Differentiation under the integral sign and uniform integrability Let $(X,\mu)$ be a measure space (if it's convenient we can assume $\mu$ is finite).  Let $[a,b]$ be an interval, and suppose we have a function $f : [a,b] \times X \to \mathbb{R}$ such that:


*

*For each $x \in X$, we have $f(\cdot, x) \in C^1([a,b])$;

*For each $t \in [a,b]$, we have $f(t, \cdot), \partial_t f(t, \cdot) \in L^1(\mu)$.
(If it helps I'm happy to also assume that $f$ and $\partial_t f$ are  jointly measurable.)
Set $F(t) = \int_X f(t,x)\, \mu(dx)$.  The classical "differentiation under the integral sign" theorem says that if we assume the hypothesis
$$\text{There exists $g \in L^1(\mu)$ such that $\sup_{t \in [a,b]} |\partial_t f(t,x)| \le g(x)$} \tag{DOM}$$
then $F$ is differentiable on $(a,b)$ and $F'(t) = \int_X \partial_t f(t,x) \,\mu(dx)$.  Indeed, it would follow that $F \in C^1([a,b])$.
Now, in many situations, one can weaken a "domination" hypothesis to uniform integrability.  (For instance, the dominated convergence theorem is extended by the Vitali convergence theorem.) So suppose we replace the hypothesis (DOM) with the following:
$$\text{$\{\partial_t f(t,\cdot) : t \in [a,b]\}$ is uniformly integrable with respect to $\mu$} \tag{UI}$$
Does the same conclusion hold?
I'd think this would be standard if it's true, but I've never seen it written down.  But I also can't think of a counterexample.
I would like to follow the proof of the classical result by proceeding as follows: Fix $t_0 \in (a,b)$ and an arbitrary sequence $t_n \to t_0$ with all $t_n \in (a,b)$.  We have $$F'(t_0) = \lim_{n \to \infty} \int_X \frac{f(t_n, x)-f(t_0, x)}{t_n - t_0}\,\mu(dx)$$
if the limit exists, and we would like to pass the limit under the integral sign.  This would be possible if the sequence of difference quotients $D_n(x) = \frac{f(t_n, x)-f(t_0, x)}{t_n - t_0}$ were uniformly integrable.  By the mean value theorem, we know that for each $x$ there is $t_n^*(x) \in (a,b)$ such that $D_n(x) = \partial_t f(t_n^*(x), x)$.  If we could choose $t_n$ independently of $x$, then $\{D_n\}$ would be dominated by a UI sequence and we would be done.  But of course that will not work in general.
Note: there are a couple of inequivalent definitions of uniformly integrable.  I would be happy to adopt the stronger one, in which we assume that $\{\partial_t f(t,\cdot) : t \in [a,b]\}$ is bounded in $L^1$ norm.
(In the specific case that I care about, $\mu$ is a probability measure and I can show $\sup_{t \in [a,b]} \|\partial_t f(t, \cdot)\|_{L^p(\mu)} < \infty$ for some $p>1$, which implies (UI) by the so-called "crystal ball condition".)
 A: 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.
A: See Chapter A16 of David Williams, Probability with Martingales for a version of this which has essentially the same proof. By the arguments in that appendix, differentiability of $f(t,x)$ in $t$ and measurability in $x$ imply both the joint measurability of $f$ and the joint measurability of $\partial_tf$.
