Proof that uniform convergence allows the sum and integral signs to be exchanged

Question

On wikipedia, the proof that all holomorphic functions on some open subset of the complex plane are analytic on said subset contains the following step: $$\frac{1}{2\pi i}\int_C\frac{1}{w-a}\cdot\sum\limits_{n=0}^{\infty}\left(\frac{z-a}{w-a}\right)^nf(w)\,\mathrm{d}w=\sum\limits_{n=0}^{\infty}\frac{1}{2\pi i}\int_C\frac{(z-a)^n}{(w-a)^{n+1}}f(w)\,\mathrm{d}w.$$ Basically, the sum and integral signs are interchanged, and it is justified in the proof by the fact that the sum is uniformly convergent on $C$. I understand this, but I want to see a proof that it implies that they can be interchanged, specifically in this case of a contour integral and a uniformly convergent sum. Do I need to know measure theory to understand how to prove this? All the results concerning the interchange of the sum and integral signs I have seen so far involve measure theory, which I am not at all familiar with.

Answer 1

With the full force of uniform convergence, no, there is no technical difficulty, because if $f_n \to f$ uniformly then given $\varepsilon > 0$ there exists $N$ such that for all $n>N$:

$$\left | \int f_n(x) dx - \int f(x) dx \right | \leq \int |f_n(x)-f(x)| dx \leq \int \varepsilon dx = \varepsilon L$$

where here $L$ is just the length of the path.

Part of the power of measure theory is not needing such a strong assumption as uniform convergence to justify interchanging limit and integration.