Let $M = (X, \mathcal{F}, \mathbb{\mu})$ 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} E_0 \subseteq E_1 \subseteq E_2 \subseteq \cdots \end{equation}

Result R984: Transforming an isotone sequence of sets into pairwise disjoint sequence with common limit shows that we can write \begin{equation} \bigcup_{n \in \mathbb{N}} E_n = E_0 \cup \left( \bigcup_{n \in \mathbb{N} + 1} E_{n + 1} \setminus E_n \right) \end{equation} where $E_0, E_2 \setminus E_1, E_3 \setminus E_2, \ldots$ is now a disjoint sequence. Applying disjoint additivity of measure \begin{equation} \begin{split} \mu \left( \bigcup_{n \in \mathbb{N}} E_n \right) & = \mu \left[ E_0 \cup \left( \bigcup_{n \in \mathbb{N} + 1} E_{n + 1} \setminus E_n \right) \right] \\ & = \mu(E_0) + \mu \left( \bigcup_{n \in \mathbb{N} + 1} E_{n + 1} \setminus E_n \right) \\ & = \mu(E_0) + \sum_{n \in \mathbb{N} + 1} \mu(E_{n + 1} \setminus E_n) \\ \end{split} \end{equation} Since $E_0, E_1, E_2, \ldots$ is an isotone sequence, then each finite union equals the set in the union with the largest index. That is, $E_0 \cup [\bigcup_{n = 1}^{N - 1} E_{n + 1} \setminus E_n] = E_N$, for each positive integer $N$. Thus \begin{equation} \begin{split} \mu \left( \bigcup_{n \in \mathbb{N}} E_n \right) & = \mu(E_0) + \sum_{n \in \mathbb{N} + 1} \mu(E_{n + 1} \setminus E_n) \\ & = \mu(E_0) + \lim_{N \to \infty} \sum_{n = 1}^{N - 1} \mu(E_{n + 1} \setminus E_n) \\ & = \lim_{N \to \infty} \left( \mu(E_0) + \sum_{n = 1}^{N - 1} \mu(E_{n + 1} \setminus E_n) \right) \\ & = \lim_{N \to \infty} \mu \left[ E_0 \cup \left( \bigcup_{n = 1}^{N - 1} E_{n + 1} \setminus E_n \right) \right] \\ & = \lim_{N \to \infty} \mu(E_N) \end{split} \end{equation} This finishes the proof. $\square$