Radon-Nikodym Theorem

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$ such that: 
:$\nu$ is absolutely continuous with respect to $\mu$.
Then there exists a $\Sigma$-measurable function $g : X \to \hointr 0 \infty$ such that: 
:$\ds \map \nu A = \int_A g \rd \mu$
for each $A \in \Sigma$.
Further, if $g_1 : X \to \hointr 0 \infty$ and $g_2 : X \to \hointr 0 \infty$ are such that: 
:$\ds \map \nu A = \int_A g_1 \rd \mu = \int_A g_2 \rd \mu$
for each $A \in \Sigma$, then: 
:$g_1 = g_2$ $\mu$-almost everywhere.
== Proof ==
=== Existence ===
We first prove the case of $\mu$ and $\nu$ finite. 
Define $\FF$ to be the set of $\Sigma$-measurable functions $f : X \to \overline \R_{\ge 0}$ with: 
:$\ds \int_A f \rd \mu \le \map \nu A$
for each $A \in \Sigma$.
We show that $\FF$ is non-empty. 
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have: 
:$\ds \int_A 0 \rd \mu = 0$
for each $A \in \Sigma$, giving: 
:$\ds \int_A 0 \rd \mu \le \map \nu A$
for each $A \in \Sigma$. 
So the constant $0$ function is contained in $\FF$. 
For each $f \in F$, we have: 
:$\ds \int f \rd \mu \le \map \nu X$
Since $\nu$ is a finite measure, we have: 
:$\ds \sup \set {\int f \rd \mu : f \in \FF} \le \map \nu X < \infty$ 
We will now show that there exists $g \in \FF$ such that: 
:$\ds \int g \rd \mu = \sup \set {\int f \rd \mu : f \in \FF}$
and that this $g$ has: 
:$\ds \map \nu A = \int_A g \rd \mu$
for each $A \in \Sigma$.
We now construct a increasing sequence of functions $\sequence {g_n}_{n \mathop \in \N}$ in $\FF$ such that: 
:$\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu = \sup \set {\int f \rd \mu : f \in \FF}$
with a view to then apply the monotone convergence theorem. 
We pick $f_n \in \FF$ such that: 
:$\ds \sup \set {\int f \rd \mu : f \in \FF} - \frac 1 n < \int f_n \rd \mu \le \sup \set {\int f \rd \mu : f \in \FF}$
==== Lemma 1 ====
For each $n \in \N$ define: 
:$g_n = \max \set {f_1, f_2, \ldots, f_n}$
From Lemma 1, we have: 
:$g_n \in \FF$ for each $n \in \N$. 
From the definition of the pointwise maximum, we have: 
So $\sequence {g_n}_{n \mathop \in \N}$ is an increasing sequence.
We have: 
and, since $g_n \in \FF$, we have: 
:$\ds \int g_n \rd \mu \le \sup \set {\int f \rd \mu : f \in \FF}$
from the definition of supremum.
Then from the Squeeze Theorem, we have: 
:$\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu = \sup \set {\int f \rd \mu : f \in \FF}$
Then $\sequence {g_n}_{n \mathop \in \N}$ is an increasing sequence in $\FF$ such that:
:$\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu = \sup \set {\int f \rd \mu : f \in \FF}$
Now let: 
:$\ds g = \lim_{n \mathop \to \infty} g_n$
From the Monotone Convergence Theorem, we have: 
:$\ds \int g \rd \mu = \sup \set {\int f \rd \mu : f \in \FF}$
Since $f_n \in \FF$ for each $n$, we have: 
:$\ds \int_A f_n \rd \mu \le \map \nu A$ 
for each $n$. 
From Lower and Upper Bounds for Sequences, we then have: 
:$\ds \int_A g \rd \mu \le \map \nu A$ 
So $g \in \FF$. 
We now verify that: 
:$\ds \map \nu A = \int_A g \rd \mu$
for each $A \in \Sigma$.
Define a function $\nu_1 : \Sigma \to \overline \R$ by: 
:$\ds \map {\nu_1} A = \int_A g \rd \mu$ 
for each $A \in \Sigma$. 
From Measure with Density is Measure, we have that: 
:$\nu_1$ is a measure.
Define a function $\nu_0 : \Sigma \to \overline \R$ by: 
:$\ds \map {\nu_0} A = \map \nu A - \map {\nu_1} A$ 
for each $A \in \Sigma$.
Since $\nu$ is a finite measure we have: 
:$\ds \infty > \map \nu A \ge \int_A g \rd \mu$
We have that: 
:$\ds \int_A g \rd \mu < \infty$ for each $A \in \Sigma$
and: 
:$g$ is $\mu$-integrable.
So the difference: 
:$\ds \map \nu A - \map {\nu_1} A$
is well-defined for each $A \in \Sigma$.
So, from Linear Combination of Signed Measures is Signed Measure, we have: 
:$\nu_0$ is a signed measure.
