Counting measure on the power set of N I need to define a sequence of functions that converge to $f$ so that I can use the monotone convergence theorem to prove. Not sure how to define the sequence of functions.
$$\int_{\mathbb{N}} f\,\text{d}\mu = \sum_{n=1}^\infty f(n)$$
 A: Let us consider themeasure space $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \sharp)$, where $\sharp$ is the caounting measure. 
Suppose $f\geqslant 0$. For each $n\in \mathbb{N}$, let $f_n =  \chi_{\{0,\dots,n\}} f$. Then
$$\int_\mathbb{N} f_n \mathrm{d}\mu=\int_\mathbb{N} \chi_{\{0,\dots,n\}} f \mathrm{d}\mu = \int_{\{0,\dots,n\}} f \mathrm{d}\mu=\sum_{i=0}^n f(i)$$
Since for all $n\in\mathbb{N}$, $f_n\geqslant 0$, and $f_n$ converges monotonically increasing to $f$, we can apply the Monotone Convergence Theorem and we get: 
$$\int_\mathbb{N} f \mathrm{d}\mu= \lim_{n \to \infty} \int_\mathbb{N} f_n \mathrm{d}\mu =  \lim_{n \to \infty} \sum_{i=0}^n f(i) = \sum_{i=0}^\infty f(i)$$
Remark 1: If $f$ is NOT a nonnegative function, then the result may be false. For instance: 
let $f$ be defined by $f(k)=(-1)^k\frac{1}{k+1}$, for each $k \in\mathbb{N}$. Then 
$$\sum_{i=0}^\infty f(i)= \ln(2)$$ But $f$ is not Lebesgue integrable in the measure space $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \sharp)$. 
Remark 2: If $f$ is not a nonnegative function, let $f^+$ and $f^-$ be defined, for each $k \in\mathbb{N}$, by $f^+(k)=\max\{f(k),0\}$ and $f^-(k)=\max\{-f(k),0\}$. Clearly $f^+$ and $f^-$ are nonnegative functions and it is easy to see that 
$$f=f^+ - f^-$$ We can apply the argument presented in this answer to $f^+$ and $f^-$ separately. If at least one of the integrals of $f^+$ and $f^-$ is finite, then we can conclude that $$\int_\mathbb{N} f \mathrm{d}\mu=\sum_{i=0}^\infty f(i)$$
(In the counterexample presented in Remark 1, the integrals of  $f^+$ and $f^-$ are both infinite).
A: $$
\int_{\mathbb{N}} f\,\text{d}\mu = \sum_{n=1}^\infty f(n) \quad \large\text{?}
$$
Suppose the series converges absolutely.  Let $A = \{n\in\mathbb N : f(n)\ge 0\}$ and $B= \mathbb N\setminus A$.  Let
$$
f^+(n) = \begin{cases} f(n) & \text{if }n\in A, \\ 0 & \text{otherwise}. \end{cases}
$$
Let
$$
f_k^+(n) = \begin{cases} f^+(n) & \text{if }n\le k, \\ 0 & \text{if }n>k. \end{cases}
$$
Then $f_k^+ \uparrow f^+$ pointwise as $k\to\infty$.  Then you can apply the monotone convergence theorem.
What to do with the complementary set $B$, I leave as an exercise.
In case of either divergence to $+\infty$ or to $-\infty$, or of conditional convergence, the function is not Lebesgue-integrable and the result is not even meaningful, let alone true.  In the case of divergence to $+\infty$ or to $-\infty$ the result remains true.
A: Lets assume that we work on the following measure space: $(\Omega, \Sigma, \mu) = (\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu = \text{counting measure})$
Then we have 
$$\int_\mathbb{N} f \,\mathrm{d}\mu = \sum_{n=1}^\infty \int_{\{n\}}f \, \mathrm{d}\mu = \sum_{n=1}^\infty f(n) \mu(\{n\}) = \sum_{n=1}^\infty f(n).$$
