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?

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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).

Answer 1

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)