The limit of an integral over a ball when the radius of the ball goes to zero

Question

Let $D(a,r)$ be an open ball in $\mathbb{R}^{k}$ ($ k\geq1 $). We know that if $f$ is a continuous function at $a$, then $$\lim_{r\to 0}\frac{1}{V_{r}}\int_{D(a,r)}f(t)dt=f(a),$$ where $V_{r}$ is the measure of the ball. Does this hold if $f$ is only locally integrable? How to prove it?

Answer 1

If $f$ is locally integrable then the limit is $f(a)$ for almost all points $a$. This is immediate from Lebesgue's Theorem. Ref: https://en.wikipedia.org/wiki/Lebesgue_point

Answer 2

No. Take $k=1$ and $f=1_{\mathbb Q}$, the char. function of $ \mathbb Q$. For $a=0$ we have $\int_{D(a,r)}f(t)dt=0$, but $f(a)=1$.