Equivalence of Definitions of Sigma-Finite Measure

Definition 1 implies Definition 2
Let $\sequence {E_n}_{n \mathop \in \N}$ be an exhausting sequence in $\Sigma$ such that:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

Then the sequence $\sequence {E_n}_{n \mathop \in \N}$ is increasing and:


 * $\ds X = \bigcup_{n \mathop = 1}^\infty E_n$

with:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

Then $\sequence {E_n}_{n \mathop \in \N}$ is a cover of $X$ in $\Sigma$ with:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

Definition 3 implies Definition 2
Let $\sequence {E_n}_{n \mathop \in \N}$ be a partition of $X$ in $\Sigma$ with:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

Then:


 * $E_n \cap E_m = \O$ for $n \ne m$

and:


 * $\ds X = \bigcup_{n \mathop = 1}^\infty E_n$

with:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

In particular, $\sequence {E_n}_{n \mathop \in \N}$ is a cover of $X$ in $\Sigma$ with:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

Definition 2 implies Definition 3
Let $\sequence {E_n}_{n \mathop \in \N}$ be a cover of $X$ in $\Sigma$ with:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

By Countable Union of Measurable Sets as Disjoint Union of Measurable Sets there exists a sequence of measurable sets $\sequence {F_n}_{n \mathop \in \N}$ such that:


 * $F_n \subseteq E_n$ for each $n \in \N$

and:


 * $\ds X = \bigcup_{n \mathop = 1}^\infty E_n = \bigcup_{n \mathop = 1}^\infty F_n$

From Measure is Monotone we have:


 * $\map \mu {F_n} < \infty$ for each $n \in \N$.

So $\sequence {F_n}_{n \mathop \in \N}$ is a partition of $X$ in $\Sigma$ with:


 * $\map \mu {F_n} < \infty$ for each $n \in \N$.

Definition 3 implies Definition 1
Let $\sequence {E_n}_{n \mathop \in \N}$ be a partition of $X$ in $\Sigma$ with:


 * $\map \mu {E_n} < \infty$ for each $n \in \N$.

Then:


 * $\ds \bigcup_{n \mathop = 1}^\infty E_n = X$

Set:


 * $\ds F_n = \bigcup_{k \mathop = 1}^n E_n$

Then $\sequence {F_n}_{n \mathop \in \N}$ is increasing from Union is Increasing Sequence of Sets.

We show:


 * $\ds \bigcup_{n \mathop = 1}^\infty F_n = \bigcup_{n \mathop = 1}^\infty E_n$

Let:


 * $\ds x \in \bigcup_{n \mathop = 1}^\infty F_n$

then:


 * $x \in F_n$ for some $n \in \N$.

So that:


 * $\ds x \in \bigcup_{k \mathop = 1}^n E_k$ for some $n \in \N$.

Then:


 * $\ds x \in E_k$ for some $k \le n$

so that:


 * $\ds x \in \bigcup_{n \mathop = 1}^\infty F_n$

giving:


 * $\ds \bigcup_{n \mathop = 1}^\infty F_n \subseteq \bigcup_{n \mathop = 1}^\infty E_n$

Now let:


 * $\ds x \in \bigcup_{n \mathop = 1}^\infty E_n$

then:


 * $x \in E_n$ for some $n \in \N$.

Then:


 * $\ds x \in \bigcup_{k \mathop = 1}^n E_n = F_n$

so:


 * $\ds x \in \bigcup_{n \mathop = 1}^\infty F_n$

giving:


 * $\ds \bigcup_{n \mathop = 1}^\infty E_n \subseteq \bigcup_{n \mathop = 1}^\infty F_n$

So that:


 * $\ds \bigcup_{n \mathop = 1}^\infty F_n = \bigcup_{n \mathop = 1}^\infty E_n = X$

Then, we have:

for each $n \in \N$.