Prove that $\int_{a}^{b} \sum_{n=1}^{\infty} f_n(x) dx=\sum_{n=1}^{\infty} \int_{a}^{b} f_n(x) dx$.

Question

Let $\{f_n\}$ be a sequence of Riemann integrable on $[a,b]$. If $f(x)=\sum_{n=1}^{\infty} f_n(x), a\leq x\leq b$, the series converging uniformly on $[a,b]$, then$$\int_{a}^{b} \sum_{n=1}^{\infty} f_n(x) dx=\sum_{n=1}^{\infty} \int_{a}^{b} f_n(x) dx.$$

Answer 1

Let $F_N=\sum_{n=1}^{N} f_n(x)$ then $$\left|\int_{a}^{b} \sum_{n=1}^{\infty} f_n(x) dx-\sum_{n=1}^{N} \int_{a}^{b} f_n(x) dx\right|=\left|\int_{a}^{b} f(x) dx-\int_{a}^{b} F_N(x) dx\right|\\\leq \int_{a}^{b}\left| f(x)-F_N(x)\right| dx\leq (b-a)\sup_{x\in [a,b]}\left| f(x)-F_N(x)\right| $$ Now use the fact that the sequence $(F_N)_N$ converges uniformly to $f$ on $[a,b]$.