Measure of Empty Set is Zero

Theorem

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

Then $\map \mu \O = 0$.

That is, $\O$ is a $\mu$-null set.

By definition of measure, there exists at least one $E \in \Sigma$ such that $\map \mu E$ is finite.

So, suppose that $E \in \Sigma$ such that $\map \mu E$ is finite.

Let $\map \mu E = x$.

Consider the sequence $\sequence {S_n}_{n \mathop \in \N} \subseteq \Sigma$ defined as:

$S_n = \begin {cases} E & : n = 1 \\ \O & : n > 1 \end {cases}$

Then:

$\ds \bigcup_{n \mathop = 1}^\infty S_n = E$

Hence:

$\ds \map \mu E$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \mu \paren {\bigcup_{n \mathop = 1}^\infty S_n}$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n}$

$=$

$\ds x$

Property $(2)$

$\ds \leadsto \ \ $

$\ds \map \mu E + \sum_{n \mathop = 2}^\infty \map \mu \O$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \sum_{n \mathop = 2}^\infty \map \mu \O$

$=$

$\ds 0$

It follows directly that $\map \mu \O = 0$.

$\blacksquare$