Show $\lim_{\varepsilon\to 0}\frac{1}{|\partial B_\varepsilon|}\int_{\partial B_\varepsilon} f(y)dy=f(x)$. Let $B_\varepsilon=B_\varepsilon(x)$ the ball of $\mathbb R^n$ centered in $x$ and of radius $\varepsilon>0$. Let $f\in C^1_c(\mathbb R^n)$. How can I show that $$\lim_{\varepsilon\to 0}\frac{1}{|\partial B_\varepsilon|}\int_{\partial B_\varepsilon}f(y)dy=f(x)\ \ ?$$
My try
$$\left|\frac{1}{|\partial B_\varepsilon|}\int_{\partial B_\varepsilon}f(y)dy-f(x)\right|\leq \int_{\partial B_\varepsilon}\frac{|f(y)-f(x)|}{|\partial B_\varepsilon|}dy,$$
and it looks to be something as a sort of derivative of $f$, and thus I can see the result, but as it's written, it's not very clear. I tried as : by mean value theorem there ix $c_{\varepsilon}\in \partial B_\varepsilon$ s.t. $$|f(x)-f(y)|\leq\|\nabla f(c_\varepsilon)\|\|x-y\|$$
and thus $$\int_{\partial B_\varepsilon}\frac{|f(x)-f(y)|}{|\partial B_\varepsilon|}\leq\frac{\varepsilon}{|\partial B_\varepsilon|}\|\nabla f(c_\varepsilon)\|=\frac{\varepsilon}{\varepsilon^{n-1}|\partial B_1|}\|\nabla f(c_\varepsilon)\|,$$
but it doesn't go to $0$.
 A: I suppose WLOG that $x=0$.
$$\frac{1}{|\partial B_\varepsilon|}\int_{\partial B_\varepsilon}f(y)dy=\frac{1}{|\partial B_\varepsilon|}\int_{\partial B_1}f(\varepsilon\sigma )\varepsilon^{n-1}d\sigma =\frac{1}{|\partial B_1|}\int_{\partial B_1}f(\varepsilon\sigma )d\sigma .$$
Using continuity and dominated convergence theorem allow you to conclude.

Notice that you don't need the fact that $f$ is compacted supported, neither that $f$ is $\mathcal C^1$.
A: You were really close, from your second line
$$
\frac{1}{|\partial B_\epsilon|}\int_{\partial B_\epsilon}|f(y)-f(x)|dy \leq \frac{1}{|\partial B_\epsilon|}\sup_{B_\epsilon}|f(y)-f(x)||\partial B_\epsilon|=\sup_{B_\epsilon}|f(y)-f(x)|,
$$
All I'm really doing here is saying the function is less than it's supremum on all $B$, which is constant, so when you integrate you just multiply by the size of the domain.
As $\epsilon \to 0$ we have $\sup_{B_\epsilon}|f(y)-f(x)|\to\sup_{x}|f(y)-f(x)|=|f(x)-f(x)|=0$.
A: Nothing is required except definition of continuity of $f$ at $x$. Given $\delta >0$ choose $\epsilon >0$ such that $|f(x)-f(y)|<\delta$ if $y \in B_\epsilon (x)$. You get the desired conclusion immediately. 
