Mean continuity of gradient 
Let $f:\mathbb R^n\longrightarrow R$ be a differentiable function, and suppose $\nabla f$ is bounded. Prove that $$\lim_{r\to 0}\frac{1}{\omega_n r^n}\int_{B_r(x)}[\nabla f(y)-\nabla f(x)] dy=\underline{0}.$$
  Show also that $$\lim_{r\to 0}\frac{1}{\omega_n r^n}\int_{B_r(x)}|\nabla f(y)-\nabla f(x)| dy=0$$ is not necessarily true.

In one variable the first statement is straightforward. I tried the spherical-coordinates-change of variables, but id doesn't seem to work easy (recall that $f$ is not necessarily $C^1$).
Like always, simple suggestions of useful tools are welcome too.
Thank you in advance.
 A: For part 1:
Revised answer (the first one was not correct - it implicitly assumed $f$ is $C^1$).
Without loss of generality assume $\mathbf{x} = \mathbf{0}$.  Suppose that there is a vector $\mathbf{v}$ such that
$$\lim_{\mathbf{h} \rightarrow 0} \frac{f(\mathbf{h})-f(\mathbf{0})-\mathbf{v}\cdot\mathbf{h}}{|\mathbf{h}|} = 0.$$
Then $\mathbf{v}$ is the gradient of $f$ at 0, i.e. $\mathbf{v} = \nabla f(\mathbf{0}).$  So we would like to show that
$$\mathbf{v} = \lim_{r \rightarrow 0} \frac{1}{|B_r(0)|} \int_{B_r(0)} \nabla f(\mathbf{y}) \, d\mathbf{y}$$
satisfies that property.
Using the fact that $\nabla f$ is bounded, we can apply the Dominated Convergence Theorem to see that:
$$\lim_{\mathbf{h} \rightarrow 0} \int_{B_r(0)} \frac{f(\mathbf{y+\mathbf{h}})-f(\mathbf{y}) - \nabla f(\mathbf{y}) \cdot \mathbf{h}}{|\mathbf{h}|} \, d\mathbf{y} = \int_{B_r(0)} \lim_{\mathbf{h} \rightarrow 0} \frac{f(\mathbf{y}+\mathbf{h})-f(\mathbf{y}) - \nabla f(\mathbf{y}) \cdot \mathbf{h}}{|\mathbf{h}|} \, d\mathbf{y} = 0.$$
Therefore,
$$\lim_{\mathbf{h} \rightarrow 0} \frac{1}{|B_r(0)|} \int_{B_r(0)} \frac{f(\mathbf{y}+\mathbf{h})-f(\mathbf{y}) - \nabla f(\mathbf{y}) \cdot \mathbf{h}}{|\mathbf{h}|} \, d\mathbf{y} = 0.$$
So we have
$$\lim_{r \rightarrow 0} \lim_{\mathbf{h} \rightarrow 0} \frac{1}{|B_r(0)|} \int_{B_r(0)} \frac{f(\mathbf{y}+\mathbf{h})-f(\mathbf{y}) - \nabla f(\mathbf{y}) \cdot \mathbf{h}}{|\mathbf{h}|} \, d\mathbf{y}$$
$$= \lim_{\mathbf{h} \rightarrow 0} \lim_{r \rightarrow 0} \frac{1}{|B_r(0)|} \int_{B_r(0)} \frac{f(\mathbf{y}+\mathbf{h})-f(\mathbf{y}) - \nabla f(\mathbf{y}) \cdot \mathbf{h}}{|\mathbf{h}|} \, d\mathbf{y}$$
$$= \lim_{\mathbf{h} \rightarrow 0}  \frac{f(\mathbf{h})-f(\mathbf{0}) - \left (\lim_{r \rightarrow 0} \frac{1}{|B_r(0)|} \int_{B_r(0)}\nabla f(\mathbf{y}) \cdot \mathbf{h} \, d\mathbf{y}\right)}{|\mathbf{h}|} = 0,$$
where the change in order of limits is justified by the continuity of the integral in both $r$ and $\mathbf{h}$.
Therefore,
$$\nabla f(\mathbf{0}) = \lim_{r \rightarrow 0} \frac{1}{|B_r(0)|} \int_{B_r(0)}\nabla f(\mathbf{y})\, d\mathbf{y}.$$
For part 2:
The point of the second limit is that if we were to assume that $f$ is $C^1$, then that limit would always be zero.  So to find an $f$ for which the limit is not zero, we consider functions that are differentiable but not $C^1$.  For example, $f(x) = x^2 \sin(1/x)$.
