Diffuse Measure of Countable Set

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Suppose that for all $x \in X$, the singleton $\left\{{x}\right\}$ is in $\Sigma$.

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


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

Then $\mu \left({E}\right) = 0$.


Proof

It holds trivially that:

$\displaystyle E = \bigcup_{e \mathop \in E} \left\{{e}\right\}$

and in particular, this union is countable.

Also, $\mu \left({\left\{{e}\right\}}\right) = 0$ for all $e \in E$ as $\mu$ is diffuse.


Hence Null Sets Closed under Countable Union applies to yield:

$\mu \left({E}\right) = 0$

$\blacksquare$


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