Measure Space Sigma-Finite iff Cover by Sets of Finite Measure

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Theorem

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


Then $\left({X, \Sigma, \mu}\right)$ is $\sigma$-finite iff there exists a sequence $\left({E_n}\right)_{n \in \N}$ in $\Sigma$ such that:

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


Proof

Necessary Condition

Let $\mu$ be a $\sigma$-finite measure.

Let $\left({F_n}\right)_{n \in \N}$ be an exhausting sequence in $\Sigma$ such that:

$\forall n \in \N: \mu \left({F_n}\right) < \infty$


Then as $\left({F_n}\right)_{n \in \N}$ is exhausting, have:

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

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

$\Box$


Sufficient Condition

Let $\mu$ be any measure.

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


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

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:

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

Hence $\left({F_n}\right)_{n \in \N}$ is an exhausting sequence in $\Sigma$.


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

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

Hence, by definition, $\mu$ is $\sigma$-finite.

Thus, $\left({X, \Sigma, \mu}\right)$ is also $\sigma$-finite.

$\blacksquare$


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