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
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:

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