Integrable if absolutely summable Let $P(\mathbb N)$ the power set of $\mathbb N$ and $f$ the counting measure on $(\mathbb N, P(\mathbb N) )$.
If $\{ a_n \}_{n \in \mathbb N}$ is a real-valued sequence, then, so the statement in my script, it should hold that $a$ is absolutely summable iff $a$ is integrable wrt $f$. 
But why? In the script, it says that "it simply follows from the definitions". But for me, this is not so clear...
 A: 
Definition:  Let $(E,X, \mu)$ be a measure space and let $f$ be a measurable function on that space. Then we say that $f$ is integrable on $S \in X$ with respect to $\mu$ if the function $|f| = f^+ + f^-$ is such that
    \begin{equation} \int_S |f| d\mu < \infty
 \end{equation}
  and it is integral is defined by
     \begin{equation} \int_S f d\mu := \int_S f^+ d\mu -\int_S f^- d\mu \;\;\; \in \mathbb{R}
 \end{equation} 

This definition gives the desirable properties of the integral $\int f d\mu$ which as a consequence of the definition is well defined.
Applying this definition to the specific measure space in the question, you see that $a$, according to the definition of integrability, is integrable if it is absolutely summable.
This can be seen by observing the following: The condition of integrability is 
\begin{equation}
\int_\mathbb{N} |a|\; d P(\mathbb{N}) = \text{\{by the definition of counting measure\}} = \sum_{n=1}^\infty |a_n|
\end{equation}
Therefore if $a$ is absolutely summable it holds that 
\begin{equation}
\int_\mathbb{N} |a|\; d P(\mathbb{N}) < \infty
\end{equation}
and therefore, from the definition we say that $a$ is integrable with respect to the counting measure.
