0
$\begingroup$

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.


Context

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.

$\endgroup$
  • 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
1
$\begingroup$

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 $$\lim_m\sum_nf_m(n)\ge s.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.