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