Frederick Kramer

Answered

2022-07-08

Given a set $\mathrm{\Omega}$ with a $\sigma $-field F defined on it. Let B be a subset of $\mathrm{\Omega}$ and define

$B\cap \mathcal{F}=\{B\cap F:F\in \mathcal{F}\}$. Call this ${\mathcal{F}}_{B}$.

The claim is that this collection of sets is a $\sigma $-field and hence (B, ${\mathcal{F}}_{B}$) is a measurable space.

I don't see why ${\mathcal{F}}_{B}$ is a $\sigma $-field just because it is a subset of one. Any insight appreciated

$B\cap \mathcal{F}=\{B\cap F:F\in \mathcal{F}\}$. Call this ${\mathcal{F}}_{B}$.

The claim is that this collection of sets is a $\sigma $-field and hence (B, ${\mathcal{F}}_{B}$) is a measurable space.

I don't see why ${\mathcal{F}}_{B}$ is a $\sigma $-field just because it is a subset of one. Any insight appreciated

Answer & Explanation

Brendan Bush

Expert

2022-07-09Added 14 answers

Suggested proof. (Axioms are $\sigma $-algebra axioms.) Notation different from question.

Let ($\star $) be

$B\cap \mathcal{E}:=\{B\cap E,E\in \mathcal{E}\}={\mathcal{E}}_{B}$

To prove the claim it is a $\sigma $-algebra on $B$ we need to show:

(i) $B\in {\mathcal{E}}_{B}$

(ii) $(B\cap E{)}^{c}$ in ${\mathcal{E}}_{B}$ $\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}E\in \mathcal{E}$.

(iii) ${\mathcal{E}}_{B}$ satisfies $\sigma $-additivity.

Proof of (i): We know that $\mathrm{\Omega}$ is in $\mathcal{E}$ by Axiom (1).

Therefore, by ($\star $), $(B\cap \mathrm{\Omega})\in {\mathcal{E}}_{B}$.

But $B\cap \mathrm{\Omega}=B$, hence $B\in {\mathcal{E}}_{B}$.

Proof of (ii): $(B\cap E{)}^{c}=({B}^{c}\cup {E}^{c})$ (de Morgan).

Now ${B}^{c}$ and ${E}^{c}$ are in $\mathcal{E}$ by Axiom (2) and their union is in $\mathcal{E}$ by Axiom (3).

Hence, by ($\star $) $B\cap ({B}^{c}\cup {E}^{c})\in {\mathcal{E}}_{B}$.

But $({B}^{c}\cup {E}^{c})=(\mathrm{\Omega}\cup \mathrm{\Omega})=\mathrm{\Omega}$, and $B\cap \mathrm{\Omega}=B$ which is in ${\mathcal{E}}_{B}$ by (i).

Proof of (iii): Let ${E}_{j}\in \mathcal{E}$.

We want to show that

$\bigcup _{j\in \mathcal{N}}(B\cup {E}_{j})\in {\mathcal{E}}_{B}.$

But,

$\bigcup _{j\in \mathcal{N}}(B\cup {E}_{j})=B\cap \bigcup _{j\in \mathcal{N}}{E}_{J}$$\bigcup _{j\in \mathcal{N}}{E}_{J}=X,\phantom{\rule{thickmathspace}{0ex}}\text{say, then}X\in \mathcal{E}$

by Axiom (3).

Hence, by $(\star )$ $(B\cap X)\in {\mathcal{E}}_{B}$

Let ($\star $) be

$B\cap \mathcal{E}:=\{B\cap E,E\in \mathcal{E}\}={\mathcal{E}}_{B}$

To prove the claim it is a $\sigma $-algebra on $B$ we need to show:

(i) $B\in {\mathcal{E}}_{B}$

(ii) $(B\cap E{)}^{c}$ in ${\mathcal{E}}_{B}$ $\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}E\in \mathcal{E}$.

(iii) ${\mathcal{E}}_{B}$ satisfies $\sigma $-additivity.

Proof of (i): We know that $\mathrm{\Omega}$ is in $\mathcal{E}$ by Axiom (1).

Therefore, by ($\star $), $(B\cap \mathrm{\Omega})\in {\mathcal{E}}_{B}$.

But $B\cap \mathrm{\Omega}=B$, hence $B\in {\mathcal{E}}_{B}$.

Proof of (ii): $(B\cap E{)}^{c}=({B}^{c}\cup {E}^{c})$ (de Morgan).

Now ${B}^{c}$ and ${E}^{c}$ are in $\mathcal{E}$ by Axiom (2) and their union is in $\mathcal{E}$ by Axiom (3).

Hence, by ($\star $) $B\cap ({B}^{c}\cup {E}^{c})\in {\mathcal{E}}_{B}$.

But $({B}^{c}\cup {E}^{c})=(\mathrm{\Omega}\cup \mathrm{\Omega})=\mathrm{\Omega}$, and $B\cap \mathrm{\Omega}=B$ which is in ${\mathcal{E}}_{B}$ by (i).

Proof of (iii): Let ${E}_{j}\in \mathcal{E}$.

We want to show that

$\bigcup _{j\in \mathcal{N}}(B\cup {E}_{j})\in {\mathcal{E}}_{B}.$

But,

$\bigcup _{j\in \mathcal{N}}(B\cup {E}_{j})=B\cap \bigcup _{j\in \mathcal{N}}{E}_{J}$

by Axiom (3).

Hence, by $(\star )$ $(B\cap X)\in {\mathcal{E}}_{B}$

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