A very nice and helpful result is that for convex functions, the integral of a subgradient is always a subgradient of the integral, without any domination assumption needed. More precisely, let $T$ be a measurable space, $\mu$ be a (positive) measure on $T$ and $f:T\times\mathbb R^d\to\mathbb R^d$ be a mapping such that:
- $f(t,\cdot)$ is convex for all $t\in T$
- $f(\cdot,x)$ is measurable and integrable with respect to $\mu$, for all $x\in\mathbb R^d$.
For all $x\in\mathbb R^d$, let $F(x)=\int_T f(t,x)d\mu(t)$, which is a convex function.
Fix $x\in\mathbb R^d$ and let $g:T\to\mathbb R^d$ be a measurable function such that $g(t)$ is a subgradient of $f(t,\cdot)$ at $x$. Then, if $g$ is integrable, then $\int_T g(t)d\mu(t)$ is a subgradient of $F$ at $x$. The proof of this fact is straightforward, following the definition of subgradients and using linearity of integral. Moreover, applying a lemma of measurable selections (e.g., Theorem 5.1 of Himmelberg's paper "Measurable selections") yields the existence of measurable subgradients of $f$ at each $x\in\mathbb R^d$. Moreover, it is easy to see that such a measurable subgradient is necessarily integrable, due to integrability of $f(\cdot,y)$ for all $y\in\mathbb R^d$.
My question is the following: Can any subgradient of $F$ at $x\in\mathbb R^d$ be written as the integral of a such a measurable subgradient?
It is known that given finitely many convex functions $f_1,\ldots,f_p$ defined on $\mathbb R^d$ (and with only finite values), then for all $x\in\mathbb R^d$, $\partial (f_1+\ldots+f_p)(x)=\partial f_1(x)+\ldots+\partial f_p(x)$, where the latter is understood as a Minkowski sum of sets (this fact can be proven by induction on $p$ and for $p=2$, the left-right inclusion is really non-trivial). I am wondering if a similar property still holds in the more general setup where we consider an integral of convex functions instead of a finite sum.