Convergence on every closed subintervals of $(a,b)$ and uniformly bounded partial sums imply uniformly convergence on $[a,b]$?

Question

I want to prove the following statement:

Let $\sum\limits_{n=1}^{\infty}u_n(x)$ be a series converging pointwise on $[a,b]$, converging uniformly on every closed subinterval of $(a,b)$. $u_n(x)\in C(a,b)$ for all $n$. The partial sum of the series $S_n=\sum\limits_{k=1}^{n}u_k(x)$ is uniformly bounded. Then $$\int_a^b\left(\sum\limits_{n=1}^{\infty}u_n(x)\right)\mathrm d x=\sum\limits_{n=1}^{\infty}\int_a^bu_n(x)\mathrm d x.$$

I know it suffices to prove that $\sum\limits_{n=1}^{\infty}u_n(x)$ converges uniformly on $[a,b]$. But I have no idea how to deal with this.

Any suggestions will be greatly appreciated.

Answer 1

Since $S_n$ converges uniformly on every closed subinterval of $(a,b)$, the function $$ S(x)=\lim_{n\to\infty}S_n(x) $$ is continuous. Moreover, for any $a<\alpha<\beta<b$, we have $$ \lim_{n\to\infty}\int_\alpha^\beta S_n(x)dx=\int_\alpha^\beta S(x)dx, $$ by uniform convergence. Now we send $\alpha\to a$ and $\beta\to b$, and show that there is a limit. For this, it suffices to show that $$ \lim_{n\to\infty}\int_\alpha^{\alpha'} S_n(x)dx\to0 $$ as $a<\alpha<\alpha'\to a$. But this is obvious, since $S_n$ is uniformly bounded.

We have treated the integral of $S$ as an improper Riemann integral, but one can see that $S$ is Riemann integrable from first principles (or by Lebesgue's criterion). Apart from two small subintervals near the endpoints, $S$ is uniformly continuous. In these two small subintervals, the oscillation is bounded and the lengths are shrinking to 0.

Answer 2

I may missing somthing here but I think that you only need that the series converges pointwise and that the partial sums are uniformly bounded. Then you can apply dominated convergence theorem. That is if you are familiar with the concept of measure.

If not here is a rough idea that I think works. Note that $x\mapsto\sum_{n=1}^{\infty} u_n(x)$ is continuous (there is some work to do here). We have that

$$\int_{a+\frac{1}{m}}^{b-\frac{1}{m}}\left(\sum\limits_{n=1}^{\infty}u_n(x)\right)\mathrm d x=\sum\limits_{n=1}^{\infty}\int_{a+\frac{1}{m}}^{b-\frac{1}{m}}u_n(x)\mathrm d x$$ Convergence comes from the uniform convergence of the series in the compact subsets of $(a,b)$. Now, letting $m$ go to $\infty$ you have indeed that $$\int_{a}^{b}\left(\sum\limits_{n=1}^{\infty}u_n(x)\right)\mathrm d x=\sum\limits_{n=1}^{\infty}\int_{a}^{b}u_n(x)\mathrm d x$$