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?
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$\begingroup$ When $k=1$, it seems wrong! For example $\lim_{t\to 0}\int^{a+r}_{a-r}f(t)dt=0$.But $\lim_{t\to 0}\frac{\int^{a+r}_{a-r}f(t)dt}{2r}=f(a).$ $\endgroup$ – Riemann Oct 26 '18 at 5:40
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$\begingroup$ You need to normalize the integral, i.e. divide by the volume of the ball you integrate over. Everything is local, you work in a neighbourhood of $a$. $\endgroup$ – Hayk Oct 26 '18 at 5:43
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$\begingroup$ Thanks Hayk. You are right. I did. $\endgroup$ – M. Rahmat Oct 26 '18 at 5:47
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$\begingroup$ Mean value theorems for definite integrals works for your question! $\endgroup$ – Riemann Oct 26 '18 at 5:50
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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
answered Oct 26 '18 at 5:48
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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 $f(a)=1$.
