HW1

Exercise 1

1. Let $E\subset X$ be any set and let $\mathcal{A}$ be some $\sigma$-algebra on $X$. Show that

$${\mathcal{A}}_E=\{E\cap A:A\in \mathcal{A}\}$$

is a $\sigma$-algebra of subsets of $E$.

\begin{proof} Since $\mathcal{A}$ is a $\sigma$-algebra of $X$, then $\varnothing ,X\in \mathcal{A}$. So $E\cap \varnothing =\varnothing$ and $E\cap X=E$ are contained in ${\mathcal{A}}_E$. If $\{E\cap A_i{\}}_{i=1}^{\infty }\subseteq {\mathcal{A}}_E$, then ${\cup }_{i=1}^{\infty }(E\cap A_i)=E\cap ({\cup }_{i=1}^{\infty }{A}_i)\in {\mathcal{A}}_E$ as ${\cup }_{i=1}^{\infty }{A}_i\in \mathcal{A}$. Therefore, ${\mathcal{A}}_E$ is a $\sigma$-algebra of subsets of $E$. \end{proof}

2. Let $\mathcal{A}$ be an algebra of subsets of $X$ and $A,B\notin \mathcal{A}$. Is the same true of $A\cup B$ ?

\begin{proof} No. Let $X=[0,1]$ and let $\mathcal{A}=\left\{\varnothing ,X,\left[0,\frac{1}{2}\right),\left[\frac{1}{2},1\right]\right\}$. Then $[0,\frac{1}{3}),[\frac{1}{3},1]\notin \mathcal{A}$ but $A\cup B=X\in \mathcal{A}$. \end{proof}

3. Let $X$ be an infinite set. Describe the $\sigma$-algebra generated by the singletons of $X$.

\begin{proof} Denote the $\sigma$-algebra generated by the singletons by $\mathcal{A}$. For any countable set $\{x_i:i\in I\}$, we have that ${\cup }_{i\in I}\{x_i\}\in \mathcal{A}$ and $({\cup }_{i\in I}\{x_i\}{)}^c\in \mathcal{A}$. Thus $\mathcal{A}\supseteq {\mathcal{A}}^{\prime }=\{A\subseteq X:A\text{mbox}iscountableor{A}^c\text{mbox}iscountable\}$. Since ${\mathcal{A}}^{\prime }$ contains all singletons of $X$ and ${\mathcal{A}}^{\prime }$ is a $\sigma$-algebra, then ${\mathcal{A}}^{\prime }=\mathcal{A}$. That is, the $\sigma$-algebra generated by the singletons of $X$ is the collection of countable subsets and their complements. (Remark: finite set is countable.) \end{proof}

4. Show that the union of algebras of subsets of $X$ is not necessarily an algebra. How about the union of an increasing collection of algebras?

\begin{proof} i) Counterexample: Let ${\mathcal{A}}_1=\left\{\varnothing ,X,\left[0,\frac{1}{2}\right),\left[\frac{1}{2},1\right]\right\}$ and let ${\mathcal{A}}_2=\left\{\varnothing ,X,\left[0,\frac{1}{3}\right),\left[\frac{1}{3},1\right]\right\}$. Then ${\mathcal{A}}_1\cup {\mathcal{A}}_2$ is not an algebra, as $[0,\frac{1}{3})\cup [\frac{1}{2},1]$ is not contained in ${\mathcal{A}}_1\cup {\mathcal{A}}_2$.

ii) Assume that ${\mathcal{A}}_1\subset {\mathcal{A}}_2\subset \cdots$ is an increasing collection of algebras, and define $\mathcal{A}$ as the union of them. Since $\varnothing ,X\in {\mathcal{A}}_1$, then $\varnothing ,X\in \mathcal{A}$. For any $A\in \mathcal{A}$, there exists $n\in \mathbb{N}$ such that $A\in {\mathcal{A}}_n$. Then $A^c\in {\mathcal{A}}_n\subseteq \mathcal{A}$. Similarly, for any $\{A_i{\}}_{i=1}^m\subseteq \mathcal{A}$, there exists $n^{\prime }\in \mathbb{N}$ such that $A_i\in {\mathcal{A}}_{n^{\prime }}$. Thus ${\cup }_{i=1}^{n^{\prime }}{A}_i\in {\mathcal{A}}_{n^{\prime }}\subseteq \mathcal{A}$. Therefore, $\mathcal{A}$ is an algebra. \end{proof}

