Diffuse Measure of Countable Set

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Suppose that for all $x \in X$, the singleton $\set x$ is in $\Sigma$.

Suppose further that $\mu$ is a diffuse measure.

Let $E \in \Sigma$ be a countable measurable set.

Then $\map \mu E = 0$.

Proof
It holds trivially that:


 * $\ds E = \bigcup_{e \mathop \in E} \set e$

and in particular, this union is countable.

Also, $\map \mu {\set e} = 0$ for all $e \in E$ as $\mu$ is diffuse.

Hence Null Sets Closed under Countable Union applies to yield:


 * $\map \mu E = 0$