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 $u$?

$\begingroup$ @PiotrHajlasz Yes. Thanks. $\endgroup$ – user124345 Jan 6 '19 at 17:58
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 $t\to 0^+$.