The classical Rademacher theorem says that any Lipschitz function on a doman in $\mathbb{R}^n$ has the first derivative almost everywhere.


I am wondering if this result can be generalized as follows. Let $X$ be an Alexandrov space with curvature bounded below of finite dimension $n$. It is well known that at almost every point $x$ (with respect to the $n$-Hausdorff measure) the tangent cone at $x$ is isometric to the Euclidean space $\mathbb{R}^{n}$. Let us call such points regular. 


Let $f\colon X\to \mathbb{R}$ be a Lipschitz function. We define its differential (if it exists) in the standard way as follows. Let $X\_N$ (resp. $\mathbb{R}\_N$) denote the space $X$ (resp. $\mathbb{R}$) with the metric multiplied by $N$. Let 
$$f\_N\colon X\_N\to \mathbb{R}\_N$$
denote the map $f$ between these rescaled spaces. $f\_N$ has the same Lipschitz constant as $f$. Fix a point $x\in X$. Then $(X\_N,x)$ converges in the Gromov-Hausdorff sense to the tangent cone $T\_xX$ at $x$, and $(\mathbb{R}\_N,f(x))$ converges to $(\mathbb{R},0)$. Then if there is a limit map $df\_x\colon T\_xX\to \mathbb{R}$ we call it the differential of $f$ (automatically it will have the same Lipschitz constant as $f$).


**Question.** Is it true that for almost every regular point $x\in X$ the differential at $x$ of a Lipschitz function $f$ exists and is a linear functional on $\mathbb{R}^n$?



Rademacher type theorem for Alexandrov spaces
I assume you are interested in the finite-dimensional case.


You can write this function in a distance chart (these charts cover almost all points). Apply the standard Rademacher theorem and notice that the distance chart is differentiable almost everywhere. Hence the result follows.


