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

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

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


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

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

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


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

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


 * $\displaystyle \bigcup_{n \in \N} A_n = X$

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

Sufficient Condition
Let $\mu$ be any measure.

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

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

Then by Sigma-Algebra Closed Under Union, $A_n \in \mathcal A$ for all $n \in \N$.

Also, $A_{n+1} = A_n \cup E_{n+1}$, hence $A_n \subseteq A_{n+1}$ by Subset of Union.

The definition of the $A_n$ assures that $X = \displaystyle \bigcup_{n \in \N} E_n = \bigcup_{n \in \N} A_n$.

Hence $\left({A_n}\right)_{n \in \N}$ is an exhausting sequence in $\mathcal A$.

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

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

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