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 20:41This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


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. 

