I was reading notes where it is used that given a Lipschitz map $u:\mathbb{R}^d\to\mathbb{R}$ and a measure $\mu$ on $\mathbb{R}^d$, then the condition
- $\mathcal{H}^{d-1}(A)<+\infty\implies \mu(A)=0$ (where $\mathcal{H}^{d-1}$ is the $d-1$-dimensional Hausdorff measure)
implies that the set on which u is not differentiable has $\mu$-measure $0$
I was wondering if this was true, if the result has a name, and if someone has an easy to read reference for the statement and the proof.
For the moment, the only result I knew, is Rademacher's Theorem which states that the set on which $u$ is not differentiable has Lebesgue measure $0$.
No, this is not true. This condition is trivially satisfied for $\mu=\mathcal{H}^{d-1}$, but the set on which a Lipschitz function is not differentiable may have positive $(d-1)$-dimensional Hausdorff measure. Take for example $u(x)=|x_1|$, which is not differentiable on the hyperplane $\{x\in\mathbb R^d: x_1=0\}$, whose $(d-1)$-dimensional Hausdorff measure is $+\infty$.