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 upper semicontinuous, or at least Borel?

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

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    $\begingroup$ I have seen done in articles with measurable $x \mapsto r_x$. The $r_x$ is a metric analogue of what some people call "stopping time" in probability. I doubt that you can do this continuously. $\endgroup$ May 19, 2019 at 14:54
  • $\begingroup$ @AdriánGonzález-Pérez thank you, interesting! I also doubt that $x \mapsto r_x$ can be chosen continuous in general, but it would be interesting to see an explicit counterexample. And if it can be chosen to be not too bad (upper semicontinuous or Borel), it would be interesting to see the proof. $\endgroup$
    – Skeeve
    May 19, 2019 at 15:01
  • $\begingroup$ I will start by taking a radius $x \mapsto r_x$ that implements the supremum up to $\epsilon > 0$ and then take the limit over a countable sequence $\epsilon_n \to 0$. $\endgroup$ May 19, 2019 at 15:04

1 Answer 1


The selection can be made lower or upper semicontinuous.

Consider any $x\in K$ and $r>0$ such that $$f_r(x):=\frac{1}{\mu(B(x,r))}\int\limits_{B(x,r)}f d\mu>a.$$ For any fixed $d\in(0,\tfrac{r}{2})$ and for any $y\in B(x,d)$ we have $B(x,r-2d)\subset B(y,r-d)\subset B(x,r)$. Therefore for any $x\in K$ and any $y\in K\cap B(x,d)$

$$f_{r-d}(y)=\frac{1}{\mu(B(y,r-d))}\int\limits_{B(y,r-d)} fd\mu\geq \frac{1}{\mu(B(x,r))}\int\limits_{B(x,r-2d)} fd\mu >a$$ if $d$ is small enough (say if $0<d\leq d_x<\tfrac{r}{2}$) just by the monotone convergence of the last integral. This means that for every $x\in K$ there exist $r_x$ and $0<d_x<\tfrac{r_x}{2}$ such that

$$f_{r_x-d_x}(y)>a\quad\text{ for every }y\in B(x,d_x)\cap K.$$ By compactness we can cover $K$ with a finite numer of balls $B_i=B(x_i,d_{i})$. Then any selection of the radii $r_{i}-d_{i}$ will work. In particular for every $y\in K$ we can take $$r(y):=\max_i\{r_{i}-d_{i}:y\in B_i\}=\max_i (r_i-d_i) \chi_{B_i}(y)$$ which gives a lower semicontinuous function (supremum of l.s.c. is l.s.c.). Choosing instead the minimum (and considering closed balls $\overline{B(x,d_x)}$ in the argument above) gives an upper semicontinuous function. Actually now that I think about it it's not necessary to extract a finite subfamily...in any case this choice gives even an essentially locally constant function.

I don't know about a continuous selection.

  • $\begingroup$ good point, but could you add some details on how you choose $r(y)$ when, say, $y\in B_1 \cap B_2$? As far as I understand, we would like to have $r(y) \in (r_1 - d_1, r_1) \cap (r_2 - d_2, r_2)$, however these intervals might be disjoint and I don't see why $r(y)=\max_{i}(r_i-d_i)$ should yield $f_{r(y)}(y)>a$. $\endgroup$
    – Skeeve
    May 22, 2019 at 19:27
  • $\begingroup$ It is just sufficient for $r(y) $ to lie in the union of those intervals - either one would work $\endgroup$
    – Del
    May 23, 2019 at 8:27

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