Null Measure is Measure

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

Then the null measure $\mu$ on $\left({X, \Sigma}\right)$ is a measure.

Proof
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.

Proof of $(1)$
Let $S \in \Sigma$.

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

Proof of $(2)$
It is to be shown that (for a sequence $\left({S_n}\right)_{n \in \N}$ of pairwise disjoint sets):


 * $\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right) = \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right)$

Now by definition of $\mu$:


 * $\mu \left({S_n}\right) = \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right) = 0$

Thus, the desired equation becomes:


 * $\displaystyle \sum_{n \mathop = 1}^\infty 0 = 0$

which trivially holds.

Proof of $(3')$
Note that $\varnothing \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.

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

Having verified the axioms, $\mu$ is a measure.