Equality of sigma-algebras using a pi-sytsem Let $(S, \Sigma)$ a measurable space and let $\mathcal{I} \subseteq \Sigma$ be a $\pi$-system such that $\sigma(\mathcal{I}) = \Sigma$. For all $A \in \Sigma$ let $\Sigma_A, \mathcal{I}_A \subseteq 2^A$ satisfy $\Sigma_A = \{ A \cap B: B \in \Sigma \}$ and $\mathcal{I}_A = \{ A \cap B: B \in \mathcal{I} \}$.
First, I verified that for all $A \in \Sigma$ it holds that $\Sigma_A$ is a $\sigma$-algebra on $A$ and that $\mathcal{I_A}$ is a $\pi$-system on $A$. 
Now, I want to prove that 
\begin{align}
\sigma(\mathcal{I_A}) = \Sigma_A.
\end{align}
The first inclusion follows easily, since $\mathcal{I} \subseteq \Sigma$, it follows that $\mathcal{I}_A \subseteq \Sigma_A$ and that $\sigma(\mathcal{I}_A) \subseteq \Sigma_A$.
The other inclusion seems to be more difficult. However, a hint is provided: 
\begin{align}
\text{consider }\ \tilde{\Sigma} = \{ E \cup F: E \in \sigma(\mathcal{I}_A), F \in \sigma(\mathcal{I}_{S\setminus A}) \}.
\end{align}
Unfortunately, I do not see why it should be useful to introduce $\tilde{\Sigma}$. The thing we can do, is to consider if $\tilde{\Sigma}$ is a $\sigma$-algebra. If so, how can we use this to prove that $\Sigma_A \subseteq  \sigma(\mathcal{I}_A)$?
 A: Pay attention to the condition that $\sigma(\mathcal{I})=\Sigma$. The statement is just taking interaction with $A$ on both side.
To make this idea clear, consider an arbitrary set $X\in\Sigma$. We see $X=(X\cap A)\cup(X\cap (S\backslash A))$. This is the motivation for constructing $\bar{\Sigma}$ to include $\sigma(\mathcal{I}_{S\backslash A})$. As $\sigma(\mathcal{I}_A)\subset\Sigma_A$, we may as well get $\sigma(\mathcal{I}_{S\backslash A})\subset\Sigma_{S\backslash A}$. So we see a set in $\bar{\Sigma}$ can be written as $X_A\cup X_{S\backslash A}$, where $X_A\in\Sigma_A$ and $X_{S\backslash A}\in\Sigma_{S\backslash A}$, both of which are in $\Sigma$. Consequently every set in $\bar{\Sigma}$ is also in $\Sigma$. Hence $\bar{\Sigma}\subset\Sigma$. But it's easy to check that $\bar{\Sigma}$ is a $\sigma$-algebra while $\mathcal{I}\subset\bar{\Sigma}$. So $\Sigma\subset\bar{\Sigma}$. This shows $\Sigma=\bar{\Sigma}$. This can only happen when $\sigma(\mathcal{I}_A)=\Sigma_A$ and $\sigma(\mathcal{I}_{S\backslash A})=\Sigma_{S\backslash A}$.  
Edit: Proof for $\mathcal{I}\subset\bar{\Sigma}$ and detailed discussion for the implication of $\sigma(\mathcal{I}_A)=\Sigma_A$ by $\Sigma=\bar{\Sigma}$.  

$(1)~\mathcal{I}\subset\bar{\Sigma}$.  

Any set $I\in\mathcal{I}$ can be expressed as $I=(I\cap A)\cup(I\cap (S\backslash A))$. Since $I\cap A\in\mathcal{I}_A\subset\sigma(\mathcal{I}_A)$, etc, we see $I\in\{E\cup F:E\in\sigma(\mathcal{I}_A),F\in\sigma(\mathcal{I}_{S\backslash A})\}=\bar{\Sigma}$. Hence $\mathcal{I}\subset\bar{\Sigma}$.  

$(2)~\Sigma=\bar{\Sigma}\Rightarrow\sigma(\mathcal{I}_A)=\Sigma_A$.  

Take $X_A\in\Sigma_A$. Since $\Sigma_A\subset\Sigma=\bar{\Sigma}$, by definition we have $X_A=E\cup F$, where $E\in\sigma(\mathcal{I}_A),F\in\sigma(\mathcal{I}_{S\backslash A})$. However, $X_A\in\Sigma_A\Rightarrow X_A\subset A$, so $X_A\cap (S\backslash A)=\emptyset$. This means $F=\emptyset$ (otherwise $X_A$ would contain some elements in $S\backslash A$). So $X_A=E\in\sigma(\mathcal{I}_A)$. Hence $\Sigma_A\subset\sigma(\mathcal{I}_A)$.