5. Let ${\mathcal{A}}_1,\cdots ,{\mathcal{A}}_n$ be $\sigma$-algebras of subsets of $X$ such that $\mathcal{A}={\cup }_{k=1}^n{\mathcal{A}}_k$ is an algebra. Show that $\mathcal{A}$ is a $\sigma$-algebra.

\begin{proof} Since $\varnothing ,X\in {\mathcal{A}}_1$, then $\varnothing ,X\in \mathcal{A}$. For any $A\in \mathcal{A}$, there exists $m\in \{1,\cdots ,n\}$ such that $A\in {\mathcal{A}}_m$. Then $A^c\in {\mathcal{A}}_m\subseteq \mathcal{A}$. For any $\{A_i{\}}_{i=1}^{\infty }\subseteq \mathcal{A}$, the set $S:=\{A_i:i\in {\mathbb{N}}_{+}\}$ can be written as $S={\cup }_{j=1}^n\{A_i^j:A_i^j\in S\cap {\mathcal{A}}_j,i\in I_j\}$. As $I_j$ are countable sets for all $j$, we have that $A^j:={\cup }_{i\in I_j}{A}_i^j\in {\mathcal{A}}_j$ and so ${\cup }_{j=1}^n{A}^j={\cup }_{i=1}^{\infty }{A}_i\in \mathcal{A}$. Therefore, $\mathcal{A}$ is a $\sigma$-algebra. \end{proof}

7. Does there exist an infinite $\sigma$-algebra $\mathcal{A}$ with countably many sets?

\begin{proof} Assume that $\mathcal{A}=\{A_i:i\in {\mathbb{N}}_{+}\}$ and $\mathcal{A}\subseteq \mathcal{P}(X)$. For any $x\in X$, define $B_x={\cap }_{i\in {\mathbb{N}}_{+},x\in A_i}{A}_i\in \mathcal{A}$. We claim that if $B_x\cap B_y\ne \varnothing$, then $B_x=B_y$. Take $z\in B_x\cap B_y$. Since $B_x\cap B_y\in \mathcal{A}$, then $B_z\subseteq B_x\cap B_y$. If $x\notin B_z$, then $(B_z^c\cap B_x\cap B_y)\in \mathcal{A}$ is a subset containing $x$ and $(B_z^c\cap B_x\cap B_y)⊊B_x$, contradiction. Therefore, $x\in B_z$ and $B_x\subseteq B_z\subseteq B_x\cap B_y\subseteq B_x$. It follows that $B_x=B_z$. Similarly, we can prove that $y\in B_z$ and $B_y=B_z$. Therefore, if $B_x\cap B_y\ne \varnothing$, then $B_x=B_y$.

Take $A_i\in \mathcal{A}$. If $x\in A_i$ then $A_i\cap B_x=B_x$. If $x\notin A_i$, then $A_i^c\cap B_x=B_x$. Therefore, for each $A_i\in \mathcal{A}$, either $A_i\supseteq B_x$ or $A_i\cap B_x=\varnothing$. Assume that there are only $n$ distinct subsets of the form $B_x$, then $|\mathcal{A}|⩽2^n$ is finite, contradiction. Thus there are infinite number of sets of the form $B_x$ and so $|\mathcal{A}|⩾2^{{\aleph }_0}$ is not countable. \end{proof}

8. An algebra $\mathcal{A}$ is a $\sigma$-algebra iff $\mathcal{A}$ is closed under countable increasing unions (i.e., if ${\left\{E_j\right\}}_{j=1}^{\infty }\subset \mathcal{A}$ and $E_1\subset E_2\subset \cdots$, then ${\cup }_{j=1}^{\infty }{E}_j\in \mathcal{A}$ ).

