# Is sup of the solutions in $t$ to $\int_{B(x,t)}f\,dV=t^n$ Borel measurable?

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.

• 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? Oct 6, 2022 at 16:13

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 $$F_\sigma$$.