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Definition of integral is given $$ \int f d\mu=\sup{\left\{\int s\ d\mu : 0 \leq s \leq f , s \ \text{ is a simple function} \right\}} $$

Now let $f: \mathbb{N} \rightarrow [0,\infty)$ be a non-negative measurable function on the natural numbers and $\mu$ is the counting measure on $\mathbb{N}$. Prove the following using definition of integral: $$ \int_{\mathbb{N}}f\ d\mu =\sum_{k=1}^{\infty} f(k) $$

I could prove that $\int_{\mathbb{N}}f\ d\mu \geq \sum_{k=1}^{\infty} f(k)$ using simple functions of the form $f_N=\sum_{k=1}^N f(k)1\{n=k\}$. How do I prove the other direction ?

Note: We shouldn't use Monotone Convergence Theorem.

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  • $\begingroup$ You can prove that $\sum_{k=1}^{\infty}f(k)$ is greater than or equal to anything in the set that you are taking the $\sup$. $\endgroup$ – An Hoa Sep 25 '15 at 15:01
  • $\begingroup$ @VuAnHoa: That's what I had trouble proving. Like taking simple functions and all. Can you give a rough outline of the procedure involved? $\endgroup$ – pikachuchameleon Sep 25 '15 at 15:39
  • $\begingroup$ Take any simple function $0 \leq s \leq f$ and let $\{y_1,...,y_n\}$ be its positive values. If $s^{-1}(y_i)$ is infinite for some $i$ then $\sum_{k=1}^{\infty}{f(k)} = \infty$ because $f$ must take values $\leq y_i > 0$ infinitely often. Else $\int s d\mu = \sum y_i \mu(s^{-1}(y_i)) = \sum_{k=1}^{\infty} s(k) \leq \sum_{k=1}^{\infty}{f(k)}$. $\endgroup$ – An Hoa Sep 26 '15 at 0:49
  • $\begingroup$ @VuAnHoa: Thanks, the argument is simple and clear :) $\endgroup$ – pikachuchameleon Sep 26 '15 at 18:41
  • $\begingroup$ Here is a link to another posts about this - although the accepted answer there uses monotone convergence theorem. $\endgroup$ – Martin Sleziak Apr 28 '18 at 14:35
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Recall that simple functions are just finite linear combinations of characteristic functions of measurable sets. In this context our measure space is discrete and all sets are measurable, so a simple function is just a linear combination of (characteristic functions of) points. More precisely, in our context every simple function takes the form $$s(n)=\sum_{k=1}^m s_k \mathbf{1}_{\{a_k\}}(n),$$ for some $m\in\mathbb{N}\cup\{\infty\}$ and $(a_k)_{k=1}^m\subset \mathbb{N}$ pairwise different numbers. Note how $m=\infty$ is also allowed, because also infinite sets of natural numbers are measurable.

Say that $s$ is one of the objects that the supremum is taken over, so a simple function with $0\le s(n)\le f(n)$ for all $n$. In particular $s_k=s(a_k)\le f(a_k)$.

Then

$$\int_{\mathbb{N}} s\,d\mu = \sum_{k=1}^m s_k\le \sum_{k=1}^m f(a_k)\le \sum_{k=1}^\infty f(k).$$

In the last step we used that $f$ is non-negative. By definition of the supremum as the least upper bound this implies

$$\int_{\mathbb{N}} f\,d\mu \le \sum_{k=1}^\infty f(k).$$

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  • $\begingroup$ $m=\infty $ is not allowed because then it won't be finite linear combination of indicator functions. $\endgroup$ – pikachuchameleon Sep 25 '15 at 21:24
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    $\begingroup$ That is not correct. Read again. $\endgroup$ – J.R. Sep 26 '15 at 8:30

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