Let $X := \mathbb R^d$, $\lambda^d$ be the $d$-dimensional Lebesgue measure on $X$, and $f:X \to \mathbb R$ convex. Then there is a Borel set $N \subset X$ such that $\lambda^d (N) = 0$ and $f$ is differentiable on $A := X \setminus N$. Clearly, $A$ is Borel measurable. It follows that the gradient $\nabla f: A \to X$ of $f$ is well-defined.

• If $f$ is differentiable on $X$, then $A = X$ and $f \in \mathcal C^1 (X)$.

• Clearly, the Borel $\sigma$-algebra $\mathcal B(A)$ is a subset of the Borel $\sigma$-algebra $\mathcal B(X)$, i.e., $\mathcal B(A) \subset \mathcal B(X)$.

My question: Is the map $\nabla f$ measurable?