\begin{proof} If $\mathcal{A}$ is a $\sigma$-algebra, then for any $\{E_j{\}}_{j=1}^{\infty }\subseteq \mathcal{A}$ we have ${\cup }_{j=1}^{\infty }{E}_j\in \mathcal{A}$. Now we assume that an algebra $\mathcal{A}\subseteq \mathcal{P}(X)$ is closed under countable increasing unions. For any $\{E_j{\}}_{j=1}^{\infty }\subseteq \mathcal{A}$, define $F_n={\cup }_{j=1}^n{E}_j$. Then $F_n\in \mathcal{A}$ by $\mathcal{A}$ algebra. Then $\{F_i{\}}_{i=1}^{\infty }\subseteq \mathcal{A}$ satisfying $F_1\subseteq F_2\subseteq \cdots$ and so ${\cup }_{j=1}^{\infty }{E}_j={\cup }_{i=1}^{\infty }{F}_i\in \mathcal{A}$. Therefore, $\mathcal{A}$ is a $\sigma$-algebra. \end{proof}

9. Let $f:X\to Y$ be a map and $\mathcal{A}$ be a $\sigma$-algebra on $Y$. Then

$${\mathcal{A}}^{\prime }=\left\{f^{-1}(B):B\in \mathcal{A}\right\}$$

is a $\sigma$-algebra on $X$.

\begin{proof} Since $\mathcal{A}$ is a $\sigma$-algebra, then $\varnothing ,Y\in \mathcal{A}$. Then $f^{-1}(\varnothing )=\varnothing$ and $f^{-1}(Y)=X$ yield that $\varnothing ,X\in {\mathcal{A}}^{\prime }$. For $\{A_i{\}}_{i=1}^{\infty }\subseteq {\mathcal{A}}^{\prime }$, there exist $\{B_i{\}}_{i=1}^{\infty }\subseteq \mathcal{A}$ such that $A_i=f^{-1}(B_i)$. As ${\cup }_{i=1}^{\infty }{A}_i=f^{-1}({\cup }_{i=1}^{\infty }{B}_i)$ and ${\cup }_{i=1}^{\infty }{B}_i\in \mathcal{A}$, then ${\cup }_{i=1}^{\infty }{A}_i\in {\mathcal{A}}^{\prime }$. Therefore, ${\mathcal{A}}^{\prime }$ is a $\sigma$-algebra. \end{proof}

10. If $\mathcal{A}$ is the $\sigma$-algebra generated by $\mathcal{M}$, then $\mathcal{A}$ is the union of the $\sigma$-algebras generated by $\mathcal{F}$ as $\mathcal{F}$ ranges over all countable subsets of $\mathcal{M}$. (Hint: Show that the latter object is a $\sigma$-algebra.)

\begin{proof} Define $\mathcal{F}=\{\mathcal{F}:\mathcal{F}\subseteq \mathcal{M},\mathcal{F}\text{mbox}iscountable\}$. Let $\mathcal{B}={\bigcup }_{\mathcal{F}\in \mathcal{F}}⟨\mathcal{F}⟩$. For any $M\in \mathcal{M}$, $\{M\}$ is countable and so $\{M\}\in \mathcal{F}$. Thus $\mathcal{B}\supseteq M$. On the other hand, since $⟨\mathcal{F}⟩\subseteq ⟨\mathcal{M}⟩$ for any $\mathcal{F}\in \mathcal{F}$, we have $\mathcal{B}\subseteq ⟨\mathcal{M}⟩$. Therefore, to show $\mathcal{A}=⟨M⟩$ equals $\mathcal{B}$, it suffices to show that $\mathcal{B}$ is a $\sigma$-algebra.

Assume that $\mathcal{M}\subseteq \mathcal{P}(X)$. Since $⟨\{M\}⟩=\{\varnothing ,X,M,M^c\}$ for any nontrivial $M\in \mathcal{M}$, we have that $\varnothing ,X\in \mathcal{B}$. For each $B_i\in \{B_i{\}}_{i=1}^{\infty }\subseteq \mathcal{B}$, $B_i={\cup }_{\mathcal{F}\in \mathcal{F}}{F}_{\mathcal{F}}^{(i)}$ with $F_{\mathcal{F}}^{(i)}\in ⟨\mathcal{F}⟩$. Then ${\cup }_{i=1}^{\infty }{B}_i={\cup }_{\mathcal{F}\in \mathcal{F}}({\cup }_{i=1}^{\infty }{F}_{\mathcal{F}}^{(i)})$. Since $⟨\mathcal{F}⟩$ is a $\sigma$-algebra, then ${\cup }_{i=1}^{\infty }{F}_{\mathcal{F}}^{(i)}\in ⟨\mathcal{F}⟩$ and so ${\cup }_{i=1}^{\infty }{B}_i\in \mathcal{B}$. Therefore, $\mathcal{B}$ is a $\sigma$-algebra, and $\mathcal{A}$ is the union of the $\sigma$-algebras generated by $\mathcal{F}$ as $\mathcal{F}$ ranges over all countable subsets of $\mathcal{M}$. \end{proof}

