Applying result
R979: Countable subadditivity of measure, we have
\begin{equation}
\mu \left( \bigcup_{n \in \mathbb{N}} E_n \right)
\leq \sum_{n \in \mathbb{N}} \mu(E_n)
= \sum_{n \in \mathbb{N}} 0
= 0
\end{equation}
Since $\mu \geq 0$, the claim follows from result
R1043: Equality from two inequalities for real numbers. $\square$
