Measure Space has Exhausting Sequence of Finite Measure iff Cover by Sets of Finite Measure

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Theorem

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


Then there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:

$\forall n \in \N: \map \mu {E_n} < \infty$

if and only if there exists a sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:

$(1): \quad \ds \bigcup_{n \mathop \in \N} E_n = X$
$(2): \quad \forall n \in \N: \map \mu {E_n} < +\infty$


Proof

Necessary Condition

Suppose that there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:

$\forall n \in \N: \map \mu {E_n} < \infty$

Let $\sequence {F_n}_{n \mathop \in \N}$ be an exhausting sequence in $\Sigma$ such that:

$\forall n \in \N: \map \mu {F_n} < +\infty$


Then as $\sequence {F_n}_{n \mathop \in \N}$ is exhausting, have:

$\ds \bigcup_{n \mathop \in \N} F_n = X$

It follows that the sequence $\sequence {F_n}_{n \mathop \in \N}$ satisfies $(1)$ and $(2)$.

$\Box$


Sufficient Condition

Let $\mu$ be any measure.

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence satisfying $(1)$ and $(2)$.


Define $F_n := \ds \bigcup_{k \mathop = 1}^n E_k$.

Then by Sigma-Algebra Closed under Union:

$F_n \in \Sigma$ for all $n \in \N$

Also, by Set is Subset of Union:

$F_{n+1} = F_n \cup E_{n+1}$, hence $F_n \subseteq F_{n + 1}$

The definition of the $F_n$ assures that:

$X = \ds \bigcup_{n \mathop \in \N} E_n = \bigcup_{n \mathop \in \N} F_n$

Hence $\sequence {F_n}_{n \mathop \in \N}$ is an exhausting sequence in $\Sigma$.


Furthermore, compute, for any $n \in \N$:

\(\ds \map \mu {F_n}\) \(=\) \(\ds \map \mu {\bigcup_{k \mathop = 1}^n E_k}\) Definition of $F_n$
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 1}^n \map \mu {E_k}\) Measure is Subadditive
\(\ds \) \(<\) \(\ds +\infty\) By $(2)$

Hence there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:

$\forall n \in \N: \map \mu {E_n} < \infty$

$\blacksquare$


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