Let $u:\mathbb{R}^n \to \mathbb{R}$ be an $L^1$ function with compact support. Let $\bar x \in \partial \mathrm{supp}\, u$ and assume that $\mathrm{supp} \, u$ satisfies the exterior cone condition at $\bar x$. Does this imply that $\bar x$ is a Lebesgue point for $u$?



Exterior cone condition for $\mathrm{supp}\, u$ and Lebesgue points of $u$
**The answer is no**. There are many counterexamples, but the following one is particularly interesting:


Let 
$$
f(x)=\begin{cases}
\sin (1/x), & 0<x<1,\\
0 & \text{otherwise.}
\end{cases}
$$
Then $f\in L^1(\mathbb{R})$ has compact support equal $[0,1]$ and $0$ is not a Lebesgue point of $f$. However,
$$
\frac{1}{t}\int\_0^t f(t)\, dt \to 0
\quad
\text{as $t\to 0^+$.}
$$
This is because of high oscillations of $f$ which cause a lot of cancellation. When you look at the definition of the Lebesgue point, then you have to take the absolute value and there is no cancellation phenomena and the integrals 
$$
\frac{1}{t}\int\_0^t |f(t)|\, dt =
\frac{1}{t}\int\_0^t |f(t)-f(0)|\, dt
$$
do not converge to zero as $t\to 0^+$.


