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.

  • $\begingroup$ 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? $\endgroup$ 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 :) $\endgroup$ 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. $\endgroup$ Apr 28 '18 at 14:35

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 $s_k=s(a_k)\le f(a_k)$.


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

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

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

  • $\begingroup$ $m=\infty $ is not allowed because then it won't be finite linear combination of indicator functions. $\endgroup$ Sep 25 '15 at 21:24
  • 3
    $\begingroup$ That is not correct. Read again. $\endgroup$
    – J.R.
    Sep 26 '15 at 8:30

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.