Integration of function with respect to counting measure 
Consider the measurable space $\left(\mathbb{N},\mathscr{P}(\mathbb{N})\right)$ with the counting measure $c$ and let $f\colon\mathbb{N}\to\mathbb{R}$ be any function. I want to show that

$$\int_{\mathbb{N}}f\,dc = \sum_{n=1}^{\infty}f(n).\tag{1}$$
I've seen the proof (on StackExchange) that $(1)$ holds given that $f\colon\mathbb{N}\to[0,\infty]$, but not for this particular problem. Further, the proof in that case uses the Monotone Convergence Theorem as $f$ is non-negative. However, how can I do the proof without the assumption that $f$ is non-negative, i.e. without the Monotone Convergence Theorem? Moreover, is there anyway to modify the proof for case that $f$ is non-negaive?
To add, I know some convergence theorem needs to hold, more than likely the Dominated Convergence Theorem, but I can figure out a sequence of functions that work. For the non-negative case, the sequence of functions given by
$$
f_n(k)=\begin{cases}f(k) & \text{if }1\leq k\leq n\\ 0 & \text{else}\end{cases},
$$
for $n\in\mathbb{N}$ works.
 A: Consider the measurable space $\left(\mathbb{N},\mathscr{P}(\mathbb{N})\right)$ with the counting measure $c$ and let $f\colon\mathbb{N}\to\mathbb{R}$ be any function.
Suppose that
$$\sum_{n=0}^{\infty}|f(n)|<+\infty \tag{1}$$
that means, the series $\sum_{n=0}^{\infty}f(n)$ is absolutely convergent.
Let $A= \{n\in \Bbb N: f(n) \geq 0\}$ and $B=  \{n\in \Bbb N: f(n) < 0\}$. Note that $\Bbb N = A \cup B$ and $A\cap B=\emptyset$.
We can define $f^+ \colon\mathbb{N}\to[0,+\infty)$ by $f^+(n) = f(n)$ if $n \in A$ and $f^+(n) = 0$ if $n \in B$ .
We can also define $f^- \colon\mathbb{N}\to[0,+\infty)$ by $f^-(n) = -f(n)$ if $n \in B$ and $f^-(n) = 0$ if $n \in A$.
Note that $f^+$ and $f^-$ are non-negative, so we have
$$ \int_{\mathbb{N}}f^+\,dc = \sum_{n=0}^{\infty}f^+(n)= \sum_{n\in A} f(n) =  \sum_{n\in A} |f(n)| <+\infty$$
and
$$ \int_{\mathbb{N}}f^-\,dc = \sum_{n=0}^{\infty}f^-(n)= \sum_{n\in B} (-f(n))=-\sum_{n\in B} f(n) = \sum_{n\in B} |f(n)| <+\infty $$
Note that $f=f^+ - f^-$. So
$$ \int_{\mathbb{N}}f\,dc = \int_{\mathbb{N}}f^+\,dc - \int_{\mathbb{N}}f^-\,dc = \sum_{n\in A} f(n) + \sum_{n\in B} f(n) = \sum_{n=0}^{\infty}f(n) $$
the last equality is true because, the series $\sum_{n=0}^{\infty}f(n)$ is absolutely convergent (see $(1)$).
Remark:
If $\sum_{n=0}^{\infty}|f(n)|=+\infty $
but
$ \int_{\mathbb{N}}f^-\,dc = \sum_{n\in B} |f(n)| <+\infty $,
we can still define
$\int_{\mathbb{N}}f\,dc = \int_{\mathbb{N}}f^+\,dc - \int_{\mathbb{N}}f^-\,dc$
and prove
$$\int_{\mathbb{N}}f\,dc =+\infty  =  \sum_{n=0}^{\infty}f(n) $$
In a similar way, if
$\sum_{n=0}^{\infty}|f(n)|=+\infty $
but
$ \int_{\mathbb{N}}f^+\,dc = \sum_{n\in A} |f(n)| <+\infty $,
we can still define
$\int_{\mathbb{N}}f\,dc = \int_{\mathbb{N}}f^+\,dc - \int_{\mathbb{N}}f^-\,dc$
and prove
$$\int_{\mathbb{N}}f\,dc =-\infty  =  \sum_{n=0}^{\infty}f(n) $$
A: Provided that $\sum_{n=1}^{+\infty}\vert f(n)\vert<+\infty$, so that $\sum_{n=1}^{+\infty}f(n)$ makes sense, then you just need to write $f=f^+-f^-$:
$$
\int_{\mathbb N}f\,dc=\int_{\mathbb N}f^+\,dc-\int_{\mathbb N}f^-\,dc=\sum_{n=1}^{+\infty}f^+(n)-\sum_{n=1}^{+\infty}f^-(n)=\sum_{n=1}^{+\infty}f(n).
$$
