Question

question 4

4. Suppose that I is the sample space and NS2. Show that = {AnN: A € F}, then F' is a d-algebra of (a) if F is a o-algebra of subsets of 12 and F subsets of N'. and : AEC}, then C' generates the o-algebra (b) if C generates the o-algebra Fin 12 and C' = {An Fin 2.

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Answer 1

(a) (I) Since ${\mathcal F}$ is $\sigma$-algebra, we have $\Omega\in {\mathcal F}$ , and therefore, $\Omega'=\Omega\cap\Omega'\in {\mathcal F}'$ .

(II) Suppose that $A\in {\mathcal F}$ and consider $A':=A\cap\Omega'$ in ${\mathcal F}'$. We have $A\in {\mathcal F}$ and $\Omega\in {\mathcal F}$ , and since $\sigma$-algebra ${\mathcal F}$ is closed under complement, we get $\Omega\setminus A\in {\mathcal F}$ , so that

$\Omega'\setminus A'=\Omega'\cap (\Omega\setminus A)\in {\cal F}'$

Thus, ${\mathcal F}'$ is closed under complement.

(III) For any countable collection $\{A'_n\}_{n=1}^\infty$ of subsets in ${\mathcal F}'$, there is a countable collection $\{A_n\}_{n=1}^\infty$ of subsets in ${\mathcal F}$ such that $A_n'=\Omega'\cap A_n$ for all $n=1,2,\cdots$. Since ${\mathcal F}'$ is closed under countable union, we get

$\bigcup_{n=1}^\infty A_n\in \cal F$

Therefore,

$\bigcup_{n=1}^\infty A_n'=\bigcup_{n=1}^\infty (\Omega'\cap A_n)=\Omega'\cap\bigcup_{n=1}^\infty A_n\in\cal F'$

showing that ${\mathcal F}$ is closed under countable union.

By (I), (II), and (III), we conclude that ${\mathcal F}'$ is $\sigma$-algebra of $\Omega'$ in ${\mathcal F}$.

b) Since $\cal C$ is the generator of $\sigma$-algebra ${\mathcal F}$, we have the following two equivalent conditions:

(I) $\cal C\subseteq\cal F$ , and

(II) if $\cal C\subseteq\cal G$ and $\cal G$ is $\sigma$-algebra on $\Omega$ then $\cal F\subseteq\cal G$ .

Now, to prove that $\cal C'$ is generator of ${\mathcal F}'$, we need to prove the following:

(I) $\cal C'\subseteq\cal F'$ , and

(II) if $\cal C'\subseteq\cal G'$ and $\cal G'$ is $\sigma$-algebra on $\Omega'$ then $\cal F'\subseteq\cal G'$ .

Proof of (I): If $C'\in \cal C'$ then $C'=C\cap \Omega'$ for some $C\in \cal C$. Since $\cal C\subseteq\cal F$ , we have $C\in \cal F$, and hence, $C'\in \cal F'$. Thus, $\cal C'\subseteq\cal F'$ .

(II) Suppose $\cal G'$ is $\sigma$-algebra on $\Omega'$ and $\cal C'\subseteq\cal G'$ . Let $\cal G$ be the $\sigma$-algebra on $\Omega$ generated by $\cal G'\cup \cal C$ . Then $\cal C\subseteq\cal G$ , and hence, $\cal F\subseteq\cal G$ . Thus, if $A\in\cal F$ then $A\in\cal G$ so that $A\cap\Omega'\in\cal G'$ . This proves $\cal F'\subseteq\cal G'$ .

Hence, $\cal C'$ is generator of ${\mathcal F}'$.