Measure of Empty Set is Zero

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

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

Proof 1
The empty set is disjoint with itself, that is: $$\varnothing \cap \varnothing = \varnothing$$.

Hence by additivity:
 * $$\mu \left({\varnothing}\right) = \mu \left({\varnothing \cup \varnothing}\right) = \mu \left({\varnothing}\right) + \mu \left({\varnothing}\right) = 2 \mu \left({\varnothing}\right)\ $$

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

Proof 2
By definition of measure, $$\mu$$ has the following properties:

$$(1)$$: For every $$S \in \mathcal A$$:
 * $$\mu \left({S}\right) \ge 0$$

$$(2)$$: For every sequence of pairwise disjoint sets $$\left \langle {S_n}\right \rangle \subseteq \mathcal A$$:
 * $$\mu \left({\bigcup_{n=1}^{\infty} S_n}\right) = \sum_{n=1}^{\infty} \mu \left({S_{n}}\right)$$

(that is, $$\mu\ $$ is a countably additive function).

$$(3):$$ There exists at least one $$A \in \mathcal A$$ such that $$\mu \left({A}\right)$$ is finite.

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

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

Consider the sequence $$\left \langle {S_n}\right \rangle \subseteq \mathcal A$$ defined as:
 * $$S_n = \begin{cases}

A & : n = 1 \\ \varnothing & : n \ne 1 \end{cases}$$

Then $$\bigcup_{n=1}^{\infty} S_n = A$$.

Hence:

$$ $$ $$ $$ $$

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