Infinite Measure is Measure

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Then the infinite measure $\mu$ on $\struct {X, \Sigma}$ is a measure.

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

We have by definition $\map \mu \O = 0$.

$\Box$

Let $S \in \Sigma$.

Then:

$\map \mu S = + \infty > 0$ for $S \ne \O$

$\map \mu S = 0 \ge 0$ for $S = \O$

So $\map \mu S \ge 0$ for all $S \in \Sigma$.

$\Box$

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

$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$

Suppose $S_n = \O$ for all $n \in \N$.

Then by definition of $\mu$:

$\ds \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = \map \mu \O = 0$

Thus, the desired equation becomes:

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

which trivially holds.

Suppose $S_n \ne \O$ for some $n \in \N$.

Then $S_n \subseteq \ds \bigcup_{n \mathop = 1}^\infty S_n \ne \O$.

$\ds \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = + \infty$

$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {S_n} = + \infty = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$

Thus $\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$.

$\Box$

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

$\blacksquare$