Null Measure is Measure

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

$\Box$


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 $\mu$:

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


Thus, the desired equation becomes:

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

which trivially holds.

$\Box$


Proof of $(3')$

Note that $\varnothing \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.

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

$\Box$


The axioms are fulfilled, and it follows that $\mu$ is a measure.

$\blacksquare$