Rademacher's theorem If $U$ is an open subset of $\mathbb{R}^n$ and $f: U \rightarrow \mathbb{R}^m$ is Lipschitz continuous, then $f$ is Fréchet differentiable almost everywhere in $U$; that is, the points in $U$ at which $f$ is not differentiable form a set of Lebesgue measure zero.
Let $F := \{x \in U : f \text{ Fréchet differentiable at }x \}$. Then $F$ is a Borel subset of $\mathbb R^n$. This can be proved by extending $f$ to the whole $\mathbb R^n$, and then applying Section 3.5 of the book Fréchet Differentiability of Lipschitz Functions and Porous Sets in Banach Spaces by Lindenstrauss/Preiss/Tišer, i.e., "In this short section we show that for an arbitrary map between Banach spaces the set of its points of Fréchet differentiability is Borel; in fact it has type $F_{\sigma \delta}$.
Now I would like to extend Rademacher's theorem to locally Lipschitz function, i.e.,
Theorem Let $U$ be an open subset of $\mathbb{R}^n$ and $f: U \to \mathbb{R}^m$ is locally Lipschitz. Let $F := \{x \in U : f \text{ Fréchet differentiable at }x \}$. Then $F$ is a Borel subset of $\mathbb R^n$ and $U \setminus F$ has Lebesgue measure zero.
Could you confirm if my below attempt is fine?
Proof For $x \in U$, there is $r_x, L_x>0$ such that $B(x, r_x) \subset U$ and $f$ is $L_x$-Lipschitz on $B(x, r_x)$. There is a countable collection $(x_n) \subset U$ such that $B(x_n, r_{x_n})$ covers $U$. For simplicity, let $L_n := L_{x_n}, r_n := r_{x_n}$, and $U_n := B(x_n, r_n)$. Let $f_n$ be the restriction of $f$ to $U_n$. Then $f_n$ is $L_n$-Lipschitz.
Let $F_n := \{x \in U_n : f \text{ Fréchet differentiable at }x \}$. By Rademacher's theorem, $F_n$ is a Borel subset of $\mathbb R^n$ and $\mathcal L^n (U_n \setminus F_n)=0$ where $\mathcal L^n$ is the Lebesgue measure on $\mathbb R^n$. The claim then follows by noticing that $$ (U \setminus F) \subset \bigcup_n (U_n \setminus F_n). $$