Continuous functions measurable in metric spaces?

Question

Let $(X,d)$ and $(Y,d')$ be metric spaces.

A function $f:X \to Y$ is continuous if $\forall x \in X \forall \epsilon>0 \exists \delta>0: \forall y \in Y d(x,y)<\delta \implies d'(f(x),f(y))<\epsilon$

A function $f:X \to Y$ is measurable if $\forall S \in \mathcal{Y}: f^{-1}(S) \in \mathcal{X}$, where $\mathcal{X}$ and $\mathcal{Y}$ are generated by the open sets.

If $f:X \to Y$ is continuous, is $f$ measurable?

What changes if we generate $\mathcal{X}$ and $\mathcal{Y}$ by the open balls instead of the open sets?

One might think that all continuous functions are measurable. However, this statement depends on the precise definition of continuous and measurable (see e.g. Example of a continuous function that is not measurable).

The fourth comment (Daniel Fischer's) on that question sums up the situation pretty well. — Ben Grossmann, Jun 02 '17 at 21:15
The "Borel $\sigma$-algebra" is the measure generated by open balls (actually it's usually defined as being generated by open rectangles, but the consequence is the same). — Ben Grossmann, Jun 02 '17 at 21:37
The two definitions are equivalent. The point of that question is that not every open set is an open ball. However, once you've decided that open balls are measurable, you can deduce that all open sets are measurable. — Ben Grossmann, Jun 02 '17 at 22:07
I have double-checked what you say, and provided an answer based on this. Feel free to provide your own answer of course. However, the two definitions are not always equivalent, and I have provided a source that shows this. — Peter, Jun 04 '17 at 11:11

score 1 · Accepted Answer · answered Jun 04 '17 at 11:08

Based on the help by @Omnomnomnom, I provide an answer for future reference:

Generated by open sets: For the $\epsilon-\delta$ style definition, an equivalent definition is that all preimages of open sets are open (see For continuous functions, preimage of open set is open.). This suffices to prove that $f$ is measurable.

Generated by open balls:

In the case where the spaces $(X,d)$ and $(Y,d')$ are separable, the definition in terms of open balls and open sets are equivalent (see If $X$ is a topological separable space then every open set is a union of a countable number of open balls?), so nothing changes.
If they are not separable, I do not know what happens. But the definitions are not equivalent in this case (see e.g. http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2007&task=show_msg&msg=3618.0001)

Continuous functions measurable in metric spaces?

1 Answers1