Let $ \mu$ be a finite Borel measure on $ \mathbb R^N$ and let $ f\in L^1(\mu)$ be a non-negative function. Let $ M_\mu f$ denote the maximal function of $ f$ relative to $ \mu$ , i.e. $ (M_\mu f)(x)=0$ if $ \mu(B(x,r))=0$ for some $ r>0$ and $ (M_\mu f)(x) = \sup_{0<r<\infty} \frac{1}{\mu(B(x,r))} \int_{B(x,r)}f \, d\mu$ otherwise. (Here $ B(x,r)$ denotes the open ball of radius $ r$ centered at $ x$ .)

Suppose that $ a>0$ and $ K\subset \mathbb R^N$ is a compact such that $ M_\mu f > a$ on $ K$ . Then for each $ x\in K$ there exists $ r_x>0$ such that $ $ \frac{1}{\mu(B(x,r_x))} \int_{B(x,r_x)}f \, d\mu > a. \tag{1}$ $

**Question.** Is it possible to choose for each $ x\in K$ the radius $ r_x>0$ in such a way that (1) holds and the mapping $ x\mapsto r_x$ is continuous or at least Borel?

This question is inspired by another recent question about Besicovich type covering theorem.