Consider the augmented sequence $E_1, E_2, E_3, \dots$ where $E_n = \emptyset$ for every $n > N$. Clearly, this augmented sequence is again disjoint. By definition, the measure of the empty set $\emptyset$ in $\mu$ is $0$. Thus, countable disjoint additivity of measure implies \begin{equation} \mu \left( \bigcup_{n = 1}^{\infty} E_n \right) = \sum_{n = 1}^{\infty} \mu(E_n) = \sum_{n = 1}^N \mu(E_n) + \sum_{n = N + 1}^{\infty} \mu(E_n) = \sum_{n = 1}^N \mu(E_n) \end{equation} $\square$