I'm reading Section 2.1 of Chapter 2 in Villani's textbook Topics in Optimal Transportation.
- Now, let us assume that $\mu$ does not give mass to small sets, and let $\varphi$ be as above. Since $\varphi$ lies in $L^{1}(d \mu)$, it is $d \mu$-almost everywhere finite: $\mu[\operatorname{Dom}(\varphi)]=1$. On the other hand, the border $\partial \operatorname{Dom}(\varphi)$ of the convex set $\operatorname{Dom}(\varphi)$ is a small set; so, $\mu[\operatorname{Int}(\operatorname{Dom}(\varphi))]=1$. Now, on $\operatorname{Int}(\operatorname{Dom}(\varphi))$, the set of nondifferentiability of $\varphi$ is a small set. On the whole, $d \mu$-almost every point of $X$ is a differentiability point for $\varphi$. So, for $d \mu$-almost all $x$, the subdifferential of $\varphi$ at the point $x$ is $\{\nabla \varphi(x)\}$. Recalling that a statement true for $d \mu$-almost all $x$ is also true for $d \pi$-almost all $(x, y)$, we obtain that $y=\nabla \varphi(x)$ for $d \pi$-almost all $(x, y)$.
I'm trying to fill in the detail by proving below theorem, i.e.,
Let $X =\mathbb R^d$ and $\mu$ be a Borel probability measures on $X$. A subset $B$ of $X$ is a small set if its Hausdorff dimension $\dim_H (B)$ is at most $d-1$. Assume that $\mu$ does not give mass to small sets, i.e., $\mu (B) =0$ for every Borel set $B$ which is also a small set. Let $f:X \to \mathbb R \cup \{+\infty\}$ be proper convex such that $f \in L_1 (\mu)$.
Theorem: The map $f$ is differentiable $\mu$-a.e. If $N$ be a $\mu$-null subset of $X$ such that $f$ is differentiable on $N^c := X \setminus N$, then the gradient $\nabla f :N^c \to X$ of $f$ is measurable.
I'm not sure if my below attempt is fine, or it contains some logical mistakes. Could you please have a check on it?
My attempt:
Let $D_1 := \operatorname{dom} f := \{x \in X \mid f(x) < +\infty\}$ and $D_2 := \operatorname{int} D_1$. Because $f$ is proper convex, $D_1 \neq \emptyset$ is convex. Because $f \in L_1 (\mu)$, $D_1$ is Borel measurable and $\mu(D_1) = 1$. Because $D_1$ is convex, its boundary $\partial D_1$ is a small set, i.e., $\dim_H (\partial D_1) \le d-1$. Then $\mu(\partial D_1)=0$. Let $$ E := \{x \in D_2 \mid f \text{ is not differentiable at }x\}. $$
Then $E$ is Borel measurable and is a small set, i.e., $\dim_H (E) \le d-1$. So $\mu (E) = 0$. Let $F := D_2 \setminus E$. Then $$ F = \{x \in D_2 \mid f \text{ is differentiable at }x\} $$ is Borel measurable and $\mu (F) =1$. Clearly, $f$ is differentiable on $F$ and thus differentiable $\mu$-a.e.
Let $N$ be a $\mu$-null subset of $X$ such that $f$ is differentiable on $N^c := X \setminus N$. Then the gradient $\nabla f : N^c \to X$ of $f$ is well-defined. Let $$ \frac{\partial f}{ \partial x_i} : N^c \to \mathbb R \quad i = 1, \ldots,d $$ be the partial derivatives of $f$ on $N^c$. We have $\nabla f$ is Borel measurable if and only if $\frac{\partial f}{ \partial x_i}$ is measurable for all $i = 1, \ldots,d$. Let $\{e_1, \ldots, e_d\}$ be an orthonormal basis of $X$. Notice that $$ \frac{\partial f}{ \partial x_i} (x) = \lim_{n \to \infty} \frac{f(x+e_i/n)-f(x)}{1/n} = \lim_{n \to \infty} n \left [ f \left (x + \frac{e_i}{n} \right) - f (x) \right ]. $$
Because $f$ is continuous, the map $$ g_{n,i}: N^c \to \mathbb R, x \mapsto n \left [ f \left (x + \frac{e_i}{n} \right) - f (x) \right ] $$ is measurable for each $n \in \mathbb N^*$ and each $i \in \{1, \ldots, d\}$. It follows that $$ \frac{\partial f}{ \partial x_i} = \lim_n g_{n, i} $$ is measurable. This completes the proof.
