Does the subdifferential of a convex function admit a measurable selection?

Question

For a convex function $\varphi: \mathbb{R}^n \to \mathbb{R},$ the subdifferential is defined as $$\partial \varphi (x) = \{z \in \mathbb{R}^n \mid \forall y \in \mathbb{R}^n, \varphi(y) \geq \varphi(x) + \langle z, y - x\rangle\}.$$ My question is under what conditions there exists a measurable function $f : \mathbb{R}^n \to \mathbb{R}^n$ such that $f(x) \in \partial \varphi(x)$ for all $x \in \mathbb{R}^n.$ I am aware of Kuratowski and Ryll-Nardzewski measurable selection theorem, but I am not sure how to apply this to the subdifferential correspondence. Any reference will be helpful!

Thanks!

Answer 1

Let $D$ be the set on which $\varphi$ is Fréchet differentiable. By this thread, $D$ is a Borel set. On the other hand, convex functions on $\mathbb R^n$ is differentiable a.e., so $D^c := \mathbb R^n \setminus D$ is a Lebesgue null set. By this thread, the restriction $\nabla \varphi:D \to \mathbb R^n$ is Borel measurable. We define $f:\mathbb R^n \to \mathbb R^n$ as follows:

• If $x \in D$ then $f(x) := \nabla \varphi (x)$.
• If $x \notin D$ then we pick any $y \in \partial f(x)$ and let $f(x) := y$.

Because the Lebesgue $\sigma$-algebra is complete, a subset of a Lebesgue null set is Lebesgue measurable. This implies arbitrary modification of a Lebesgue measurable function on a Lebesgue null set remains Lebesgue measurable. Hence $f$ is Lebesgue measurable.