Here's an example that came up in practice. 

> **Theorem: ([Cauchy-Pompeiu][1])**
Let $U \subset \mathbb{C}$ be a bounded open set with piecewise- $C^1$ boundary $\partial U$ oriented positively (see appendix B ), and let $f: \bar{U} \rightarrow \mathbb{C}$ be continuous with bounded continuous partial derivatives in $U$. Then for $z \in U$ :
>$$
f(z)=\frac{1}{2 \pi i} \int_{\partial U} \frac{f(\zeta)}{\zeta-z} d \zeta+\frac{1}{2 \pi i} \int_U \frac{\frac{\partial f}{\partial \bar{\zeta}}(\zeta)}{\zeta-z} \, \text{Leb}_{\mathbb C}(d \zeta)
$$

In the linked source (a libretexts book written by Jiří Lebl), there is the following question:
> Why can we not differentiate in $\bar z$ under the integral in the second term of the Cauchy–Pompeiu formula? Notice that it would lead to an impossible result. 

[Namely, applying $\partial_{\bar z}$ to both sides, if it was possible to move $\partial_{\bar z}$ from the outside to the inside of the integral in the 2nd term, then we would get that 2nd term $=0$ and so we would conclude $\partial_{\bar z} f=0$, i.e. every $C^1$ function $f$ on $\bar U$ is holomorphic. Which is of course totally false.]

The [standard][2] differentiation under the integral result is 

> Let $X$ be an open subset of $\mathbb R$, and $\Omega$ be a measure
> space. Suppose ${\displaystyle g\colon X\times \Omega \to \mathbb R}$
> satisfies the following conditions:
> 
>   1. ${\displaystyle g(x,\omega )}$ is a Lebesgue-integrable function of ${\displaystyle \omega }$ for each ${\displaystyle x\in X}$.
> 
>   2. For almost all ${\displaystyle \omega \in \Omega }$, the partial derivative ${\displaystyle g_{x}}$ exists for all ${\displaystyle x\in X}$.
>   3. There is an integrable function ${\displaystyle \theta \colon \Omega \to \mathbb {R} }$ such that ${\displaystyle |g_{x}(x,\omega)|\leq \theta (\omega )}$ for all ${\displaystyle x\in X}$ and almost every ${\displaystyle \omega \in \Omega }$.
> 
> Then, for all ${\displaystyle x\in X}$,  $${\frac d{dx} \int _{\Omega }g(x,\omega )\,d\omega =\int _{\Omega}g_{x}(x,\omega )\,d\omega .}$$

The same proof (Dominated Convergence Theorem) shows the same thing if we replace $\mathbb R$ by $\mathbb C$ and $\partial_x$ by $\partial_{\bar z}$. 

So in our Cauchy-Pompeiu setting, we have both $X,\Omega:= U$; and $g(z,\zeta):= \frac{\partial_{\bar \zeta} f(\zeta)}{\zeta-z} : U \times U \to \mathbb C$. 

(1.) is satisfied, (2.) is *almost* satisfied: for *all* fixed $\zeta \in U$, the derivative $\partial_{\bar z} g$ exists (and equals $0$, and is in fact (3.) dominated by the integrable function $\theta \equiv 0$), for all $z\in U$ *except* at just one point $z=\zeta$! 

So the Leibniz rule fails to apply for our Cauchy-Pompeiu integral, because a "for all" condition fails at **just one point**! I find this to be quite subtle: there are "for all" and "for almost all" conditions in the theorem, and it is of critical importance to remember which variable requires the "for all" condition! 

(Just remember: the variable being differentiated needs "for all", and the variable being integrated just needs "for almost all". Which makes total sense.)

  [1]: https://math.libretexts.org/Bookshelves/Analysis/Tasty_Bits_of_Several_Complex_Variables_(Lebl)/04%3A_The_-Problem/4.01%3A_The_Generalized_Cauchy_Integral_Formula
  [2]: https://en.wikipedia.org/wiki/Leibniz_integral_rule