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

Question

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

Answer 1

$$ \begin{align*} \int_a^b \left(\sum_{k=1}^\infty g_k(x)\right)dx &= \int_a^b f(x)dx \\ &= \lim_{n \to \infty} \int_a^b f_n(x)dx\\ &= \lim_{n \to \infty} \int_a^b \left(\sum_{k=1}^n g_k(x)\right)dx\\ &= \lim_{n \to \infty} \sum_{k=1}^n \left(\int_a^b g_k(x)dx\right)\\ &= \sum_{k=1}^\infty \left(\int_a^b g_k(x)dx\right) \end{align*} $$

--- rk

Answer 2

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.