Countable partition additivity of unsigned basic measure

Formulation 0

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

(i)	$E, F_0, F_1, F_2, \dots \in \mathcal{F}$ are each a D1109: Measurable set in $M$
(ii)	$F_0, F_1, F_2, \ldots$ is a D5143: Set partition of $X$

Then \begin{equation} \mu(E) = \sum_{n \in \mathbb{N}} \mu(E \cap F_n) \end{equation}

Subresults

Subresult 0
Countable partition additivity of probability measure

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$E, F_0, F_1, F_2, \dots \in \mathcal{F}$ are each an D1716: Event in $P$
(ii)	$F_0, F_1, F_2, \ldots$ is a D5143: Set partition of $\Omega$

Then \begin{equation} \mathbb{P}(E) = \sum_{n \in \mathbb{N}} \mathbb{P}(E \cap F_n) \end{equation}

Subresult 1
Finite partition additivity of unsigned basic measure

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

(i)	$E, F_1, \ldots, F_N \in \mathcal{F}$ are each a D1109: Measurable set in $M$
(ii)	$F_1, \ldots, F_N$ is a D5143: Set partition of $X$

Then \begin{equation} \mu(E) = \sum_{n = 1}^N \mu(E \cap F_n) \end{equation}

Proofs

Proof 0

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

(i)	$E, F_0, F_1, F_2, \dots \in \mathcal{F}$ are each a D1109: Measurable set in $M$
(ii)	$F_0, F_1, F_2, \ldots$ is a D5143: Set partition of $X$

Since $F_0, F_1, F_2, \dots$ partitions $X$, then applying disjoint additivity and result R237: Intersection distributes over union, one has \begin{equation} \begin{split} \mu (E) = \mu (E \cap X) & = \mu \left( E \cap \bigcup_{n \in \mathbb{N}} F_n \right) \\ & = \mu \left( \bigcup_{n \in \mathbb{N}} (E \cap F_n) \right) = \sum_{n \in \mathbb{N}} \mu (E \cap F_n) \end{split} \end{equation} $\square$