Integrable if absolutely summable

Question

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...

Answer 1

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.