Let $f_m:\mathbb N\to \mathbb R^+\cup\{+\infty \}$ an increasing sequence. Set $$f(n):=\lim_{m\to \infty }f_m(n)$$

(it can possibly be $\infty $).

I want to prove that $$\lim_{m\to \infty }\sum_{n\in\mathbb N} f_m(n)=\sum_{n\in\mathbb N}f(n).$$ First, $$\sum_{n\in\mathbb N}f_m(n)\leq \sum_{n\in\mathbb N}f(n)$$ is clear, and thus $$\lim_{m\to \infty }\sum_{n\in\mathbb N}f_m(n)\leq \sum_{n\in\mathbb N}f(n).$$

I have problem for the reverse inequality. I tried to use the fact that $f(n)=\sup_{m\in\mathbb N}f_m(n).$ So if $\varepsilon>0$, there is $M_n$ s.t. $$f(n)-\varepsilon\leq f_{m}(n)\leq f(n),$$ for all $m\geq M_n$. Now since $\sum_{n\in\mathbb N}\varepsilon$ doesn't converge, I can't conclude.


On $(\mathbb N,\mathbb P(\mathbb N),\mu)$ where $\mu(A)=\# A$, I want to show that $$\int_{\mathbb N}f(n)d\mu(n)=\sum_{n\in\mathbb N}f(n).$$

If $f$ is simple it work. Now let $f$ measurable. There is a sequence of simple function s.t. $f_m(n)\nearrow f(n)$. Then, using Monotone convergence, $$\int_{\mathbb N} f(n)d\mu(n)=\lim_{m\to \infty }\int_{\mathbb N}f_m(n)d\mu(n)=\lim_{m\to \infty }\sum_{n\in\mathbb N}f_m(n).$$ So now, I have to prove that $$\lim_{m\to \infty }\sum_{n\in\mathbb N}f_m(n)=\sum_{n\in\mathbb N}f(n),$$ to conclude.

  • 2
    $\begingroup$ It's a quick result from Lebesgue's monotone convergence theorem. $\endgroup$ – Jakobian Jul 22 '18 at 14:24
  • $\begingroup$ @Adam: I add more context. So as you see, I can't use it like this. $\endgroup$ – user380364 Jul 22 '18 at 14:31
  • $\begingroup$ If you want an inequality from below you can truncate the sum: $\sum_{n<M}f_m(n)\leq\sum_{n\in\mathbb{N}}f_m(n)$. Take limit $\lim_{m\to\infty}$ and get $\sum_{n<M}f(n)\leq \lim_{m\to\infty}\sum_{n\in\mathbb{N}}f_m(n)\leq\sum_{n\in\mathbb{N}}f(n)$. Finally, take limit $M\to\infty$. $\endgroup$ – user574889 Jul 22 '18 at 14:32
  • $\begingroup$ Use the fact that Lebesgue's integral is countably additive, you will have $\int_{\mathbb{N}}f(n)d\mu(n)=\sum_{i\in\mathbb{N}} \int_{\{i\}} f(n)d\mu(n) = \sum_{i\in\mathbb{N}} f(i) $ $\endgroup$ – Jakobian Jul 22 '18 at 14:34
  • $\begingroup$ @DavidC.Ullrich: Yes, sorry, I wanted to write $\mathbb N\to \mathbb R\cup\{\pm\infty \}$ $\endgroup$ – user380364 Jul 22 '18 at 15:43

First, the last section seems a little silly. If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if $f_{m+1}\ge f_m\ge0$ then $\lim\int f_m=\int\lim_m f_m$, qed. (Hint for showing that if $\mu$ is counting measure and $f\ge0$ then $\int_{\Bbb N}f\,d\mu=\sum_nf(n)$: By definition $\sum_nf(n)=\lim_{N\to\infty}\sum_{n=1}^Nf(n)$.)

There's a very simple elementary proof. Let $$\sum_n f(n)=s\in[0,\infty].$$ Monotonicity makes it clear that the limit exists and that $$\lim_m\sum_nf_m(n)\le s.$$

For the other direction, suppose that $\alpha<s$. (Note I'm talking about a general $\alpha<s$ instead of talking about $s-\epsilon$ in order to include the possibility that $s=\infty$.) Choose $N$ so that $$\sum_{n=1}^Nf(n)>\alpha.$$ Then it's clear that there exists $M$ so $$\sum_{n=1}^Nf_m(n)>\alpha\quad(m>M).$$Hence $$\lim_m\sum_nf_m(n)>\alpha,$$and since this holds for every $\alpha<s$ we must have $$\lim_m\sum_nf_m(n)\ge s.$$

  • $\begingroup$ Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$\lim_{m\to \infty }\int_{\mathbb N}f_m=\int_{\mathbb N}f.$$ But to prove that $\int_{\mathbb N}f(n)d\mu=\sum_{n\in\mathbb N}f(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ? $\endgroup$ – user380364 Jul 22 '18 at 15:48
  • $\begingroup$ Let's get the hypothesis straight first. The post says $f_m\to \mathbb N\to \mathbb R\cup\{\pm\infty \}$ , which is gibberish. If we change that to $f_m:\mathbb N\to \mathbb R\cup\{\pm\infty \}$ it makes sense, but then none of the sums you mention need exist. $\endgroup$ – David C. Ullrich Jul 22 '18 at 15:59
  • $\begingroup$ @user380364 Under the current hypothesis it's impossible to prove that $\int f(n)\,d\mu=\sum f(n)$, in fact neither the integral nor the sum need even exist. $\endgroup$ – David C. Ullrich Jul 22 '18 at 16:36

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