Measure of Empty Set is Zero

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Theorem

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

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


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


Proof

By definition of measure, there exists at least one $E \in \Sigma$ such that $\mu \left({E}\right)$ is finite.

So, suppose that $E \in \Sigma$ such that $\mu \left({E}\right)$ is finite.

Let $\mu \left({E}\right) = x$.


Consider the sequence $\left \langle {S_n}\right \rangle_{n \in \N} \subseteq \Sigma$ defined as:

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

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

Hence:

\(\displaystyle \mu \left({E}\right)\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right)\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right)\) \(=\) \(\displaystyle x\) Property $(2)$
\(\displaystyle \implies \ \ \) \(\displaystyle \mu \left({E}\right) + \sum_{n \mathop = 2}^\infty \mu \left({\varnothing}\right)\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 2}^\infty \mu \left({\varnothing}\right)\) \(=\) \(\displaystyle 0\)

It follows directly that $\mu \left({\varnothing}\right) = 0$.

$\blacksquare$


Sources

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