Integral w.r.t counting measure 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.
 A: 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
$$\int_{\mathbb{N}} f\,d\mu \le \sum_{k=1}^\infty f(k).$$
