# Proof that integral of uniformly convergent series converges to sum of the integrals

I'm confused about the proof of a simple corollary.

We're given that if $f_n \to f$ uniformly then $$\lim_{n \to \infty} \int_a^b f_n(x)\,dx = \int_a^b f(x)\,dx.$$

As a corollary, my textbook considers the case that if

• the sums $\sum_{k=1}^n g_k(x)$ converge uniformly
• the $g_k$ are Riemann integral on $[a,b]$

then $$\int_a^b \left(\sum_{k=1}^\infty g_k(x)\right)\,dx = \sum_{k=1}^\infty \left(\int_a^b g_k(x)\,dx\right).$$

The proof defines $f_n(x) \triangleq \sum_{k=1}^n g_k(x)$ and states that the conclusion follows from the given since $$\int_a^b f_n(x)\,dx \to \int_a^b f(x)\,dx.$$

This proof seems incomplete to me, however. How do we know that $$\int_a^b f(x)\,dx = \sum_{k=1}^\infty \left(\int_a^b g_k(x)\,dx\right) \text{ ?}$$

• For finite sums you have $\int \sum_k = \sum_k \int$. So, if $s_n$ is the finite sum and $s$ the sum, you have $\int s_n \to \int s$. Jun 2, 2018 at 22:47
• Can you explain exactly what part of the proof you think is incomplete? Jun 2, 2018 at 22:49
• $S_n(x)=\sum_{k=1}^n g_k(x)$. Take $S_n(x)$ as a uniformly convergent sequence. Jun 2, 2018 at 22:49
• I'll give you a hint: write the result of the corollary exactly the other way around. Jun 2, 2018 at 22:54

\begin{align*} \int_a^b \left(\sum_{k=1}^\infty g_k(x)\right)dx &= \int_a^b f(x)dx \\ &= \lim_{n \to \infty} \int_a^b f_n(x)dx\\ &= \lim_{n \to \infty} \int_a^b \left(\sum_{k=1}^n g_k(x)\right)dx\\ &= \lim_{n \to \infty} \sum_{k=1}^n \left(\int_a^b g_k(x)dx\right)\\ &= \sum_{k=1}^\infty \left(\int_a^b g_k(x)dx\right) \end{align*}
For each $n$, we have $$\int_a^b f_n(x)dx=\int_a^b \sum_{k=1}^n g_k(x)dx=\sum_{k=1}^n\int_a^b g_k(x)dx.$$ Now we take the limit as $n\to\infty$. The limit of the left side of this equation is $\int_a^b f(x)dx$ and the limit of the right side is $\sum_{k=1}^\infty\int_a^b g_k(x)dx$, so those two must be equal.