Before asking, here I bring Theorem 10.7, Rudin's RCA.
(Theorem begins.)
Let $X$ be a measure space with a complex measure $\mu$.
Let $\varphi$ be a complex measurable function on $X$.
Let $\Omega$ be an open set in the complex plane.
Here, $\Omega$ does not meet $\varphi(X)$.
Define $f:\Omega\to\mathbb{C}$ by $$f(z)=\int_{X} \frac{d\mu (\zeta)}{\varphi(\zeta)-z}.$$
Then $f$ is representable by power series, that is, for each open disc $D(a,r)$ in $\Omega$, there is a sequence $\{c_n\}_{n=0}^{\infty}$ of complex numbers such that the series $\sum_{n=0}^{\infty} c_n z^n $ converges to $f(z)$ for all $z$ in $D(a,r)$.
(Theorem ends.)
The proof says the geometric series $$\sum_{n=0}^\infty \frac{(z-a)^n}{(\varphi(\zeta)-a)^{n+1}} = \frac{1}{\varphi(\zeta)-z}$$ converges uniformly on $X$, which I understood. But I don't know why I can exchange the summation and integral symbols in the following. $$\int_X \sum_n \frac{(z-a)^n}{(\varphi(\zeta)-a)^{n+1}} d\mu(\zeta) = \sum_n \int_X \frac{(z-a)^n}{(\varphi(\zeta)-a)^{n+1}} d\mu(\zeta) $$ Why is this possible? Can I apply Lebesgue's dominated convergence theorem?
Thank you.
For fixed $z \in \Omega$, the series converges uniformly. If $S_n$'s are the partial sums then $S_n$ tends to the sum $S$ uniformly and $|\int S_n d\mu -Sd\mu| \leq \int |S_n-S| d|\mu|<\epsilon |\mu| (X)$ for $n$ sufficiently large.