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.