# Proving that one can integrate a uniformly convergent series of functions term by term

I aim to understand the following proof from Serge Lang's Introduction to Complex Analysis at a graduate level

[ and I have the following definitions

My question is: What does the last paragraph of the theorem in question actually mean? Does it mean that every $$f_i$$ is convergent, and if yes, how is this derived from the definition?

The last paragraph means that if for each $$n$$ we set $$s_n:=\sum_{k=1}^n f_k,$$ then the sequence $$(s_n)_{n=1}^\infty$$ converges uniformly on $$U$$ to $$s:=\sum_{k=1}^\infty f_k$$. Then, by the first part of the Theorem $$\lim_n\int_{\gamma} s_n = \int_\gamma s$$ which translates as $$\lim_n\int_{\gamma} \sum_{k=1}^n f_k = \int_\gamma \sum_{k=1}^\infty f_k$$ and since $$\lim_n\int_{\gamma} \sum_{k=1}^n f_k =\lim_n \sum_{k=1}^n \int_{\gamma}f_k = \sum_{k=1}^\infty \int_{\gamma} f_k$$ the desired result follows.
• thanks for the asnwer. I have two questions. 1) Does it mean that for each $n$ we have $s_n$ uniformaly convergent (and is this from the definition above or a derived result)? 2) When you interchange the integral and sumation aren't you already using the result of what you are trying to prove? Jul 31 '20 at 0:27
• You're welcome! (1) Each $s_n$ is a function so it doesn't make sense to say that a function is uniformly convergent. However, $(s_n)_n$ is a sequence of continuous functions that converges uniformly to $s$. (2) At the very last line when I exchanged the limit with the sum, notice that the sum is finite, so $\int \sum_1^n = \sum_1^n\int$ by linearity of the integral. The result proves that you can still perform this exchange for the infinite sum case provided that you have uniform convergence. Hope that this answers your questions. Jul 31 '20 at 0:43
• What do exactly do you mean for every term to be convergent? By definition a sequence $(g_n)_n$ of functions on $U$ is uniformly convergent if $\| g_n -g \|_\infty \to 0$ as $n \to \infty$. In particular this implies pointwise convergence, that is for any $z \in U$, $|g_n(z)-g(z)| \to 0$ as $n \to \infty$. In other words, if $g_n \to g$ uniformly, then $g_n(z) \to g(z)$ for each $z \in U$. Jul 31 '20 at 1:00
• Hi @Alonso Delfín, I didn't express my self correctly. What I meant is, does the uniform convergence of a sequence $(g_n)_n$ imply that each term is well defined or only the limit term? Jul 31 '20 at 10:36
"$$\sum_n f_n$$ is a series of continuous functions converging uniformly on $$\Omega$$" means each $$f_n$$ is a continuous function on $$\Omega$$, and the series $$\sum_n f_n$$ converges uniformly on $$\Omega$$.