\subseteq X$, we have that $M\cap A^c=(M\cap E^c)\cup (M\cap E\cap A^c)$ and so ${\mu }^{\ast }(M\cap A^c)⩽{\mu }^{\ast }(M\cap E^c)+{\mu }^{\ast }(M\cap E\cap A^c)$. Since $\mu \ast (M\cap E\cap A^c)⩽{\mu }^{\ast }(E\cap A^c)=0$, there is ${\mu }^{\ast }(M\cap A^c)⩽{\mu }^{\ast }(M\cap E^c)$. Note that

$${\mu }^{\ast }(M\cap A)+{\mu }^{\ast }(M\cap A^c)⩽{\mu }^{\ast }(M\cap E)+{\mu }^{\ast }(M\cap E^c)={\mu }^{\ast }(M)$$

for all $M\subseteq X$. It follows that $A$ is measurable, contradiction. Therefore, ${\mu }^{\ast }(E-A)>0$. \end{proof}

3. Let $\left\{A_n\right\}$ be a sequence of subsets of $X$. Assume that there exists a disjoint sequence $\left\{B_n\right\}$ of measurable sets such that $A_n\subset B_n$ holds for each $n$. Show that

$${\mu }^{\ast }\left({\cup }_{n=1}^{\infty }{A}_n\right)=\sum _{n=1}^{\infty }{\mu }^{\ast }\left(A_n\right)$$

\begin{proof} Since $B_1$ is measurable, then ${\mu }^{\ast }({\cup }_{n=1}^{\infty }{A}_1)={\mu }^{\ast }(({\cup }_{n=1}^{\infty }{A}_n)\cap B_1)+{\mu }^{\ast }(({\cup }_{n=1}^{\infty }{A}_n)\cap B_1^c)={\mu }^{\ast }(A_1)+{\mu }^{\ast }({\cup }_{n=2}^{\infty }{A}_n)$. Repeat this procedure, for any $N\in {\mathbb{N}}_{+}$, we have that

$${\mu }^{\ast }({\cup }_{n=1}^{\infty }{A}_n)=\sum _{n=1}^N{\mu }^{\ast }(A_n)+{\mu }^{\ast }({\cup }_{n=N+1}^{\infty }{A}_n).\text{\hspace{1em}(*)}$$

If ${\mu }^{\ast }({\cup }_{n=1}^{\infty }{A}_n)=\infty$, then ${\sum }_{n=1}^{\infty }{\mu }^{\ast }(A_n)⩾{\mu }^{\ast }({\cup }_{n=1}^{\infty }{A}_n)=\infty$, ${\lim }_{N\to \infty }{\sum }_{n=1}^N{\mu }^{\ast }(A_n)=\infty$ and so $(\ast )$ holds. If ${\mu }^{\ast }({\cup }_{n=1}^{\infty }{A}_n)<\infty$, then ${\mu }^{\ast }({\cup }_{n=N+1}^{\infty }{A}_n)\to 0$ as $N\to 0$, which yields that $(\ast )$ holds. \end{proof}

4. Show that a subset $E$ of $X$ is measurable if and only if for each $ϵ>0$ there exists a measurable set $F$ such that $F\subset E$ and ${\mu }^{\ast }(E-F)<ϵ$.

\begin{proof} If $E$ is measurable, then take $F=E$. Since ${\mu }^{\ast }(E-F)=0$, for each $ϵ>0$, there is $0={\mu }^{\ast }(E-F)<ϵ$.

Now assume that for each $ϵ>0$ there exists a measurable set $F$ such that $F\subset E$ and ${\mu }^{\ast }(E-F)<ϵ$. We aim to show for any $M\subseteq X$,

$${\mu }^{\ast }(M\cap E)+{\mu }^{\ast }(M\cap E^c)⩽{\mu }^{\ast }(M).$$

For any $ϵ>0$, assume that measurable $F_{ϵ}\subseteq E$ satisfying ${\mu }^{\ast }(E-F_{ϵ})<ϵ$. Note that ${\mu }^{\ast }(M\cap E)⩽{\mu }^{\ast }(M\cap F_{ϵ})+{\mu }^{\ast }(M\cap F_{ϵ}^c\cap E)⩽{\mu }^{\ast }(M\cap F_{ϵ})+ϵ$ and so

$${\mu }^{\ast }(M\cap E)+{\mu }^{\ast }(M\cap E^c)⩽{\mu }^{\ast }(M\cap F_{ϵ})+{\mu }^{\ast }(M\cap F_{ϵ}^c)+ϵ={\mu }^{\ast }(M)+ϵ.$$

