I want to prove the following:
Theorem: Let $(f_n)$ be a sequence of monotone (integrable) functions $f_n : [0,\infty) \rightarrow \mathbb{R}$ (for $t_i \leq t_j$ we have $f_n(t_i) \leq f_n(t_j)$ such that the sequence is monotone increasing ($f_n \leq f_{n+1}$) and $f_n \rightarrow f$. Then $$\lim_{n\rightarrow \infty}\int_0^\infty f_n(x)dx = \int_0^\infty f(x)dx.$$
This is for a proof of the monotone convergence theorem, we can't use it and neither can we use the dominated convergence theorem. The author claims that the proof is easy and all that we need to do is show that the upper and lower Riemann sums converge. I am not quite understanding what he means with this but let me write a sketch of my work.
sketch: We start with the right-hand side so that $$\lim_{n\rightarrow \infty}\int_0^\infty f_n(x)dx = \lim_{n\rightarrow \infty}\lim_{c \rightarrow \infty}\int_0^cf_n(x)dx,$$and now we apply the Riemann sums: $$\lim_{n\rightarrow \infty}\lim_{c \rightarrow \infty}\int_0^cf_n(x)dx = \lim_{n\rightarrow \infty}\lim_{c \rightarrow \infty}\sum_{\substack{i=1,\ t_m = c}}^mf_n(t_i)(t_i-t_{i-1}).$$Now we are essentially dealing with two limits and if we manage to exchange them, then the result will follow because we have convergence of $f_n$ to $f$. The exchange of limits is the part that I am not sure, according with this post, we can exchange the limits because if the function f is integrable then the right side of what we want to prove exists. But what about if it diverges? We shouldn't be able to switch the limits, right? I don't think we can use Dini's theorem because our functions are defined in a non-compact interval (?).
Maybe I am overcomplicating this but I am looking for a solid reason as to why the convergence of $f_j$ to $f$ justifies the statement.
Thank you very much!
