Prove that $A_B=\left\{B \cap A \,; A \in \mathbb{A} \right\}$ is a σ-algebra

Question

I found the following statement and I don't see exactly why it's true so if any of you guys could help , thanks in advance.

So we have $(E,\mathbb{A}) \;$ a measurable space and $B \subset E $

and define the injective function $\gamma :$

$$\gamma : (B,A_B) \rightarrow (E,\mathbb{A})$$ $$x \mapsto x $$

Then :

i) $A_B=\left\{B \cap A \,; A \in \mathbb{A} \right\}$ is a σ-algebra on B that makes the injection $\gamma\,$ measurable .

ii) If $B\in \mathbb{A}$ , we have $A_B =\left\{ A \in \mathbb{A}\,;\,A\subset B \right\}.$

Also I found here something that might help for i) but I don't see how it makes $\gamma$ measurable , but still have no clue for ii)

Answer 1

$\gamma$ is measurable because for each $A\in\mathbb A$, either $A\in A_B$ in which case $\gamma^{-1}(A)=A\in A_B$, or $A\notin A_B$, in which case $\gamma^{-1}(A) = \varnothing\in A_B$.

Let $A^B = \{A\in\mathcal A:A\subset B\}$. If $C\in A_B$ then $C=B\cap A$ for some $A\in\mathbb A$, so $C\in\mathbb A$ and $C\subset B$ so that $C\in A^B$. Similarly, if $C\in A^B$ then $C\in\mathbb A$ and $C\subset B$, so $C = B\cap C\in A_B$. It follows that $A^B=A_B$.