Take $ϵ\to 0$, then we get ${\mu }^{\ast }(M\cap E)+{\mu }^{\ast }(M\cap E^c)⩽{\mu }^{\ast }(M)$. Since ${\mu }^{\ast }(M\cap E)+{\mu }^{\ast }(M\cap E^c)⩾{\mu }^{\ast }(M)$, we get ${\mu }^{\ast }(M\cap E)+{\mu }^{\ast }(M\cap E^c)={\mu }^{\ast }(M)$ and so $E$ is measurable. \end{proof}

Exercise 5

1. Let $(X,\mathcal{A},\mu )$ be a measure space and let $f_n:X\to \mathbb{R}$ be any sequence of measurable functions. Show that

$$\begin{array}{rl}E:& =\left\{x\in X：\left\{f_n(x)\right\}\text{ is a Cauchy sequence }\right\}\\ & ={\cap }_{k=1}^{\infty }{\cup }_{n_0=1}^{\infty }{\cap }_{n,m\ge n_0}\left\{x\in X:\left|f_n(x)-f_m(x)\right|<\frac{1}{2^k}\right\}\end{array}$$

and so $E$ is measurable.

\begin{proof} i) Define $L=\left\{x\in X：\left\{f_n(x)\right\}\text{ is a Cauchy sequence }\right\}$ and $R={\cap }_{k=1}^{\infty }{\cup }_{n_0=1}^{\infty }{\cap }_{n,m\ge n_0}\left\{x\in X:\left|f_n(x)-f_m(x)\right|<\frac{1}{2^k}\right\}$.

For any $x\in R$, we prove that $\{f_n(x)\}$ is a Cauchy sequence. For any $ϵ>0$, there exists $K$ such that $\frac{1}{2^K}<ϵ$ and $x\in {\cap }_{n,m⩾n_0}\{x\in X:|f_n(x)-f_m(x)|<\frac{1}{2^K}<ϵ\}$ for some $n_0\in {\mathbb{N}}_{+}$. Therefore, for any $ϵ>0$, there exists $n_0$ such that $|f_n(x)-f_m(x)|<ϵ$ for any $n,m⩾n_0$. Hence, $\{f_n(x)\}$ is a Cauchy sequence and so $x\in L$.

Now we assume that $x\in L$. For any $k\in {\mathbb{N}}_{+}$, there exists $N_k$ such that $|f_n(x)-f_m(x)|<\frac{1}{2^k}$ for any $n,m⩾N_k$. It follows that

$$x\in \bigcap _{n,m⩾N_k}\{x\in X:|f_n(x)-f_m(x)|<\frac{1}{2^k}\}\subseteq \bigcup _{n_0=1}^{\infty }\bigcap _{n,m\ge n_0}\left\{x\in X:\left|f_n(x)-f_m(x)\right|<\frac{1}{2^k}\right\}$$

for all $k\in {\mathbb{N}}_{+}$ and so $x\in R$. Therefore, we proved that $L=R$.

ii) Since $f_n$ are measurable functions, $|f_n-f_m|$ is also measurable for any $n,m\in {\mathbb{N}}_{+}$. Therefore, $\{x\in X:|f_n(x)-f_m(x)|<\frac{1}{2^k}\}$ is measurable and so for $E$. \end{proof}

2. A measure space $(X,\mathcal{A},\mu )$ is complete iff given $f$ and $g$ on $X$ with $f$ measurable and $g=f$ $\mu$ a.e., $g$ is measurable.

\begin{proof} Assume that $f,g$ are maps from $X$ to $Y$, where $(Y,\mathcal{B})$ is a measurable space.

