Prove that $\{A\cap E:A\in\mathcal A\}$ is a $\sigma$-algebra of $E$

Question

Is my proof correct?

Let $(X,\mathcal A)$ be a measure space and let $E\subseteq X$. Prove that $\mathcal A_1=\{A\cap E:A\in\mathcal A\}$ is a $\sigma$-algebra of $E$.

Proof

1. $\varnothing\in\mathcal A_1$ because $\varnothing\in \mathcal A.$

2. $E\in\mathcal A_1$ because $E\subseteq X\subseteq\mathcal A.$

3. Suppose $\{A_n\}\subset \mathcal A_1$

Then $A_n\cap E:A_n\in\mathcal A$

Then $\bigcup A_n\subset\mathcal A$

Taking the union $A_n\cap E$ yields $\bigcup A_n\cap E$ and such that $\bigcup A_n\subset\mathcal A.$

4. Let $B\in\mathcal A_1 $

Then $B\cap E:B\in\mathcal A$

Then $B^c\in\mathcal A$

So $B^c\cap E:B^c\in\mathcal A$

$\therefore B^c\in\mathcal A_1 $

Answer 1

No, your proof has errors. A set belongs to $\mathcal A_1$ if it has the form $A\cap E$ for some $A \in \mathcal A$.

$\emptyset=\emptyset\cap E$ and $\emptyset \in \mathcal A$, so $\emptyset \in \mathcal A_1$. $E=X\cap E$ and $X \in \mathcal A$, so $E \in \mathcal A_1$.

If $A_n \in \mathcal A_1$ for all $n$ then there exist sets $B_n \in \mathcal A$ such that $A_n=B_n\cap E$ and $\cup_n A_n=(\cup_n B_n) \cap E$ so $\cup_n A_n \in \mathcal A_1$. Finally, if $A \in \mathcal A_1$ then $A=B\cap E$ for some $B \in \mathcal A$ and the complement of $A$ in $E$ is $E\cap A^{c}$ which can be written as $B^{c}\cap E$. Hence the complement belongs to $\mathcal A_1$