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
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 4$: Problem $10 \ \text{(i)}$