suppose on a set X,
$\{p_{j}, j \geq 1\} \subseteq X$, $j \in \mathbb{N} $ denoted as a set of distinct points.
$\{a_{j}, j \geq 1\}$ $j \in \mathbb{N}$ denoted as a sequence of real numbers.
$C \subseteq P(X)$,
then Discrete Measure is defined as: $\forall A \in C: \mu(A)= \sum_{j \geq1} a_{j}*1\{p_{j}\in A\}$.
Hence $\mu$ is an additive set function.
It is easy to check $\mu(\emptyset)=0$, such as When $A=\emptyset$, then $p_{j} \notin \emptyset $, $\forall j \in \mathbb{N}$, Hence $\mu(A) = 0$,
However, I am having trouble to check the second condition for $\mu$ being additive, Suppose we have a sequence of disjoint ${E_{i}},i \in \mathbb{N}, E_{i} \in C$, and $\cup_{i\geq1}E_{i} \in C$, then $\mu(\cup_{i \geq 1}E_{i})$ = $\sum_{i\geq 1}\sum_{j\geq 1} a_{j} * 1 \{{p_j} \in E_{i}\}$. Is this enough? I think I have missed something, because I have just stated what the second condition is.
We can always interchange the order of summation when the terms are non-negative. Hence $ \sum_i \mu (E_i)=\sum_i \sum_j a_j I_{\{p_j \in E_i\}} =\sum_j \sum_i a_j I_{\{p_j \in E_i\}}=\sum_j a_j I_{\{p_j \in \cup_i E_i\}} =\mu (\cup_i E_i)$. [$\sum_i I_{\{p_j \in E_i\}}= I_{\{p_j \in \cup_i E_i\}}$ because of disjointness].