$f$ integrable iff $\sum\limits_{n=1}^{\infty} f(n)$ converges absolutely

Question

Given a measure space $(\Bbb R, \mathcal P(\Bbb R), \mu_\Bbb N)$ where $\mathcal P(\Bbb R)$ denotes the power set of $\Bbb R$ and $\mu_\Bbb N$ is defined by $\mu_\Bbb N(A)= \vert {A \cap \Bbb N}\vert$= quantity of natural numbers in $A$,

how can I show that the function $f:\Bbb R \to \Bbb R$ is $\mu_\Bbb N$-integrable if and only if the series $\sum\limits_{n=1}^{\infty} f(n)$ absolutely converges , where $\int f d\mu_\Bbb N$ =$\sum\limits_{n=1}^{\infty} f(n)$.

Any ideas?

Answer 1

$(\Bbb R, \mathcal P(\Bbb R), \mu_\Bbb N)$ where $\mu_\Bbb N(A)= \vert {A \cap \Bbb N}\vert$= quantity of natural numbers in $A$, $\int f d\mu_\Bbb N$ =$\sum_{n=1}^{\infty} f(n)$.

Define the space $L^1(\Bbb R, \mathcal P(\Bbb R),\mu_\Bbb N)=\{f:\Bbb R \to \Bbb R, \int_{\Bbb R} \vert f \vert d\mu_\Bbb N <\infty \}$ that is the space of the integrable functions

$\sum_{n=1}^{\infty} f(n)$ absolutely converges, means $\sum_{n=1}^{\infty} \vert f(n) \vert <\infty$ which means exactly $\int_{\Bbb R} \vert f \vert d\mu_\Bbb N <\infty$.

Now you need to prove $f$ integrable entails $\sum_{n=1}^{\infty} f(n)$ absolutely converges, which is trivial as well by passing through the definition

$f$ integrable = $f \in L^1(\Bbb R, \mathcal P(\Bbb R),\mu_\Bbb N) \rightarrow \infty >\int_{\Bbb R} \vert f \vert d\mu_\Bbb N = \sum_{n=1}^{\infty} \vert f(n) \vert$