# Invariance of domain theorem tells us that if a subset V of <mrow class="MJX-TeXAtom

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 Rn is homeomorphic to a Borel subset of ${\mathbb{R}}^{n}$, must $V$ be Borel ?
Recall Borel(${\mathbb{R}}^{n}$) is defined to be the σ-algebra generated by the topology of ${\mathbb{R}}^{n}$.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

The answer to the question is yes.
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 $f\left(B\right)\subseteq {\mathbb{R}}^{n}$ is Borel.

We also have the measurable analog of the Invariance of Domain:

2. 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 $f\left(B\right)\subseteq {\mathbb{R}}^{n}$ is Borel and $f{|}_{B}:B\to f\left(B\right)$ is a Borel isomorphism.

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