Let (X,d) be a polish space. Does the metric d have to be measurable (regarding the Borel sigma-algebra in the product space)?
2 Answers
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If you just say "complete metric" but not "separable", then the answer can be negative. Take $X$ of power greater than the continuum, and the discrete metric (all nontrivial distances are $1$) so that the algebra of Borel sets is the full power set. But the diagonal in $X \times X$ is not measurable for the product sigma-algebra. And thus the metric is not a Borel function.
answered Aug 15, 2011 at 13:49
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$\begingroup$ The diagonal is trivially Borel on account of its being closed. Am I missing something? $\endgroup$ Aug 15, 2011 at 14:13
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1$\begingroup$ The argument above ($d$ is [Lipschitz] continuous, and therefore Borel-measurable) works for arbitrary metric spaces, no assumption of completeness or separability is needed. $\endgroup$ Aug 15, 2011 at 14:21
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1$\begingroup$ @EJ: That is true if you endow the product with the sigma-algebra generated by the open sets in the product topology. But the product sigma-algebra is generated by the product of the Borel sigma-algebras. For non-separable metric spaces these sigma-algebras differ. Arguably, the question was about the first notion, but I'm not sure about that. $\endgroup$ Aug 15, 2011 at 14:56
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The metric function is continuous, therefore measurable.
One can show that measurable functions can be built with a hierarchy similar to the Borel hierarchy. You start with continuous functions (into $\mathbb R$), then take pointwise limits, and reiterate $\omega_1$ many steps (each time taking pointwise limits of previously defined stages).
This is done by taking functions that preserve $\Sigma^0_\alpha$ and $\Pi^0_\alpha$ sets (i.e. a preimage of a $\Sigma^0_\alpha(\mathbb R)$ is $\Sigma^0_\alpha(X)$, similarly for $\Pi$. Continuous functions are indeed the first level, as preimage of open/closed set is an open/closed set), and prove by induction that the pointwise limits behave as we would like.
answered Aug 15, 2011 at 12:05