Integrating over the naturals. What does it mean?

Question

Let $F$ be the power set of $\Bbb{N}$ and consider the measurable space $(\Bbb{N}, F)$. Then what does it mean to take the integral with respect to the measure $\mu(A) = \sum_{a \in A} \frac{1}{a}$. What would $\int f \ d\mu$ represent, where $f$ is some function $f: \Bbb{N} \to \Bbb{R}$?

My attempt. Take the simplest integrand which is usually $1$ and integrate to get $\int 1 \ d\mu = 1 \mu(\Bbb{N}) = \infty$. This means what?

Answer 1

$\int f d\mu = \sum_a f(a)\mu(\{a\}) = \sum_a \frac{f(a)}{a}$

Answer 2

Consider $f: A \to \Bbb N$. Now write: $$ \int_A f(n) \,{\rm d}\mu(n) = \sum_{n \in A}\int_{\{n\}}f(n)\,{\rm d}\mu(n) = \sum_{n \in A}f(n)\mu(\{n\}) = \sum_{n \in A} \frac{f(n)}{n}. $$