2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ yes, every measurable subset have a induced Lebesgue measure in the same sense that a subset have the induced topology. $\endgroup$ – Masacroso Nov 11 '18 at 0:56
  • $\begingroup$ @Masacroso why are you talking about measures? $\endgroup$ – mathworker21 Nov 11 '18 at 1:07
  • $\begingroup$ and, indeed, the Borel $\sigma$-algebra of some measurable subset of $\Bbb R$ are Lebesgue measurable in $\Bbb R$ $\endgroup$ – Masacroso Nov 11 '18 at 1:10
2
$\begingroup$

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 $\mathcal{B}$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.