Integration of function nonnegative at $\mathbb{N}$ I'm having a little trouble to write down my solution of exercise 4G of Bartle's book The Elements of Integration and Lebesgue Measure. Here is the problem:

Let  X = $\mathbb{N}$ and $\mathcal{A} = 2^X$ be $\sigma$-Algebra and define the measure to be the counting measure. If $f:X \rightarrow \mathbb{R}$ is nonnegative then

*

*$f$ is measurable

*$\int_{\mathbb{N}}f = \sum_{k=1}^{\infty}f(k)$

So what I think to item (1) how the function is from naturals to naturals, then the preimage of f is a subset on $2^{\mathbb{N}}$, I think that this is the idea.
To the item (2) I used the item (1) and the following lemma:
If $(E_n)$ is a sequence of disjoints sets, considering $E = \cup_{n=1}^{\infty}E_n$ then
\begin{equation}
\int_{E}fd\mu = \sum_{k=1}^{\infty}\int_{E_k}fd\mu
\end{equation}
To the exercise consider $E_k = \{k\}$ with $k \in \mathbb{N}$, this sets are disjoints, with $\mathbb{N} = \cup_{n=1}^{\infty}E_n$ then we can use the lemma above, thus:
\begin{equation}
 \int_\mathbb{N} f = \sum_{k=1}^{\infty}\int_{E_k}fd\mu \overset{\mathrm{?}}{=} \sum_{k=1}^{\infty}\sum_{j=1}^{k} a_j \mu(E_k)
\end{equation}
I'm considering that $\int fd\mu = sup \int sd\mu$ where $s$ are the simple functions and $\int s d\mu = \sum a_j \mu(E_j)$, but I'm not sure about the second equality above (?). Now I used that $\mu(E_k) = 1$ and $\sum_{j=1}^{k} a_j = f(k)$. This is the supose end of proof, but I don't think so.
Can you appoint my mistakes and show the path to make rigth?

Denote $v$ the counting measure.
Then  $$\int_{\{n\}}f(k)dv=\int_{\{k \in \Bbb{N}:k=n\}}f(k)dv=f(n)\int_{\{n\}}dv=f(n)v(\{n\})=f(n)$$
