If $A$ be a Borel subset of $\mathbb{R}$,show that the smallest sigma field of the subsets of $A$ containing the sets open in $A$ is $\{B \in B_{\mathbb{R}}: B \subset A\}$ .
I was taking $C = \{\text{all open sets in }A\}$.
We need to show $\sigma(C)= \{B \in B_{\mathbb{R}}: B \subset A\}$ .
I think we might need the principle of good sets here.
Any leads?
Let $\mathcal A$ be the sigma algebra generated by open sets in $A$ and $\mathcal C$ be the class $\{B\in B_{\mathbb R}: B \subset A\}$. We have to show that $\mathcal A=\mathcal C$. First verify that $\mathcal C$ is a sigma algebra of subsets of $A$. This is straightforward. Next observe that any open set in $A$ is of the form $A \cap U$ where $U$ is open in $\mathbb R$. Celarly. $A\cap U$ is a Borel set in $\mathbb R$ contained in $A$, so it belongs to $\mathcal C$. This proves that $\mathcal A \subset \mathcal C$.
For the reverse inclusion consider $\{E\in B_{\mathbb R}: E \cap A \in \mathcal A\}$. Verify that this is a sigma algebra containing all open sets in $\mathbb R$. Hence it contains all Borel sets. In partcicular we have proved that any Borel set in $\mathbb R$ already contained in $A$ belongs to $\mathcal A$.