6

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 $\mathbb{R}^n$.

Alp Uzman
  • 10,742
Amr
  • 20,030
  • 2
    Homeomorphisms are continuous, and in particular measurable. If $f\colon V \to B$ is a homeomorphism with $B$ measurable, then so is $f^{-1}(B) = V$ – Didier Jan 17 '22 at 19:47
  • @Didier I am not sure this argument is valid, am I missing something obvious ? – Amr Jan 17 '22 at 19:55
  • With respect to the Borel $\sigma$-algebras, there is nothing more to say. – Didier Jan 17 '22 at 19:55
  • If $f:X\rightarrow Y$ is continuous map between topological spaces, then it is a measurable map between $X,Y$ as measurable spaces when both are equipped with Borel $\sigma-algebras$. Fine, I agree about that. (I will add a continuation due to word limit) – Amr Jan 17 '22 at 19:57
  • @Didier This argument just tells you $V$ is measurable as a subset of itself with respect to its own Borel-algebra, which is trivial. It says nothing about measurability of $V$ as a subset of $\mathbb{R}^n$. – Thorgott Jan 17 '22 at 19:57
  • @Thorgott Exactly, Confusing intrinsic vs. Extrinsic – Amr Jan 17 '22 at 19:58
  • So $f:V\rightarrow B$ is a measurable map between measurable spaces $V,B$ when equipped with their Borel algebras, This gives that $B=f^{-1}(B)$ is measurable subset of the measurable space $B$, but we still dont know if its a measrable subset of $\mathbb{R}^n$ – Amr Jan 17 '22 at 20:00
  • @Didier In other words, why cant one use your argument to deduce invariance of domain trivially – Amr Jan 17 '22 at 20:01
  • 2
    Yes, I agree there's something missing, my bad – Didier Jan 17 '22 at 20:04

1 Answers1

5

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.)

Alp Uzman
  • 10,742