Show that function is Lebesgue integrable if and only if series is absolutely convergent

Question

I'm trying to solve the following problem:

Let $\Omega$ $=$ $\mathbb{N}$ and $A$ = $P$($\mathbb{N}$). We define the counting measure µ($A$) on $P$($\mathbb{N}$). Let $f$ : $\Omega$ $\rightarrow$ $\mathbb{R}$ be a function. Show that $f$ is integrable if and only if the series $\sum_{n=1}^{\infty}$ $f(n)$ is absolutely convergent. Then show $\int_\mathbb{N}$ $f$ dµ = $\sum_{n=1}^{\infty}$ $f(n)$.

Remark by me: definiton of $f$ integrable: $f$ is integrable if $f$ is measurable and $\int$ $f^+$ dµ < $\infty$ and $\int$ $f^-$ dµ < $\infty$ . We define $\int$ $f$ dµ $=$ $\int$ $f^+$ dµ $-$ $\int$ $f^-$ dµ.

Answer 1

Let $\Omega^+ = \{ n \in \mathbb{N} : f(n) > 0 \}$, and $\Omega^- = \{ n \in \mathbb{N} : f(n) < 0 \}$. If $f$ is integrable, this means that the nonnegative series

$$\int\limits_{\Omega^+}f = \sum\limits_{n \in \Omega^+} f(n)\space; \space \int\limits_{\Omega^-}-f = \sum\limits_{n \in \mathbb{N}}-f(n)$$ are both finite. Then their sum is also finite:

$$\sum\limits_{n=1}^{\infty} |f(n)| = \sum\limits_{n \in \Omega^+}f(n) + \sum\limits_{n \in \Omega^-} f(n) < \infty$$

Going the other direction, you assume the above series is finite. Then the two subseries coming from $\Omega^+$ and $\Omega^-$ are also finite, so $f$ is integrable.