Let (X,d) be a polish space. Does the metric d have to be measurable (regarding the Borel sigmaalgebra in the product space)?
closed as too localized by Andres Caicedo, Todd Trimble♦, Emil Jeřábek, Andreas Blass, Yemon Choi Aug 15 '11 at 20:41
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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 sigmaalgebra. And thus the metric is not a Borel function. 


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. 

