Is every Borel subset of a measurable set measurable? Let $A$ be a Lebesgue measurable subset of $\mathbb{R}$.  Let us consider the subspace topology on $A$, and let us consider the Borel sigma algebra under that topology.  My question is, is every Borel set under that topology a Lebesgue measurable subset of $\mathbb{R}$?
I think the answer is probably yes, but I’m not sure.
 A: Yes. Let $\mathcal{B}$ denote the borel sigma-algebra induced by $A$ and $\mathcal{L}$ the collection of Lebesgue-measurable sets. Let $\Sigma = \{B \in \mathcal{B} : B \in \mathcal{L}\}$. Clearly $\Sigma$ is a $\sigma$-algebra. Note that it contains all open subsets of $A$: if $B \subseteq A$ is open relative to $A$, then $B = A \cap U$ for some open $U \subseteq \mathbb{R}$, so in particular, $B \in \mathcal{L}$. It follows that $\Sigma$ contains $\mathcal{B}$ and is thus $\mathcal{B}$.
A: Let $A$ be a topological space and let $i:A\to\mathbb{R}$ be an embedding.
Since $i$ is an embedding then we get that all open $U$ of $A$ we have an open set $V$ of $\mathbb{R}$ such that $i^{-1}(V)=U$.
We know that the inverse image commutes with unions and intersections and as such the borel sets of  $A$ are the preimage of some borel set of the reals.
From this we get that the borel subsets $B$ of $A$ satisfy the relation $i(B)=C\cap i(A)$ for some Borel set $C$.
Assuming now that $i(A)$ is Lebesgue measurable we get that all Borel subsets of $A$ are Lebesgue measurable inside the reals.
