$X$ a topological space, what are the Borel sets of a closed subset $Y$? Let $Y$ be a closed subset of $X$.  Then $\mathcal B(X)$, the set of Borel sets of $X$, is the $\sigma$-algebra generated by all closed sets of $X$.  Let $E$ be the intersection of $\mathcal B(X)$ with the power set of $Y$.  Is it true that $E = \mathcal B(Y)$?
First, it's clear that $E$ is a $\sigma$-algebra of $Y$: both the power set of $Y$ and $\mathcal B(X)$ are closed under countable union and complements with respect to $Y$.  Also, a subset of $Y$ which is closed in $Y$, is also closed in $X$.  So $E$ contains the closed sets of $Y$, hence $E$ contains $\mathcal B(Y)$.  Not so sure about the other inclusion.
 A: This is a consequence of a more general claim. First some notation: If $X$ is a set and $A \subset \mathcal{P}(X)$, define $\Sigma(A)$ to be the $\sigma-$algebra generated by $A$. If $f:Y \rightarrow X$ is any function and $A \subset \mathcal{P}(X)$, define $f^{-1}(A) \subset \mathcal{P}(Y)$ by $\{f^{-1}(S)|S \in A\}$. Then the claim is:
$$\Sigma(f^{-1}(A)) = f^{-1}(\Sigma(A))$$
In your case, let $A$ denote the closed sets of $X$, and $i:Y \rightarrow X$ inclusion. Then the left side of the above equality is $\mathcal{B}(Y)$ and the right side is $\mathcal{B}(X) \cap \mathcal{P}(Y)$.
As you've noticed, the hard part of the claim is the implication that $f^{-1}(\Sigma(A)) \subseteq \Sigma(f^{-1}(A))$. To prove this, define:
$$\mathcal{S} = \{S \in \mathcal{P}(X) \mid f^{-1}(S) \in \Sigma(f^{-1}(A))\}$$
You can prove pretty easily that $\mathcal{S}$ is a $\sigma-$algebra, and $A \subseteq \mathcal{S}$. That means $\Sigma(A) \subseteq \mathcal{S}$, and so $f^{-1}(\Sigma(A)) \subseteq f^{-1}(\mathcal{S})$. By the definition of $\mathcal{S}$ it is clear that $f^{-1}(\mathcal{S}) \subseteq \Sigma(f^{-1}(A))$, and the claim follows.
