 Frederick Kramer

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
$B\cap \mathcal{F}=\left\{B\cap F:F\in \mathcal{F}\right\}$. 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 Brendan Bush

Expert

Suggested proof. (Axioms are $\sigma$-algebra axioms.) Notation different from question.
Let ($\star$) be
$B\cap \mathcal{E}:=\left\{B\cap E,E\in \mathcal{E}\right\}={\mathcal{E}}_{B}$
To prove the claim it is a $\sigma$-algebra on $B$ we need to show:
(i) $B\in {\mathcal{E}}_{B}$
(ii) $\left(B\cap E{\right)}^{c}$ 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 ($\star$), $\left(B\cap \mathrm{\Omega }\right)\in {\mathcal{E}}_{B}$.
But $B\cap \mathrm{\Omega }=B$, hence $B\in {\mathcal{E}}_{B}$.
Proof of (ii): $\left(B\cap E{\right)}^{c}=\left({B}^{c}\cup {E}^{c}\right)$ (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 ($\star$) $B\cap \left({B}^{c}\cup {E}^{c}\right)\in {\mathcal{E}}_{B}$.
But $\left({B}^{c}\cup {E}^{c}\right)=\left(\mathrm{\Omega }\cup \mathrm{\Omega }\right)=\mathrm{\Omega }$, 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
$\bigcup _{j\in \mathcal{N}}\left(B\cup {E}_{j}\right)\in {\mathcal{E}}_{B}.$

But,
$\bigcup _{j\in \mathcal{N}}\left(B\cup {E}_{j}\right)=B\cap \bigcup _{j\in \mathcal{N}}{E}_{J}$
by Axiom (3).
Hence, by $\left(\star \right)$ $\left(B\cap X\right)\in {\mathcal{E}}_{B}$

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