Does the mean integral over B(x,r) of a L1 function u converge a.e. to u(x)?

Question

Suppose $u\in L^1(\Omega )$. Let $u_{x,r}$ be the mean of $u$ over the ball $B(x,r)$ (s.t. $B(x,r) \subset \Omega$), i.e. $ u_{x,r} := \frac{1}{|B(x,r)|} \int_{B(x,r)} u(y) dy$.

1. Is it true that $u_{x,r} \stackrel{r\to 0^+}{\longrightarrow} u(x)$ for almost all $x\in \Omega$?
2. If 1 is true then is it also true that $u_{x,r} \stackrel{r\to 0^+}{\longrightarrow} u(x)$ in $L^1 (\Omega ')$, where $\Omega ' \subset \subset \Omega$? (the problem here is essentially bounding $u_{x,r}$ by an integrable function independently of $r$)

Best wishes.

Answer 1

As I said in my comment, the answer to the first part of your question is "yes". This is known as the Lebesgue differentiation theorem, see e.g. http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem.

The answer to the second part of your question is also "yes". Seen in the right way, this becomes a question about approximation by convolution (with mollifiers).

For convenience, I will give the argument specialized to the current setting. By extending $f$ on $\Bbb{R}^n \setminus \Omega$ by zero, we can assume $\Omega = \Bbb{R}^n$. We then have

\begin{eqnarray*} \Vert f(x) - f_{x,r} \Vert_{L^1} & = & \int_{\mathbb{R}^{d}}\left|f\left(x\right)-\frac{1}{\lambda\left(B_{r}\right)}\int_{B_{r}\left(x\right)}f\left(y\right)\, dy\right|\, dx\\ & = & \int_{\mathbb{R}^{d}}\left|\frac{1}{\lambda\left(B_{r}\right)}\int_{B_{r}\left(x\right)}f\left(x\right)-f\left(y\right)\, dy\right|\, dx\\ & \leq & \int_{\mathbb{R}^{d}}\frac{1}{\lambda\left(B_{r}\right)}\int_{B_{r}\left(x\right)}\left|f\left(x\right)-f\left(y\right)\right|\, dy\, dx\\ & \overset{z=x-y}{=} & \int_{\mathbb{R}^{d}}\frac{1}{\lambda\left(B_{r}\right)}\int_{B_{r}\left(0\right)}\left|f\left(x\right)-f\left(x-z\right)\right|\, dz\, dx\\ & \overset{\text{Fubini/Tonelli}}{=} & \frac{1}{\lambda\left(B_{r}\right)}\int_{B_{r}\left(0\right)}\int_{\mathbb{R}^{d}}\left|f\left(x\right)-f\left(x-z\right)\right|\, dx\, dz\\ & = & \frac{1}{\lambda\left(B_{r}\right)}\int_{B_{r}\left(0\right)}\left\Vert f-T_{z}f\right\Vert _{L^{1}\left(\mathbb{R}^{d}\right)}\, dz\\ & \leq & \sup_{\left|z\right|<r}\left\Vert f-T_{z}f\right\Vert _{L^{1}\left(\mathbb{R}^{d}\right)}\\ & \xrightarrow[r\downarrow0]{} & 0, \end{eqnarray*} where the last step used that translation (of a single $L^1$ function) is continuous w.r.t. the $L^1$-norm, see e.g. proof that translation of a function converges to function in $L^1$, the argument given there for $n=1$ generalizes easily to dimensions $n>1$.