Differentiation under the integral sign and uniform integrability

Solution 1:

Thinking a little more, I believe this is true.

Let's adopt the strong sense of uniform integrability, and further assume that $\partial_t f$ is jointly measurable.

Set $G(t) = \int_X \partial_t f(t,x) \mu(dx)$. The assumption (UI) ensures that $G$ is continuous on $[a,b]$. Fixing $t_0$ in $[a,b]$ and a sequence $t_n \to t_0$ within $[a,b]$, we have assumed $\partial_t f(t_n,x) \to \partial_t f(t_0, x)$ for each $x$, so by the Vitali convergence theorem we have $G(t_n) \to G(t_0)$.

Now if we adopt the strong sense of uniform integrability, then (UI) implies $$M := \sup_{t \in [a,b]} \int |\partial_t f(t,x)|\,\mu(dx) < \infty.$$ As such we have, for any $t \in [a,b]$, $$\int_a^t \int_X |\partial_t f(s,x)|\,\mu(dx)\,ds \le M|t-a| < \infty$$ So Fubini's theorem and the first fundamental theorem of calculus gives $$\begin{align*} \int_a^t G(s)\,ds &= \int_a^t \int_X \partial_t f(s,x)\,\mu(dx)\,ds \\ &= \int_X \int_a^t \partial_t f(s,x)\,ds\,\mu(dx) \\ &= \int_X (f(t,x) - f(a,x)) \,\mu(dx) \\ &= F(t) - F(a). \end{align*} $$ So $F(t) = F(a) + \int_a^t G(s)\,ds$, for any $t \in [a,b]$. The second fundamental theorem of calculus says that $F$ is differentiable on $(a,b)$ and $F' = G$. Since $G$ is continuous, $F$ is $C^1$.

I never did find a proof in the literature, so I wrote it out as Lemma 6.1 of Eldredge, Nathaniel, Strong hypercontractivity and strong logarithmic Sobolev inequalities for log-subharmonic functions on stratified Lie groups, Nonlinear Anal., Theory Methods Appl., Ser. A, Theory Methods 168, 1-26 (2018). ZBL1382.35084, arXiv:1706.07517.