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.
Best Answer
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 $\mathcal{B}$.