# Exterior cone condition for $\mathrm{supp}\, u$ and Lebesgue points of $u$

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 $$u$$?

• @PiotrHajlasz Yes. Thanks. – user124345 Jan 6 '19 at 17:58

Let $$f(x)=\begin{cases} \sin (1/x), & 0 Then $$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^+$$.