User:Caliburn/s/mt/Radon-Nikodym Theorem/Signed Measure

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

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

Let $\nu$ be a finite signed measure on $\struct {X, \Sigma}$ such that:


 * $\nu$ is absolutely continuous with respect to $\mu$.

Then there exists a $\Sigma$-measurable function $h : X \to \hointr 0 \infty$ such that:


 * $\ds \map \nu A = \int_A h \rd \mu$

for each $A \in \Sigma$.

Further, if $h_1 : X \to \hointr 0 \infty$ and $h_2 : X \to \hointr 0 \infty$ are such that:


 * $\ds \map \nu A = \int_A h_1 \rd \mu = \int_A h_2 \rd \mu$

for each $A \in \Sigma$, then:


 * $h_1 = h_2$ $\mu$-almost everywhere.

Existence
Let $\tuple {\nu^+, \nu^-}$ be the Jordan decomposition of $\nu$.

From Absolute Continuity of Signed Measure in terms of Jordan Decomposition, we have:


 * $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.

From the Radon-Nikodym Theorem, there exists $\Sigma$-measurable function $g_1 : X \to \hointr 0 \infty$ such that:


 * $\ds \map {\nu^+} A = \int_A g_1 \rd \mu$

for each $A \in \Sigma$.

Applying the Radon-Nikodym Theorem again, there exists $\Sigma$-measurable function $g_2 : X \to \hointr 0 \infty$ such that:


 * $\ds \map {\nu^-} A = \int_A g_2 \rd \mu$

for each $A \in \Sigma$.

From Jordan Decomposition of Finite Signed Measure, we have:


 * $\nu^+$ and $\nu^-$ are finite measures.

So, we have:


 * $\ds \map {\nu^+} X = \int g_1 \rd \mu < \infty$

and:


 * $\ds \map {\nu^-} X = \int g_2 \rd \mu < \infty$

so $g_1$ and $g_2$ are $\mu$-integrable functions.

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

Essential Uniqueness
Suppose that $\Sigma$-measurable $h_1 : X \to \hointr 0 \infty$ and $h_2 : X \to \hointr 0 \infty$ are such that:


 * $\ds \map \nu A = \int_A h_1 \rd \mu = \int_A h_2 \rd \mu$

for each $A \in \Sigma$.

Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$.

From the definition of the Jordan decomposition and Uniqueness of Jordan Decomposition, we have:


 * $\map {\nu^+} A = \map \nu {A \cap P}$

and:


 * $\map {\nu^-} A = -\map \nu {A \cap N}$

for each $A \in \Sigma$.

We therefore have:

and:

So $h_1 \times \chi_P : X \to \hointl 0 \infty$ and $h_2 \times \chi_P : X \to \hointl 0 \infty$ are such that:


 * $\ds \map {\nu^+} A = \int_A \paren {h_1 \times \chi_P} \rd \mu = \int_A \paren {h_2 \times \chi_P} \rd \mu$

From the essential uniqueness part of the Radon-Nikodym Theorem, we therefore have:


 * $h_1 \times \chi_P = h_2 \times \chi_P$ $\mu$-almost everywhere.

Similarly, we have:

and:

So $h_1 \times \chi_N : X \to \hointl 0 \infty$ and $h_2 \times \chi_N : X \to \hointl 0 \infty$ are such that:


 * $\ds \map {\nu^-} A = \int_A \paren {h_1 \times \chi_N} \rd \mu = \int_A \paren {h_2 \times \chi_N} \rd \mu$

From the essential uniqueness part of the Radon-Nikodym Theorem, we therefore have:


 * $h_1 \times \chi_N = h_2 \times \chi_N$ $\mu$-almost everywhere.

From the definition of Hahn decomposition, we have that $P$ and $N$ are disjoint with:


 * $X = P \cup N$

so that:

and:

From Pointwise Addition preserves A.E. Equality: General Result, we have:


 * $h_1 \times \chi_P + h_1 \times \chi_N = h_2 \times \chi_P + h_2 \times \chi_N$ $\mu$-almost everywhere

so that:


 * $h_1 = h_2$ $\mu$-almost everywhere.