Note that on Wikipedia, the Lebesgue version of Leibniz rule states that
Let $X$ be an open subset of $\mathbf{R}$, and $\Omega$ be a measure space. Suppose $f: X \times \Omega \rightarrow \mathbf{R}$ satisfies the following conditions
- $f(x, \omega)$ is a Lebesgue-integrable function of $\omega$ for each $x \in X$.
- For almost all $\omega \in \Omega$, the partial derivative $f_x$ exists for all $x \in X$.
- There is an integrable function $\theta: \Omega \rightarrow \mathbf{R}$ such that $\left|f_x(x, \omega)\right| \leq \theta(\omega)$ for all $x \in X$ and almost every $\omega \in \Omega$. Then, for all $x \in X$, $$ \frac{d}{d x} \int_{\Omega} f(x, \omega) d \omega=\int_{\Omega} f_x(x, \omega) d \omega . $$
Note that if we let $X = \Omega = [0,1]$, with $f(\omega,x) = \mathbb{1}\{\omega < x\}$, then we know that $f_x(\omega,x) \stackrel{\text{a.s}}{=}0$ and so $$ \int_{\Omega} f_x(\omega,x) d\omega = 0.$$ However,
$$ \frac{d}{dx} \int_{\Omega} f(\omega,x) d\omega = \frac{d}{dx}x = 1. $$
My guess is that I need to use the distributional derivative of the indicator function rather than the normal derivative, but I have not studied distributions or delta functions formally before, and don't see why the regular almost-everywhere derivative does not apply here.