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:


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

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 $\mu$.