most recent 30 from http://mathoverflow.net 2013-05-22T00:36:49Z http://mathoverflow.net/feeds/question/72922 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72922/are-metrics-borel-measurable-functions BBB 2011-08-15T12:03:29Z 2011-08-15T13:49:25Z

<p>Let (X,d) be a polish space. Does the metric d have to be measurable (regarding the Borel sigma-algebra in the product space)?</p>

http://mathoverflow.net/questions/72922/are-metrics-borel-measurable-functions/72923#72923 Asaf Karagila 2011-08-15T12:05:35Z 2011-08-15T12:14:01Z

<p>The metric function is continuous, therefore measurable.</p> <p>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).</p> <p>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.</p>

http://mathoverflow.net/questions/72922/are-metrics-borel-measurable-functions/72927#72927 Gerald Edgar 2011-08-15T13:49:25Z 2011-08-15T13:49:25Z

<p>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.</p>