Countable infinite set So I have some problems with the following problem:

*

*Let $X=\{x_1, x_2,…\}$ be a countable infinite set and let $\mu$ be a measure on $(X,\mathbb{P}(X))$ (where $\mathbb{P}(X)$ is the powerset of X) 
Show that $\mu (X) = \sum^{\infty}_{j=1} \mu (\{x_j\})$
I am a little unsure of how to work with infinite sets, since all sets I have worked with before is finite.
Can anyone give a hint on how to start with this?
 A: This follows directly from the definition of a measure. All the singletons $\{x_i\}$ are disjoint, and as the $\sigma$-algebra is the powerset they are all measurable. Thus, by the $\sigma$-additivity of $\mu$ we have
$$\mu(X)=\mu(\bigcup_{i=1}^\infty \{x_i\})=\sum_{i=1}^\infty \mu(\{x_i\}).$$
A: Countable additivity is a part of the definition of measure. By the way, if you actually want to know about countable additivity about outer measure, I show the proof.
Step 1: Aim to prove $\mu(X)\geq\sum_{j=1}^{\infty}\mu(\{x_j\})$.
By order-preserving, $\forall m \in \mathbb{N} $, since $\bigcup_{j=1}^{m}\{x_j\}\subset X$, we have:
$$\begin{aligned}
\mu(X)\geq \sum_{j=1}^{m}\mu(\{x_j\}), \forall m \in \mathbb{N}
\end{aligned}$$
Since m is arbitrary, we have:
$$\begin{align}
\mu(X)&\geq \operatorname{sup}_{m \in \mathbb{N}} \sum_{j=1}^{m}\mu(\{x_j\})\\
&=\sum_{j=1}^{\infty} \mu(\{x_j\}) \tag{1}
\end{align}$$
Step 2: Aim to prove $\mu(X)\leq\sum_{j=1}^{\infty}\mu(\{x_j\})$.
By countable sub-additivity of out measure, since $X=\bigcup_{j=1}^{\infty}\{x_j\}$, we have:
$$\begin{equation}
\mu(X)\leq \sum_{j=1}^{\infty} \mu(\{x_j\}) \tag{2}
\end{equation}$$
After combining (1) and (2), we can gain the result: $\mu(X)=\sum_{j=1}^{\infty}\mu(\{x_j\})$.
