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:

$\map \mu E = 0$

$\blacksquare$


Sources

  • 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 6$: Problem $5 \ \text{(iii)}$
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