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Suppose that $X$ is a topological space equipped with the Borel $\sigma$-algebra $\mathcal{B}_X$.

Let $Y$ be a Borel subset of $X$, $Y\in \mathcal{B}_X$. In particular, $Y$ is a topological space with respect to the induced topology and therefore it has it's own Borel $\sigma$-algebra $\mathcal{B}_Y$.

My question: Let $A\in\mathcal{B}_X$, is it necessarily true that $A\cap Y\in \mathcal{B}_Y$?

Some observations: Clearly, $A\cap Y\in \mathcal{B}_X$. Also if $A$ is closed/ open, then $A\cap Y$ is closed/open and the claim is trivial.

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Let$$\mathcal{B}=\left\{A\in\mathcal{B}_X\,\middle|\,A\cap Y\in\mathcal{B}_Y\right\}.$$What you want to know is whether or not the statement $\mathcal{B}_X=\mathcal{B}$ holds. But, as you know, every open subset of $X$ belongs to $\mathcal B$. And $\mathcal B$ is a $\sigma$-algebra. But $\mathcal{B}_X$ is the smallest $\sigma$-algebra which contains every open subset of $X$. Therefore, $\mathcal{B}\supset\mathcal{B}_X$ and so $\mathcal{B}=\mathcal{B}_X$.

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Let $\tau_X$ denote the topology on $X$ and let $i:Y\to X$ denote the inclusion.

Then $\tau_Y:=i^{-1}(\tau_X)=\{i^{-1}(U)\mid U\in\tau_X\}=\{U\cap Y\mid U\in\tau_X\}$ is the subspace topology on $Y$.

Then: $$\mathcal B_Y=\sigma(\tau_Y)=\sigma(i^{-1}(\tau_X))=i^{-1}(\sigma(\tau_X))=i^{-1}(\mathcal B_X)\tag1$$

So this tells us that: $$\mathcal B_Y=\{A\cap Y\mid A\in\mathcal B_X\}$$

The third equality in $(1)$ is a consequence of the general rule:$$f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)=\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)$$

You can find a proof of that in this answer.

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There is a theorem in Theory of Measure and Integration, J Yeh, page 8:

Let $\mathfrak{C}$ be an arbitrary collection of subsets of a set $X$ and let $A\subseteq X$. Then $\sigma_{A}(\mathfrak{C}\cap A)=\sigma(\mathfrak{C})\cap A$.

Where $\sigma(\mathfrak{C})$ means the sigma-algebra generated by $\mathfrak{C}$ in $X$, and $\sigma_{A}(\mathcal{C}\cap A)$ is the sigma-algebra generated by $\mathfrak{C}\cap A=\{C\cap A: C\in\mathfrak{C}\}$ in $A$.

Now $Y\in\mathcal{B}_{X}=\sigma(\mathfrak{O})$, where $\mathfrak{O}$ is the collection of all open sets in $X$. Now $\mathfrak{O}\cap Y$ is the collection of all open sets in $Y$ so $\sigma_{Y}(\mathfrak{O}\cap Y)=\mathcal{B}_{Y}$. By $\sigma_{Y}(\mathfrak{O}\cap Y)=\sigma(\mathfrak{O})\cap Y$ and $A\in\sigma(\mathfrak{O})$ then $A\cap Y\in\sigma(\mathfrak{O})\cap Y$, so $A\cap Y\in\sigma_{Y}(\mathfrak{O}\cap Y)=\mathcal{B}_{Y}$.

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