Measure of Empty Set is Zero

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 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 $\left \langle {S_n}\right \rangle \subseteq \mathcal A$:
 * $\displaystyle \mu \left({\bigcup_{n=1}^{\infty} S_n}\right) = \sum_{n=1}^{\infty} \mu \left({S_{n}}\right)$

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

Hence:

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