Exercise 3.D. in Robert G. Bartle's book Let us consider the problem 3.D. in the book 
$\textbf{The Elements of Integration and Lebesgue Measure}$ 
of Robert G. Bartle
Let $X=\mathbb{N}$ and $\mathcal{A}$ be the $\sigma-$algebra of all subsets of $\mathbb{N}$. If $(a_n)$  is a sequence of nonnegative real numbers and if we define $\mu$ by
$$
\mu(\emptyset)=0; \quad \mu(E)=\sum_{n\in E}a_n, \quad E\ne\emptyset,
$$
then $\mu$ is a measure on $\mathcal{A}$. Conversely, every measure on $\mathcal{A}$ is obtained in this way for some sequence $(a_n)$ in $\overline{R}^+$.
I am stuck in proving the countably additive property of $\mu$.
My attempt. I intend to use the theory of double index series to prove countably additive property.
Thank you for all solutions.
 A: The proof of countable additivity:
\begin{align*}
\mu\left( \bigcup_{i=1}^\infty E_i\right)&=\sum_{n\in\bigcup _{i=1}^\infty E_i}a_n
\end{align*}
Notice that $(E_i)$ is a disjoint sequence, then n belongs to one and only one $E_i$, which means:
$$\sum_{n\in\bigcup _{i=1}^\infty E_i}a_n = \sum_{i=1}^{\infty}\sum_{n\in E_i}a_n = \sum_{i=1}^{\infty} \mu(E_i)$$
The proof of every measure on A is obtained in this way for some sequence $(a_n)$ in $\overline{R}^+$:
For every $E\subset N$, $E=\bigcup_{i=1}^{\infty}{\{n_i\}}$, where $n_i\in N$ or $\{n_i\}=\emptyset$. From the countable additivity of $\mu$ we know that:
$$\mu(E) = \sum_{n \in E}\mu(\{n\})$$
We set $\mu(\{n\}):=a_n$, which establishs the proposition.
A: We have to check the following


*

*$\mu(\emptyset)=0$, by definition

*$\mu(E)=\sum_{n\in E}a_n\geq 0$, since $a_n\geq 0$, for all $n$.

*Let $\{E_k\}$ be disjoint, then\begin{align*}
\mu\left( \bigcup_{k=1}^\infty E_k\right)&=\sum_{n\in\bigcup E_k}a_n=\sum_{k=1}^\infty \sum_{n\in E_k}a_n=\sum_{k=1}^\infty\mu (E_k).
\end{align*}
Thus, $\mu$ is a measure on X. Conversely, for $(a_n)$ in $\overline{\mathbb{R}}^+$, we can obtain any measure on X using the enunciated definition of $\mu$.
