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I have no idea how to find an answer to the following question:

Let $f:\mathbb{R}^{n} \rightarrow [0,\infty)$ be a smooth, non-negative function such that for any $x\in\mathbb{R}^{n}$, there exists $t_x\in (0,\infty)$ such that $$\int_{B(x,t_x)}fdV=t^{n}_x.$$ Define the function $r:\mathbb{R}^n \rightarrow (0,\infty]$ by $r(x) = \sup \left\{t\in(0,\infty)\,:\,\int_{B(x,t)}fdV=t^{n}\right\}$.

My question: Is $r$ Borel measurable?

Any hint would be appreciated.

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  • $\begingroup$ If I am not mistaken, there could be many candidates $t_1(x),\cdots,t_n(x),\cdots$ that satisfy the property for a single point $x$. What is the criteria to define the map $x\mapsto t(x)$ then? Or is the supremum taken over all these possible candidates? $\endgroup$ Oct 6, 2022 at 16:13

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In general, consider a closed set $C\subseteq \Bbb R^n\times\Bbb R$ and $r:\Bbb R^n\to[-\infty,\infty]$, $r(x)=\sup\{t\in\Bbb R\,:\, (x,t)\in C\}$. In your case, $C=g^{-1}(0)$, where $g$ is the continuous function $g(x,t)=t^n-\int_{B(x,t)} f(y)\,dy$, for some measurable $f$ that is bounded on bounded sets.

I claim that $r$ is Borel. In fact, notice that, for any $\alpha\in\Bbb R$ and $x\in\Bbb R^n$, $$r(x)>\alpha\Leftrightarrow \exists y>\alpha, (x,y)\in C\Leftrightarrow x\in \pi_{\Bbb R^n}\left[C\cap (\Bbb R^n\times (\alpha,\infty))\right]$$

Where $\pi_{\Bbb R^n}(s,t)=s$. Now, $C\cap (\Bbb R^n\times(\alpha,\infty))$ is a countable union of compact sets. Therefore its image by the continuous function $\pi_{\Bbb R^n}$ is $F_\sigma$.

The property of $C$ that makes this work is $\sigma$-compactness, i.e. being union of countably many compact sets. In $\Bbb R^m$, that is the same as being $F_\sigma$.

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  • $\begingroup$ Thank for a very clear answer. $\endgroup$
    – hana
    Oct 6, 2022 at 22:11

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