11. Let $\mathcal{A}$ be the collection of all subsets of $\mathbb{R}$ that are unions of finitely many intervals of the form $(a,b],(a,\infty )$, or $(-\infty ,b]$. Show that $\mathcal{A}$ is an algebra but is not a $\sigma$-algebra.

\begin{proof} Note that $(-\infty ,0]\cup (0,\infty )=\mathbb{R}\in \mathcal{A}$ and $(1,0]=\varnothing \in \mathcal{A}$. Also we have that $(a,b{]}^c=(-\infty ,a]\cup (b,\infty )$ and $(a,\infty {)}^c=(-\infty ,a]$. For $\{A_i{\}}_{i=1}^n\subseteq \mathcal{A}$, as $\mathcal{A}$ is the collection of unions of finitely many intervals of the form $(a,b],(a,\infty )$, or $(-\infty ,b]$, then each $A_i$ is a union of finitely many intervals and so ${\cup }_{i=1}^n{A}_i\in \mathcal{A}$. Therefore, $\mathcal{A}$ is an algebra.

Since ${\cup }_{i=1}^{\infty }(-\infty ,b-\frac{1}{n}]=(-\infty ,b)\notin \mathcal{A}$, $\mathcal{A}$ is not a $\sigma$-algebra. \end{proof}

12. Let $A\subset {\mathbb{R}}^n$ be a Borel set. Show that for $h\in {\mathbb{R}}^n$ and $t\in \mathbb{R}$ the sets $A+h:=\{a+h:a\in A\},tA:=\{ta:a\in A\}$ are Borel as well.

\begin{proof} Let $\mathcal{B}$ be the Borel $\sigma$-algebra of ${\mathbb{R}}^n$. Given a fixed $h\in {\mathbb{R}}^n$, let

$${\mathcal{A}}_h=\{B\subseteq {\mathbb{R}}^n:h+B\in \mathcal{B}\}$$

We claim that ${\mathcal{A}}_h$ is a $\sigma$-algebra. Since $\varnothing +h=\varnothing \in \mathcal{B}$ and ${\mathbb{R}}^n+h={\mathbb{R}}^n\in \mathcal{B}$, then $\varnothing ,{\mathbb{R}}^n\in {\mathcal{A}}_h$. If $B\in {\mathcal{A}}_h$, then $h+B,(h+B{)}^c\in \mathcal{B}$. By $(h+B{)}^c=h+B^c\in \mathcal{B}$, we have that $B^c\in {\mathcal{A}}_h$. For $\{B_i{\}}_{i=1}^{\infty }\subseteq {\mathcal{A}}_h$, there is $h+B_i\in \mathcal{B}$ and $h+{\cup }_{i=1}^{\infty }{B}_i={\cup }_{i=1}^{\infty }(h+B_i)\in \mathcal{B}$. Thus ${\cup }_{i=1}^{\infty }{B}_i\in {\mathcal{A}}_h$. It follows that ${\mathcal{A}}_h$ is a $\sigma$-algebra.

Notice that for all open sets $O$, $h+O$ is also an open set and $h+O\in \mathcal{B}$. So for all open set $O,O\in {\mathcal{A}}_h$ and so $\mathcal{B}\subseteq {\mathcal{A}}_h$. Therefore, for any $A\in \mathcal{B}\subseteq {\mathcal{A}}_h$, we have that $A+h\in \mathcal{B}$, i.e., $A+h$ is a Borel set.

If $t=0$, $tA=\{0\}$ is a Borel set. Now we assume that $t\ne 0$. Similarly, define ${\mathcal{C}}_t=\{B\subseteq {\mathbb{R}}^n:tB\in \mathcal{B}\}$ and we can show that ${\mathcal{C}}_t$ is a $\sigma$-algebra which contains $\mathcal{B}$. Then for any $A\in \mathcal{B}\subseteq {\mathcal{C}}_t$, $tA\in \mathcal{B}$ and so $tA$ is a Borel set. \end{proof}