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

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?

• When $k=1$, it seems wrong! For example $\lim_{t\to 0}\int^{a+r}_{a-r}f(t)dt=0$.But $\lim_{t\to 0}\frac{\int^{a+r}_{a-r}f(t)dt}{2r}=f(a).$ – Riemann Oct 26 '18 at 5:40
• 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 $a$. – Hayk Oct 26 '18 at 5:43
• Thanks Hayk. You are right. I did. – M. Rahmat Oct 26 '18 at 5:47
• Mean value theorems for definite integrals works for your question! – Riemann Oct 26 '18 at 5:50

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
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 $$f(a)=1$$.
• I think it should be $\lim_{r\to 0}\frac{\int_{D(a,r)}f(t)dt}{Vol(D(a,r))}=f(a).$ – Riemann Oct 26 '18 at 5:44