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Akira
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Let $f$ be convex and $A$ a Borel subset of $\mathbb R^d$ on which $f$ is differentiable. Is the gradient $\nabla f: A \to \mathbb R^d$ measurable?

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?

Akira
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