Measurable invariance of domain

Question

Invariance of domain theorem tells us that if a subset $V$ of $\mathbb{R}^n$ is homeomorphic to an open subset of $\mathbb{R}^n$, then $V$ must be open itself.

Question: If a subset $V$ of $\mathbb{R}^n$ is homeomorphic to a Borel subset of $\mathbb{R}^n$, must $V$ be Borel ?

Recall $Borel(\mathbb{R}^n)$ is defined to be the $\sigma$-algebra generated by the topology of $\mathbb{R}^n$.

Answer 1

The answer to the question is yes. This is a consequence of the Lusin-Souslin Theorem (see e.g. Kechris' Classical Descriptive Set Theory, p. 89 or https://math.stackexchange.com/a/56061/169085):

• If $B\subseteq \mathbb{R}^n$ is Borel and $f:\mathbb{R}^n\to \mathbb{R}^n$ is continuous such that $f|_B$ is injective, then the image $f(B)\subseteq \mathbb{R}^n$ is Borel.

We also have the measurable analog of the Invariance of Domain as stated on Wikipedia:

• If $B\subseteq \mathbb{R}^n$ is Borel and $f:\mathbb{R}^n\to \mathbb{R}^n$ is Borel such that $f|_B$ is injective, then the image $f(B)\subseteq \mathbb{R}^n$ is Borel and $f|_B: B\to f(B)$ is a Borel isomorphism.

(In general one can replace $\mathbb{R}^n$ with a standard Borel space.)