Measure of Empty Set is Zero

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Theorem

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

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


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


Proof

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:

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

Hence:

\(\displaystyle \map \mu E\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \mu \paren {\bigcup_{n \mathop = 1}^\infty S_n}\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \map \mu {S_n}\) \(=\) \(\displaystyle x\) Property $(2)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \mu E + \sum_{n \mathop = 2}^\infty \map \mu \O\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 2}^\infty \map \mu \O\) \(=\) \(\displaystyle 0\)

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

$\blacksquare$


Sources