I came by the following theorem on a book that did not provide any proof.
Theorem: Suppose that $\sum_\limits{n=k}^{\infty}f_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n\:\:,n\geqslant k$, are integrable on $[a,b]$. Then:
$\int_\limits{a}^{b}F(x)dx=\sum_\limits{n=k}^{\infty}\int_\limits{a}^{b}f_n(x)dx$
Since it is admitted that each $f_n$ is integrable I guess that each function could be integrated in the sum. However I fail to see a proof of the theorem.
Question:
How should I prove the claim?
Thanks in advance!
For $K \ge k$, define $F_K: [a,b] \to \mathbb R$ by $$F_K(x) = \sum^K_{n=k} f_n(x), \,\,\,\, x \in [a,b].$$ We're assuming $F_K\to F$ uniformly. Then for any $\epsilon > 0$, there is $K^* \in \mathbb N$ such that $K > K^*$ ensures that $$\lvert F(x) - F_K(x) \rvert < \epsilon / (b-a), \,\,\,\, \text{ for all } x \in [a,b].$$ Now for $K > K^*$, \begin{align*} \left \lvert \int^b_a F(x) dx - \int^b_a F_K(x) dx \right \rvert &= \left \lvert \int^b_a [F(x) - F_K(x)] dx \right \rvert \\ &\le \int^b_a \lvert F(x) - F_K(x) \rvert dx \\ & \le \int^b_a \frac{\epsilon}{b-a} dx = \epsilon. \end{align*} This shows that $$\lim_{K\to\infty} \int^b_a F_K(x) dx = \int^b_a F(x) dx.$$ But by linearity of the integral, $$\int^b_a F_K(x) dx = \sum^K_{n=k} \int^b_af_n(x) dx,$$ so $$\lim_{K\to\infty} \sum^K_{n=k}\int^b_a f_n(x) dx = \int^b_a F(x) dx.$$ Or in other words, $$\sum^\infty_{n=k} \int^b_a f_n(x) dx = \int^b_a F(x) dx.$$