Differentiation under the integral sign and uniform integrability

Question

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".)

Answer 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 $C^1$.