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:
\(\ds \map \nu A\) | \(=\) | \(\ds \map \nu {\paren {A \cap N} \cup \paren {A \cap N^c} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \nu {A \cap N} + \map \nu {A \cap N^c}\) | countable additivity of the measure $\nu$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\nu_a} A + \map {\nu_s} A\) |
so:
- $\nu = \nu_a + \nu_s$
verifying $(3)$.
We have:
\(\ds \map {\nu_s} {N^c}\) | \(=\) | \(\ds \map \nu {N \cap N^c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \nu \O\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Empty Set is Null Set |
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:
\(\ds \map \nu {N \cup B}\) | \(=\) | \(\ds \map \nu N + \map \mu B\) | since $N$ is disjoint from $B$ | |||||||||||
\(\ds \) | \(>\) | \(\ds \map \nu N\) |
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.