Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is integrable. Is it true that $$ \lim_{r\to 0}\frac{\displaystyle\int_{B_r(0)}f(y)~\mathrm dy}{r^{n-1}}=0 \quad ? $$ This is obvious if $0$ is a Lebesgue point of $f$ or if $n=1$, but I would like to know if it's true in general.

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    $\begingroup$ $|x|^{-\alpha}$ with $1 \leq \alpha <n$ is an example when $n>1$. $\endgroup$ Commented Sep 5, 2022 at 11:10
  • $\begingroup$ Thanks, it was easier than I thought. You should post your comment as an answer so I can accept it. $\endgroup$
    – No-one
    Commented Sep 5, 2022 at 11:16

1 Answer 1


Metafune has given an example of the limit failing to be $0$ at a particular point - namely for $n > 1$, the function $|x|^{-\alpha}$, with $1 \leq \alpha < n$ has that limit equal to $\infty$ at $0$.

However, you can still get some kind of affirmative result.

In general the limit in question is zero $\mathcal H^{n-1}$-a.e, where $\mathcal H^{n-1}$ denotes the $n-1$ dimensional Hausdorff measure.

This is Theorem 2.10 in Measure Theory and Fine Properties of Functions by Evans and Gariepy (2015 version).


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