Showing a limit for the mean value property I need to understand a proof and i have a problem with the limit process at a specific point in this proof. I want to know, what is needed to get the following result:
Let $f \in C^1(G)$, $\bar{x} \in G \subset \mathbb{R}^3$, then it holds:
$\lim_{a\rightarrow 0} \frac{1}{4\pi a^2} \int_{\partial B_a(\bar{x})} f(x) = f(\bar{x})$.
What arguments are needed for the rigorous proof ? I think, it has to do with continuity of $f$ and i can transform it to polar coordinates...but i have no idea what exactly happens here. Any suggestions? 
Thank you very much,
Maria
 A: Some hints.
Hint 1: Since $f$ is continuous, for any $\epsilon\gt0$, there is a $\delta\gt0$ so that
$$
|x-\bar{x}|\le\delta\implies|f(x)-f(\bar{x})|\le\epsilon
$$
Hint 2: $x\in\partial B_a(\bar{x})\implies|x-\bar{x}|=a$.
Hint 3:
$$
\left|\int_{\partial B_a(\bar{x})}f(x)\,\mathrm{d}x-\int_{\partial B_a(\bar{x})}f(\bar{x})\,\mathrm{d}x\right|\le\int_{\partial B_a(\bar{x})}|f(x)-f(\bar{x})|\,\mathrm{d}x
$$
Hint 4:
$$
\int_{\partial B_a(\bar{x})}1\,\mathrm{d}x=4\pi a^2
$$
A: I won't provide you with a finished proof, but I'll provide you with an outline of such a proof.
First, look at the geometric situation. The left-hand side integrates $f$ over the surface of a ball with radius $x$ around $\bar{x}$, and scales the result by the area of that surface.
Now, assume for  for a second that $f$ is constant within $B_a(\bar{x})$. The theorem is then very easy to prove - you won't be needed to take the limit at all. Can you write down the formal proof for this case?
For the non-constant case, the continuity of $f$ around $\bar{x}$ enters the picture. Instead of being constant, we'll now want $f$ to be nearly constant within $B_a(\bar{x})$, i.e. we'll want that $\left|f(x) - f(\bar{x})\right| < \epsilon$ for all $x\in B_a(\bar{x})$. If you compare that requirement to the definition of continuity, you'll find that no matter how small you pick epsilon, you always find an $a > 0$ which satisfies that requirement.
Finally, rewrite $f(x)$ as the sum of a constant part $f(\bar{x})$ and an "error" term $e(x)$, i.e. find $e(x)$ such that $f(x) = f(\bar{x}) + e(x)$, and put that into your integral. Integrating the constant part is easy - you dealt with that case already. For the non-constant part, you'll now have to argue that it goes to zero as $a$ goes to zero. For that, note that, per the previous paragraph, you can make $|e(x)|$ arbitrarily small by picking a sufficiently small $a > 0$...
