User:Caliburn/s/mt/Lebesgue Decomposition Theorem/Finite Measure

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

Let $\mu$ be a measure on $\struct {X, \Sigma}$. Let $\nu$ be a finite measure on $\struct {X, \Sigma}$.

Then there exists finite measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that:


 * $(1) \quad$ $\nu_a$ is absolutely continuous with respect to $\mu$
 * $(2) \quad$ $\nu_s$ and $\mu$ are mutually singular
 * $(3) \quad$ $\nu = \nu_a + \nu_s$.

Proof
Define the set ${\mathcal N}_\mu$ by:


 * ${\mathcal N}_\mu = \set {B \in \Sigma : \map \mu B = 0}$

Since $\nu$ is a finite measure, there exists $M \ge 0$ such that:


 * $\map \nu A \le M$

for all $A \in \Sigma$.

So by the Continuum Property:


 * the supremum of $\set {\map \nu B : B \in {\mathcal N}_\mu}$ exists as a real number $L$.

By the definition of the supremum, for each $n \in \N$ there exists $B_n \in {\mathcal N}_\mu$ such that:


 * $\ds L - \frac 1 n \le \map \nu {B_n} \le L$

Then, from the Squeeze Theorem:


 * $\ds \lim_{n \mathop \to \infty} \map \nu {B_n} = L$

Now set:


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

From Null Sets Closed under Countable Union, we have that:


 * $N$ is $\mu$-null.

So:


 * $N \in {\mathcal N}_\mu$

Further, from Set is Subset of Union, we have:


 * $B_n \subseteq N$ for each $N \in \N$

giving:


 * $\map \nu {B_n} \le \map \nu N \le L$ for each $n \in \N$

from the definition of the supremum and Measure is Monotone.

Taking $n \to \infty$, we obtain:


 * $\map \nu N = L$

by Limits Preserve Inequalities.

Explicitly, we have found that:


 * $\map \nu N = \sup \set {\map \nu B : B \in {\mathcal N}_\mu}$

Let $\nu_a$ be the intersection measure of $\nu$ by $N^c$.

Let $\nu_s$ be the intersection measure of $\nu$ by $N$.

Since $\nu$ is finite, so are $\nu_a$ and $\nu_s$ from Intersection Measure of Finite Measure is Finite Measure.

Then, for each $A \in \Sigma$ we have:

so:


 * $\nu = \nu_a + \nu_s$

verifying $(3)$.

We have:

and $\map \mu N = 0$.

So $\nu_s$ is concentrated on $N$ and $\mu$ is concentrated on $N^c$.

So $\mu$ and $\nu_s$ are mutually singular, verifying $(2)$.

We finish by showing that $\nu_a$ is absolutely continuous with respect to $\mu$.

To simplify notation, first consider a $\Sigma$-measurable $B \subseteq N^c$ such that $\map \mu B = 0$.

That is, $B \in {\mathcal N}_\mu$.

that $\map \nu B \ne 0$.

In particular, $\map \nu B > 0$.

Then, we have:

However, from Null Sets Closed under Countable Union, we have:


 * $N \cup B$ is $\mu$-null.

So:


 * $\map \nu {N \cup B} \in \set {\map \nu B : B \in {\mathcal N}_\mu}$, contradicting that $\map \nu N$ is the supremum of this set.

So we must have $\map \nu B = 0$.

Now suppose that general $B \in \Sigma$ has $\map \mu B = 0$.

Then, from Intersection is Subset, we have:


 * $B \cap N^c \subseteq B$

From Null Sets Closed under Subset, we have:


 * $\map \mu {B \cap N^c} = 0$

while $B \cap N^c$ is a $\Sigma$-measurable subset of $N^c$, so:


 * $\map \nu {B \cap N^c} = 0$

so:


 * $\map {\nu_a} B = 0$

So $\nu_a$ is absolutely continuous with respect to $\mu$, and we are done.