Theorem: Lebesgue Decomposition Theorem/Finite Measure

Let $\nu$ be a finite measure on $\struct {X, \Sigma}$. 
Then there exist 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 $\NN_\mu$ by: 
:$\NN_\mu = \set {B \in \Sigma : \map \mu B = 0}$
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 $\set {\map \nu B : B \in \NN_\mu}$ exists as a real number $L$.
By the definition of the supremum, for each $n \in \N$ there exists $B_n \in \NN_\mu$ such that: 
:$L - \dfrac 1 n \le \map \nu {B_n} \le L$
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:
:$N \in \NN_\mu$ 
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: 
:$\map \nu N = \sup \set {\map \nu B : B \in \NN_\mu}$
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:
{{begin-eqn}}
{{eqn	| l = \map \nu A
		| r = \map \nu {\paren {A \cap N} \cup \paren {A \cap N^c} }
}}
{{eqn	| r = \map \nu {A \cap N} + \map \nu {A \cap N^c}
		| c = countable additivity of the measure $\nu$
}}
{{eqn	| r = \map {\nu_a} A + \map {\nu_s} A
}}
{{end-eqn}}
so:
:$\nu = \nu_a + \nu_s$ 
verifying $(3)$.
We have: 
{{begin-eqn}}
{{eqn	| l = \map {\nu_s} {N^c}
		| r = \map \nu {N \cap N^c}
}}
{{eqn	| r = \map \nu \O
}}
{{eqn	| r = 0
		| c = Empty Set is Null Set
}}
{{end-eqn}}
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, $B \in \NN_\mu$. 
{{AimForCont}} that $\map \nu B \ne 0$.
In particular, $\map \nu B > 0$. 
Then, we have:
{{begin-eqn}}
{{eqn	| l = \map \nu {N \cup B}
		| r = \map \nu N + \map \mu B
		| c = since $N$ is disjoint from $B$
}}
{{eqn	| o = >
		| r = \map \nu N
}}
{{end-eqn}}
However, from Null Sets Closed under Countable Union, we have: 
:$N \cup B$ is $\mu$-null.
So: 
:$\map \nu {N \cup B} \in \set {\map \nu B : B \in \NN_\mu}$, 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 $\mu$.
Hence the result.
{{qed}}
Category:Lebesgue Decomposition Theorem
Category:Measures
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