# 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.

• yes, every measurable subset have a induced Lebesgue measure in the same sense that a subset have the induced topology. – Masacroso Nov 11 '18 at 0:56
• @Masacroso why are you talking about measures? – mathworker21 Nov 11 '18 at 1:07
• and, indeed, the Borel $\sigma$-algebra of some measurable subset of $\Bbb R$ are Lebesgue measurable in $\Bbb R$ – Masacroso Nov 11 '18 at 1:10

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 $$\mathcal{B}$$.
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.