Let $f$ be a continuous function with compact support in $\mathbb{R}^n$. Show that \begin{equation} \lim_{r\to 0} \frac{1}{|B_r(x)|} \int_{B_r(x)} f(y)\,dy = f(x), \end{equation} where $B_r(x)$ is the ball of radius $r$ centered at $x$.
In 1d, it looks like the Fundamental Theorem of Calculus. However, I'm not sure how to prove it in higher dimensions. How can I use the "compact support"?
Thank you.
Let's make a sketch. Given na $\epsilon > 0$ there is a $\delta >0$ such that $$|\frac{1}{|B_r(X)|}\int_{B_r(x)}f(y)dy - f(x)| < \epsilon$$ if $|x-y| < \delta$. Then $$|\frac{1}{|B_r(X)|}\int_{B_r(x)}f(y)dy - f(x)| \le \frac{1}{|B_r(X)|}\int_{|B_r(X)|}|f(x)-f(y)|dy.$$ Since we have compact support and $f$ is continuous the $f$ is uniformly continuous. Then given $\epsilon >0$ there is $\delta >0$ such that $|f(x) - f(y)| < \epsilon$ if $|x-y| < \delta$. Then $$\frac{1}{|B_r(X)|}\int_{B_r(X)}|f(x)-f(y)|dy < \epsilon.$$