Assume $\Omega\subset \mathbb{R}^n$ is a bounded open set with Lipschitz boundary; let $d H^{n-1}$ be the Hausdorff measure on $\partial \Omega$.
Since $\Omega$ has Lipschitz boundary, for $d H^{n-1}$-a.e. point $x\in\partial\Omega$, there exists an outward unit normal vector $\vec{\nu}(x)$. Fix such a point $x\in \partial \Omega$.
I want to prove that $$ \lim_{\varepsilon\to 0} \frac{dH^{n-1}(\Omega\cap\partial B(x,\varepsilon))}{dH^{n-1}(\partial B(x,\varepsilon))} = \frac{1}{2}. $$
I can prove this if $\Omega$ has $C^{1,\alpha}$ boundary without much trouble, but the argument I used there fails in the Lipschitz case. The idea is that on very small scales, the intersection $ \Omega\cap\partial B(x,\varepsilon) $ looks like the top half of a sphere.
If it matters, this is essentially the last step in a theorem of Gauss in potential theory which I cannot find details for - I have looked in every book I can find to no avail and can't get it myself. I attached two pictures of this claim in books, but their proofs lack a lot of detail in my opinion.
First by Medkova 
and then by Verchota