Prove that $f(x) = \lim_{\epsilon\rightarrow 0}\frac{1}{Vol(B_\epsilon(x))}\int_{B_\epsilon(x)}f(y)dV$ Let $f:\mathbb{R}^3\rightarrow\mathbb{R}$ be a continuous function. Prove that
$f(x) = \lim_{\epsilon\rightarrow 0}\frac{1}{Vol(B_\epsilon(x))}\int_{B_\epsilon(x)}f(y)dV$
where $B_\epsilon(x)$ denotes the ball of radius $\epsilon$ centered at $x$, and $Vol(B_\epsilon(x))$ denotes its volume.
My attempt: Intuitively, this makes sense, as when you decrease the radius of the ball to $0$, you wind up at the point $x$ anyways. However, I'm having difficulties proving this rigorously. Would taking the derivative help at all?
 A: Let $\epsilon'>0$ and choose $\delta$ such that if $y \in B(x,\delta)$ then
$f(y) \in B(f(x),\epsilon')$.
Suppose $\epsilon \le \epsilon'$, then $\int_{B(x,\epsilon)} \|f(x)-f(y)\| dy \le \epsilon m B(x,\epsilon)$. It follows that $\lim_{\epsilon \to 0} {1 \over m B(x,\epsilon)} \int_{B(x,\epsilon)} \|f(x)-f(y)\| dy  = 0$.
Since 
\begin{eqnarray}
\| {1 \over m B(x,\epsilon)}\int_{B(x,\epsilon)} (f(x)-f(y)) dy \| &=&\| f(x)- {1 \over m B(x,\epsilon)} \int_{B(x,\epsilon)} f(y) dy \|  \\
&\le& {1 \over m B(x,\epsilon)} \int_{B(x,\epsilon)} \|f(x)-f(y)\| dy
\end{eqnarray}
 we have the desired result.
Note that this holds in much greater generality, see the Lebesgue differentiation theorem.
A: First of all, even though $f$ may take values in $\Bbb R^3$ and this wouldn't change anything significant in the proof, I am almost sure that $f$ is supposed in fact to take values in $\Bbb R$ - it's probably a typo.
Next, consider the change of variable $y = x + \epsilon u$ with $u \in B_1(0)$. This gives us
$$\int \limits _{B_\epsilon(x)} f(y) \ \Bbb dy = \int \limits _{B_1(0)} f(x + \epsilon u) \epsilon^n \ \Bbb du$$
and
$$\textrm {Vol} (B_\epsilon(x)) = \int \limits _{B_\epsilon(x)} 1 \ \Bbb dy = \int \limits _{B_1(0)} 1 \cdot  \epsilon^n \ \Bbb du = \epsilon^n \ \textrm {Vol} (B_1(0)) ,$$
which taken together give
$$\lim_{\epsilon \to 0} \frac 1 {\textrm {Vol} (B_\epsilon(x))} \int \limits _{B_\epsilon(x)} f(y) \ \Bbb dy = \lim_{\epsilon \to 0} \frac 1 {\textrm {Vol} (B_1(0))} \int \limits _{B_1(0)} f(x + \epsilon u) \ \Bbb du = \frac 1 {\textrm {Vol} (B_1(0))} \int \limits _{B_1(0)} f(x) \ \Bbb du = \\
\frac 1 {\textrm {Vol} (B_1(0))} \textrm {Vol} (B_1(0)) \ f(x) = f(x) .$$
In the above I have used the fact that $f$ is continuous, therefore $\lim_{\epsilon \to 0} f(x + \epsilon u) = f(x)$ and a variation of Lebesgue's dominated convergence theorem (the functions $x \mapsto f(x + \epsilon u)$ are dominated by $\sup _{y \in B_r(x)} |f(y)|$ for all $0 < \epsilon <r$, and this supremum is finite because $f$ is continuous on $\Bbb R^3$ and $B_r(x)$ is relatively compact.)
A: Consider $f\colon\Bbb R^3\to\Bbb R$ (a scalar function)
Fix $x\in\Bbb R^3$ and $\varepsilon>0$. It follows by the intermediate value property that there exists $x_{\varepsilon}\in B_{\varepsilon}(x)$ s.t. $$\frac{1}{Vol(B_\varepsilon(x))}\int_{B_\varepsilon(x)}f(y)dV=f(x_{\varepsilon}).$$ Try to show this. Now tending with $\varepsilon$ to $0$ causes $x_{\varepsilon}\to x$, which finishes our proof.
If, as you claim, $f\colon\Bbb R^3\to\Bbb R^3$ (the vector-valued function), integrate the components separately.
A: Hint: Since $f$ is continuous, so are the component functions $f_i;\ 1\le i\le 3.\ $ Then, $m^i_{\epsilon }(vol \overline B_{\epsilon (x)})\le \int_{B_\epsilon(x)}f_i(y)dV\le M^i_{\epsilon }(vol \overline B_{\epsilon (x)}),\ $where $M^i_{\epsilon }$ and $m^i_{\epsilon }$ are the maximum, minimum, respectively, of $f_i$ on the compact set $\overline B_{\epsilon (x)}$.
