Question about differentiation under integral sign

Question

In Folland's book Real Analysis, he proves the following theorem (Theorem 2.27) concerning differentiation under integral sign.

Suppose $f:X \times [a,b] \to C$ is a complex valued function such that $f(\cdot,t)$ is integrable for each $t \in [a,b]$. Let $F(t)=\int_X f(x,t) d\mu(x)$. Then, if $\partial t f$ is dominated by an integrable function, we can differentiate $F(t)$ under the integral sign.

He then goes on to saying that if the parameter $t$ varies over a non-compact interval, we can still apply this theorem to any compact subinterval. But, my question is why is the theorem stated for compact interval $[a,b]$ in the first place? I do not see where/how the compactness of the interval is used in the proof

Answer 1

Since differentiability is only a local property, you can replace $[a,b]$ by any open set, or $\mathbb{R}$. However, the key-point is that $\partial_t f$ must be dominated by an integrable function independent of $t$. In this formulation the interval $[a,b]$ indicates that this property only need to be satisfied locally.

If you replace $[a,b]$ by $\mathbb{R}$, this restriction is too strong: It is enough that for any open neighborhoud of any point $t$ there exists a uniformly integrable majorant.

Answer 2

I had the same question about that theorem. After some thought, and after reading the accepted answer, here is my interpretation: Loosely speaking, $F$ is differentiable at some $t_0 \in \mathbb{R}$ so long as we can find an interval $I$ containing $t_0$ and an integrable function $g: I \to \mathbb{C}$ which satisfies $|\partial f/\partial t(x,t)| \leq g(x)$ for all $(x,t) \in X \times I$. To be precise:

Theorem. Let $(X,\mathcal{M},\mu)$ be measure space, let $A$ be an arbitrary subset of $\mathbb{R}$, and let $f: X \times A \to \mathbb{C}$ be a function. Suppose that $f(\cdot,t): X \to \mathbb{C}$ is integrable for every $t \in A$. Define $F: A \to \mathbb{C}$ by $F(t) := \int_X f(x,t) \,d\mu(x).$ Let $t_0 \in A$ be fixed, and suppose there exists an interval $I \subset A$ containing $t_0$ such that:

1. $\frac{\partial f}{\partial t}(x,t)$ exists for all $(x,t) \in X \times I$, and
2. there is an integrable function $g:I \to \mathbb{C}$ such that $\left|\frac{\partial f}{\partial t}(x,t) \right| \leq |g(x)|$ for all $(x,t) \in X \times I$.

Then $F$ is differentiable at $t_0$ and \begin{align*} F'(t_0) = \int_X \frac{\partial f}{\partial t}(x,t_0) \,d\mu(x). \end{align*}

We then immediately get the following corollary.

Corollary. Let $(X,\mathcal{M},\mu)$ be measure space, let $A$ be an arbitrary subset of $\mathbb{R}$, and let $f: X \times A \to \mathbb{C}$ be a function. Suppose that $f(\cdot,t): X \to \mathbb{C}$ is integrable for every $t \in A$. Define $F: A \to \mathbb{C}$ by $F(t) := \int_X f(x,t) \,d\mu(x).$ Suppose also that:

1. $\frac{\partial f}{\partial t}(x,t)$ exists for all $(x,t) \in X \times A$, and
2. there is an integrable function $g:A \to \mathbb{C}$ such that $\left|\frac{\partial f}{\partial t}(x,t) \right| \leq |g(x)|$ for all $(x,t) \in X \times A$.

Then $F$ is differentiable and $$ F'(t) = \int_X \frac{\partial f}{\partial t}(x,t) \,d\mu(x) \quad \text{for all } t \in A. $$

Taking $A = [a,b] \; (-\infty < a < b < \infty)$ in the corollary gives the exact statement of Theorem 2.27 in Folland.