Radon-Nikodym Theorem/Lemma 2

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

Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$.

Then t T here exists a pairwise disjoint sequence of sets $\sequence {X_n}_{n \in \mathop \N}$ such that:


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

where:


 * $\map \mu {X_n} < \infty$ and $\map \nu {X_n} < \infty$ for each $n$.

Proof
Since $\mu$ is $\sigma$-finite, by Measure Space Sigma-Finite iff Cover by Sets of Finite Measure, there exists a sequence of $\Sigma$-measurable sets $\sequence {A_n}_{n \mathop \in \N}$ such that:


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

with:


 * $\map \mu {A_n} < \infty$ for each $n$.

From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma$-measurable sets $\sequence {E_n}_{n \mathop \in \N}$ with:


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

with:


 * $E_n \subseteq A_n$ for each $n$.

From Measure is Monotone, we have that:


 * $\map {\mu} {E_n} \le \map {\mu} {A_n}$ for each $n$.

So:


 * $\map {\mu} {E_n}$ is finite for each $n$.

Since $\nu$ is $\sigma$-finite, by Measure Space Sigma-Finite iff Cover by Sets of Finite Measure, there exists a sequence of $\Sigma$-measurable sets $\sequence {B_n}_{n \mathop \in \N}$ such that:


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

with:


 * $\map \nu {B_n} < \infty$ for each $n$.

From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma$-measurable sets $\sequence {F_n}_{n \mathop \in \N}$ with:


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

with:


 * $F_n \subseteq B_n$ for each $n$.

From Measure is Monotone, we have that:


 * $\map \nu {F_n} \le \map \nu {B_n}$ for each $n$.

So:


 * $\map \nu {F_n}$ is finite for each $n$.

Note that for $\tuple {i_1, j_1} \ne \tuple {i_2, j_2}$, we have:

So $\set {E_i \cap F_j : \tuple {i, j} \in \N^2}$ is a pairwise disjoint family of sets.

We have:

For each $i, j$, we now have:

and:

From Cartesian Product of Countable Sets is Countable, we have:


 * $\set {E_i \cap F_j : \tuple {i, j} \in \N^2}$ is countable.

So we can find a bijection:


 * $f : \N \to \set {E_i \cap F_j : \tuple {i, j} \in \N^2}$

Writing:


 * $X_n = \map f n$

for each $n \in \N$, we have:


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

with $\sequence {X_n}_{n \mathop \in \N}$ pairwise disjoint:


 * $\map \mu {X_n} < \infty$ and $\map \nu {X_n} < \infty$ for each $n$.

So we are done.