Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is integrable. Is it true that $$ \lim_{r\to 0}\frac{\displaystyle\int_{B_r(0)}f(y)~\mathrm dy}{r^{n1}}=0 \quad ? $$ 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.

5$\begingroup$ $x^{\alpha}$ with $1 \leq \alpha <n$ is an example when $n>1$. $\endgroup$– Giorgio MetafuneSep 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. $\endgroup$– TittiSep 5, 2022 at 11:16
1 Answer
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 $\mathcal H^{n1}$a.e, where $\mathcal H^{n1}$ denotes the $n1$ dimensional Hausdorff measure.
This is Theorem 2.10 in Measure Theory and Fine Properties of Functions by Evans and Gariepy (2015 version).