Assume that $(X,\mathcal{A},\mu )$ is complete, and assume that $f=g$ a.e. with $f$ measurable. It yields that $f=g$ on $X-X_0$ where $X_0$ is a measurable subset of $X$ and $\mu (X_0)=0$. For any $B\in \mathcal{B}$, $f^{-1}(B)\in \mathcal{A}$ by $f$ measurable. Note that $g^{-1}(B)\subseteq f^{-1}(B)\cup X_0$. As $D_1:=f^{-1}(B)-g^{-1}(B)$ and $D_2=g^{-1}(B)-f^{-1}(B)$ are subsets of $X_0$ and $(X,\mathcal{A})$ is complete, we have that $D_1$ and $D_2$ are measurable. It follows that $f^{-1}(B)-D_1$ is measurable and so $(f^{-1}(B)-D_1)\cup D_2=g^{-1}(B)$ is measurable. Therefore, $g$ is measurable.

Conversely, assume that for any $f$ and $g$, if $f$ is measurable and $f=g$ a.e., then $g$ is measurable. Note that the conclusion does not hold if $Y$ is a singleton or $\mathcal{B}=\{\varnothing ,Y\}$. Hence from now on we assume that $Y$ contains a proper measurable set $B$. Take $y_1\in B$ and $y_0\in B^c$. For any $E\in \mathcal{A}$ with $\mu (E)=0$ and any subset $F\subseteq E$, define

g_F(x)=\begin{cases} y_1, &\mbox{ if }x\notin F\\ y_0, &\mbox{ if } x\in F. \end{cases}$$ Since $f$ is measurable and $g=f$ a.e., we have that $g_F$ is measurable. It follows that $g^{-1}(B^c)=F$ is measurable. Therefore, for any $E\in\mathcal{A}$ and $F\subseteq E$, $F$ is also a measurable set. Hence $(X,\mathcal{A},\mu)$ is complete. `\end{proof}` **3. Let $\mathcal{A}$ be a $\sigma$-algebra of subsets of $X$ and $f$ a function on $X$ such that $|f|$ is measurable. Find an additional condition that ensures that $f$ is measurable.** `\begin{proof}` An additional condition is, $f^+$ is a measurable function. Define $$f^+(x)=\begin{cases} f(x),&\mbox{if } f(x)\geqslant 0, \\ 0,&\mbox{if }f(x)<0. \end{cases}$$ If $f^+$ and $|f|$ are measurable, then $f^+-|f|$ is measurable and so $f=2f^+-|f|$ is also measurable. `\end{proof}` **4. Given two measurable spaces $(X, \mathcal{A})$ and $(X, \mathcal{B})$. Let $f$ be a measurable mapping of $X$ into $Y$. Let $\mu$ be a measure on $\mathcal{A}$. Show that the set function defined by $\nu=\mu \circ f^{-1}$ on $\mathcal{B}$, that is, $\nu(B)=\mu\left(f^{-1}(B)\right)$ for $B \in \mathcal{B}$ is a measure on $\mathcal{B}$. This measure $\nu$ is called the image measure or push forward of $\mu$ under $f$ and is denoted by $f(\mu)$.** `\begin{proof}` As $f$ is a measurable function, for any $B\in\mathcal{B}$, $f^{-1}(B)\in\mathcal{A}$. Since $f^{-1}(\emptyset)=\emptyset\in\mathcal{A}$, there is $\nu(\emptyset)=\mu(\emptyset)=0$. For pairwise disjoint $\{B_i\}_{i=1}^\infty\subseteq\mathcal{B}$, we have that $f^{-1}(B_i)\cap f^{-1}(B_j)=\emptyset$ for any $i\neq j$. Note that

\nu(\cup_{i=1}^\infty B_i)=\mu(f^{-1}(\cup_{i=1}^\infty B_i))=\mu(\cup_{i=1}^\infty f^{-1}(B_i))=\sum_{i=1}^\infty \mu(f^{-1}(B_i))=\sum_{i=1}^\infty\nu(B_i).

Therefore, $\nu$ is a measure. `\end{proof}`