This may be a very basic question. Let $X$ be a complete metric space and let $T$ be a compact subset of $X$. Say that a function $\pi: X \to T$ is a projection if $$ d(x, \pi(x)) = d(x, T) \quad \forall x \in X\,. $$ If $T$ is a closed convex subset of $\mathbb{R}^d$ then the (unique) projection $\pi$ onto $T$ is always continuous; of course this will not happen in general. I am interested in the following weaker question:

Does there always exist a projection that is Borel measurable?

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    $\begingroup$ Did you try to use the measurable selection theorem of Kuratowski and Ryll-Nardzewski? $\endgroup$ – Jochen Wengenroth Apr 21 '17 at 8:55

Consider the closed multimap $\Pi:X\to 2^T$, whose graph is the closed set $\mathrm{graph}(\Pi):=\big\{(x,t)\in X\times T\, : d(x,t)=\min_{s\in T}d(x,s)\big\}\subset X\times T$. You want a measurable selection $\pi:X\to T$ of $\Pi$, that is $\mathrm{graph}(\pi)\subset\mathrm{graph}(\Pi)$. Since $K$ is a compact metric space, thus a Polish space, the classical Kuratowski–Ryll-Nardzewski measurable selection theorem does the job.

rmk. The Kuratowski–Ryll-Nardzewski measurable selection theorem admits an easy proof in the particular case of a closed, non-empty set valued multimap $\Pi:X\to2^T$ with $T$ a compact metric space. In this case (by the Alexandroff-Hausdorff theorem) $T$ is a continuous image $T=\kappa( C)$ of the Cantor set $C$ via some continuous surjective map $\kappa:C\to T$. The pre-image of the closed set $\mathrm{graph}(\Pi)$ via $\mathrm{id}_X\times\kappa:X\times C\to X\times T$ is therefore the graph of a closed multimap $X\to 2^C$, that is the multimap $x\mapsto \kappa^{-1}\Pi(x)\neq\emptyset;$ a lower semi-continuous, hence measurable selection of the latter, is $x\mapsto \min\kappa^{-1}\Pi(x) $, and a measurable selection of $\Pi$ is therefore


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    $\begingroup$ Ah! I was unaware of (or had forgotten) the term "selection," which made it much harder to search for useful results. In my application $X$ is separable, so this does the job. Thanks! $\endgroup$ – J.. Apr 21 '17 at 9:02
  • $\begingroup$ Actually we don't need any further hypotheses on the metric space $(X,d)$ because $T$ is already a Polish space, being a compact metric space. (edited) $\endgroup$ – Pietro Majer Apr 22 '17 at 15:23

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