ThmDex – An index of mathematical definitions, results, and conjectures.
Result R976 on D1158: Measure space
Finite disjoint additivity of unsigned basic measure

Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space.
Let $E_1, \dots, E_N \in \mathcal{F}$ each be a D1109: Measurable set in $M$ such that
 (i) $E_1, \dots, E_N$ is a D1681: Disjoint set collection
Then $$\mu \left( \bigcup_{n = 1}^N E_n \right) = \sum_{n = 1}^N \mu(E_n)$$
Proofs
Proof 0
Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space.
Let $E_1, \dots, E_N \in \mathcal{F}$ each be a D1109: Measurable set in $M$ such that
 (i) $E_1, \dots, E_N$ is a D1681: Disjoint set collection
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 $$\mu \left( \bigcup_{n = 1}^{\infty} E_n \right) = \sum_{n = 1}^{\infty} \mu(E_n) = \sum_{n = 1}^N \mu(E_n) + \sum_{n = N + 1}^{\infty} \mu(E_n) = \sum_{n = 1}^N \mu(E_n)$$ $\square$