Let $\Omega\subset \mathbb{R}^n$ an open set and $u:\Omega\to \mathbb{R}$ be a (locally) $L^1$function. Then it is well known that the Lebesgue differentiation theorem holds: For almost every $x\in \Omega$, $$\frac{1}{B_r(x)} \int_{B_r(x)} (u(y)  u(x)) d y \to 0$$ as $r\to 0$. My question is if it is true that $$\frac{1}{B_r(x)} \int_{B_r(x)} u(y)  u(x) d y \to 0$$ as $r\to 0$ for almost every $x\in \Omega$. I am doubtful about this, but the universal counterexample in measure theory  the (characteristic function of the) fat Cantor set  doesn't work here. If it is actually true, then it could probably further be generalized to $$\frac{1}{B_r(x)} \int_{B_r(x)} u(y)  u(x)^p d y \to 0$$ if $u\in L^p(\Omega)$ and $p\ge 1$. Actually that is what I'm interested in.
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It's true that $$\frac{1}{B_r(x)}\int_{B_r(x)} u(y)u(x)^p\to 0$$ for $r\to 0$. For $p=1$ this is standard (Somehow I always overlooked the absolute value signs in the definition of Lebesgue points), and the same proof carries over for $p>1$ (at least the proof used in Ziemer, "Weakly differentiable functions", where variants of this theorem for Sobolev spaces are discussed as well). 

