Null Measure is Measure
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Then the null measure $\mu$ on $\struct {X, \Sigma}$ is a measure.
Proof
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.
Proof of $(1)$
Let $S \in \Sigma$.
Then $\map \mu S = 0 \ge 0$.
$\Box$
Proof of $(2)$
It is to be shown that (for a sequence $\sequence {S_n}_{n \in \N}$ of pairwise disjoint sets):
- $\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$
Now by definition of $\mu$:
- $\ds \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = 0$
Thus, the desired equation becomes:
- $\ds \sum_{n \mathop = 1}^\infty 0 = 0$
which trivially holds.
$\Box$
Proof of $(3')$
Note that $\O \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.
Hence $\map \mu \O = 0$.
$\Box$
The axioms are fulfilled, and it follows that $\mu$ is a measure.
$\blacksquare$