Let $\Omega$ be a nonempty set and $\mathcal{B}$ be a $\sigma$-algebra on $\Omega$. Let $A\subset\Omega$ and $\mathcal{B}_{A} = \{B\cap A:B\in\mathcal{B}\}$. Show that $\mathcal{B}_{A}$ is a $\sigma$-algebra on $A$.
MY ATTEMPT
To begin with, notice that $A\in\mathcal{B}_{A}$. Since $\Omega\in\mathcal{B}$, we conclude that $A = \Omega\cap A\in\mathcal{B}_{A}$.
Let us suppose that $S_{1},S_{2},\ldots, \in\mathcal{B}_{A}$. Then one has that $S_{i} = B_{i}\cap A$ for some $B_{i}\in\mathcal{B}$, where $i\geq 1$. Consequently,
\begin{align*}
S = \bigcup_{i=1}^{\infty}S_{i} = \bigcup_{i=1}^{\infty}(B_{j}\cap A) = \left(\bigcup_{i=1}^{\infty}B_{j}\right)\cap A = B\cap A \Rightarrow S\in\mathcal{B}_{A}
\end{align*}
Could someone help me to prove that $\mathcal{B}_{A}$ is closed under complementation?
Any help is appreciated.
Best Answer
Suppose $S\in \mathcal{B_A}$. Then there is some $B\in\mathcal{B}$ such that $S=B\cap A$. Since $\mathcal{B}$ is a sigma-algebra we have $\Omega\setminus B\in\mathcal{B}$. So now if we can show that $A\setminus S=(\Omega\setminus B)\cap A$ then we are done. This is just a two sided inclusion.
If $x\in (\Omega\setminus B)\cap A$ then in particular $x\in A$, and since $x\notin B$ we also have $x\notin S$. So $x\in A\setminus S$.
On the other hand, if $x\in A\setminus S$ then $x\in A$, and in particular $x\in\Omega$. Also, $x\notin B$. (because if $x$ was in $B$ then we would get $x\in B\cap A=S$, a contradiction). So $x\in (\Omega\setminus B)\cap A$.