Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space such that

(i)	$E_0, E_1, E_2, \ldots \in \mathcal{F}$ are each a D1109: Measurable set in $M$
(ii)	\begin{equation} \forall \, n \in \mathbb{N} : \mu(E_n) = 0 \end{equation}

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. $\square$