Leibniz Integral Rule (Differentiating under Integral) + Proof

“Differentiating under the Integral” is a useful trick, and here we describe and prove a sufficient condition where we can use the trick. This is the Measure-Theoretic version, which is more general than the usual version stated in calculus books.

Let $X$ be an open subset of $\mathbb{R}$ , and $\Omega$ be a measure space. Suppose $f:X\times\Omega\to\mathbb{R}$ satisfies the following conditions:
1) $f(x,\omega)$ is a Lebesgue-integrable function of $\omega$ for each $x\in X$ .
2) For almost all $w\in\Omega$ , the derivative $\frac{\partial f}{\partial x}(x,\omega)$ exists for all $x\in X$ .
3) There is an integrable function $\Theta: \Omega\to\mathbb{R}$ such that $\displaystyle \left|\frac{\partial f}{\partial x}(x,\omega)\right|\leq\Theta(\omega)$ for all $x\in X$ .

Then for all $x\in X$ , $\displaystyle \frac{d}{dx}\int_\Omega f(x,\omega)\,d\omega=\int_\Omega\frac{\partial}{\partial x} f(x,\omega)\,d\omega.$

Proof:
By definition, $\displaystyle \frac{\partial f}{\partial x}(x,\omega)=\lim_{h\to 0}\frac{f(x+h,\omega)-f(x,\omega)}{h}.$

Let $h_n$ be a sequence tending to 0, and define $\displaystyle \phi_n(x,\omega)=\frac{f(x+h_n,\omega)-f(x,\omega)}{h_n}.$

It follows that $\displaystyle \frac{\partial f}{\partial x}(x,\omega)=\lim_{n\to\infty}\phi_n(x,\omega)$ is measurable.

Using the Mean Value Theorem, we have $\displaystyle |\phi_n(x,\omega)|\leq\sup_{x\in X}|\frac{\partial f}{\partial x}(x,\omega)|\leq\Theta(w)$ for each $x\in X$ .

Thus for each $x\in X$ , by the Dominated Convergence Theorem, we have $\displaystyle \lim_{n\to\infty}\int_\Omega \phi_n(x,\omega)\,d\omega=\int_\Omega\lim_{n\to\infty}\phi_n(x,\omega)\,dw$ which implies $\displaystyle \lim_{h_n\to 0}\frac{\int_\Omega f(x+h_n,\omega)\,d\omega-\int_\Omega f(x,\omega)\,d\omega}{h_n}=\int_\Omega \frac{\partial f}{\partial x}(x,\omega)\,d\omega.$