Radon-Nikodym Theorem/Lemma 2

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Lemma

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

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


Then there exists a pairwise disjoint sequence of $\Sigma$-measurable 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, 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, 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:

\(\ds \paren {E_{i_1} \cap F_{j_1} } \cap \paren {E_{i_2} \cap F_{j_2} }\) \(=\) \(\ds \paren {E_{i_1} \cap E_{i_2} } \cap \paren {E_{i_2} \cap F_{j_2} }\)
\(\ds \) \(=\) \(\ds \O\) since either $i_1 \ne i_2$ or $j_1 \ne j_2$

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

We have:

\(\ds \bigcup_{i \mathop = 1}^\infty \bigcup_{j \mathop = 1}^\infty \paren {E_i \cap F_j}\) \(=\) \(\ds \bigcup_{i \mathop = 1}^\infty \paren {E_i \cap \paren {\bigcup_{j \mathop = 1}^\infty F_j} }\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds \paren {\bigcup_{i \mathop = 1}^\infty E_i} \cap \paren {\bigcup_{j \mathop = 1}^\infty F_j}\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds X \cap X\)
\(\ds \) \(=\) \(\ds X\)

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

\(\ds \map \mu {E_i \cap F_j}\) \(\le\) \(\ds \map \mu {E_i}\) Intersection is Subset, Measure is Monotone
\(\ds \) \(<\) \(\ds \infty\)

and:

\(\ds \map \nu {E_i \cap F_j}\) \(\le\) \(\ds \map \nu {F_j}\) Intersection is Subset, Measure is Monotone
\(\ds \) \(<\) \(\ds \infty\)

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.

$\blacksquare$

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