# 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.