A sufficient condition for a Lebesgue point Let $f\in L^1(\Bbb R^n)$ and let $x\in \Bbb R^n$. $x$ is said to be a Lebesgue point of $f$ if $\lim_{r\to 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} |f(y)-f(x)|~dm(y)=0$ where $m$ is Lebesgue measure on $\Bbb R^n$. Clearly the condition $\lim_{r\to 0} \int_{B(x,r)}f(y)~dm(y)=f(x)$ is a weaker condition. Is there a counterexample of $f$ such that the latter holds but the former does not?

Take any hyperplane $\pi$ passing through $x$. Then define $f(y) = 0$ for $y \notin B(x, 1)$, otherwise put $f(y) = f(x)+1$ on one side of $\pi$ and $f(y) = f(x)-1$ on the other. It is easy to see that this $f$ works. 