Since: 
:$\map {\nu_0} A \ge 0$ for each $A \in \Sigma$
we have that: 
:$\nu_0$ is a measure
from Non-Negative Signed Measure is Measure.
We want to show that: 
:$\map {\nu_0} A = 0$ for all $A \in \Sigma$.
, suppose that: 
:$\map {\nu_0} A \ne 0$ for some $A \in \Sigma$. 
Then from Measure is Monotone, we have: 
:$\map {\nu_0} X \ne 0$
Since $\mu$ is a finite measure, we have: 
:$\map \mu X < \infty$
So, we can pick $\epsilon > 0$ such that: 
:$\map {\nu_0} X > \epsilon \map \mu X$
From Linear Combination of Signed Measures is Signed Measure, we have that: 
:$\nu_2 = \nu_0 - \epsilon \mu$ is a signed measure.
Let $\struct {P, N}$ be the Hahn decomposition of $\nu_2$.
Since $P$ is a $\nu_2$-positive set, we have: 
:$\map {\nu_2} {A \cap P} \ge 0$ 
for each $A \in \Sigma$, since $A \cap P \subseteq P$ from Intersection is Subset. 
So:
:$\map {\nu_0} {A \cap P} \ge \epsilon \map \mu {A \cap P}$
Then for each $A \in \Sigma$, we have:
So:
:$g + \epsilon \chi_P \in \FF$. 
We show that $\map \mu P > 0$. 
 suppose that $\map \mu P = 0$.
Then since $\nu$ is absolutely continuous with respect to $\mu$, we have $\map \nu P = 0$.
Since: 
:$\ds \int_P g \rd \mu \le \map \nu P$
we then have: 
:$\ds \int_P g \rd \mu = 0$
and so:
:$\map {\nu_0} P = 0$
We then have: 
which contradicts: 
:$\map {\nu_0} X > \epsilon \map \mu X$
so we have: 
:$\map \mu P > 0$
From Integral of Characteristic Function: Corollary, we have: 
:$\ds \int \chi_P \rd \mu = \map \mu P > 0$
So, from Integral of Positive Measurable Function is Positive Homogeneous, we then have: 
:$\ds \int \epsilon \chi_P \rd \mu > 0$ 
Then: 
So:
:$\ds \int \paren {g + \epsilon \chi_P} \rd \mu > \sup \set {\int f \rd \mu : f \in \FF}$
contradicting the definition of supremum. 
So:
:$\map {\nu_0} A = 0$ for all $A \in \Sigma$.
So we have: 
:$\ds \map \nu A = \int_A g \rd \mu$
for each $A \in \Sigma$.
Since $g$ is $\mu$-integrable, we have that: 
:there exists a real-valued $\mu$-integrable function $h : X \to \R$ such that $g = h$ $\mu$-almost everywhere
from Measurable Function is Integrable iff A.E. Equal to Real-Valued Integrable Function.
From A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 2, we have:
:$\ds \map \nu A = \int_A h \rd \mu$
for each $A \in \Sigma$. 
So we are done in the case that $\mu$ and $\nu$ are finite. 
Now suppose that $\mu$ and $\nu$ are $\sigma$-finite.
==== Lemma 2 ====
Let $\Sigma_n$ be the trace $\sigma$-algebra of $X_n$ in $\Sigma$. 
From Trace Sigma-Algebra of Measurable Set, this consists precisely of the $\Sigma$-measurable subsets of $X_n$. 
Let $\mu \restriction_{\Sigma_n}$ and $\nu \restriction_{\Sigma_n}$ be the restrictions of $\mu$ and $\nu$ to $\Sigma_n$.
We show that for each $n \in \N$: 
:$\nu \restriction_{\Sigma_n}$ and $\mu \restriction_{\Sigma_n}$ are finite and $\nu \restriction_{\Sigma_n}$ is absolutely continuous with respect to $\mu \restriction_{\Sigma_n}$.
We have: 
:$\map {\nu \restriction_{\Sigma_n} } {X_n} = \map \nu {X_n} < \infty$
and: 
:$\map {\mu \restriction_{\Sigma_n} } {X_n} = \map \mu {X_n} < \infty$ 
so $\mu \restriction_{\Sigma_n}$ and $\nu \restriction_{\Sigma_n}$ are finite measures for each $n \in \N$. 
Since $\nu$ is absolutely continuous with respect to $\mu$, we have: 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.
So, in particular: 
:for all $A \in \Sigma$ with $A \subseteq X_n$ and $\map \mu A = 0$, we have $\map \nu A = 0$.
That is: 
:for all $A \in \Sigma_n$ with $\map {\mu \restriction_{\Sigma_n} } A = 0$, we have $\map {\nu \restriction_{\Sigma_n} } A = 0$.
Applying the finite case to each $\nu \restriction_{\Sigma_n}$:
:for each $n \in \N$ there exists a $\Sigma_n$-measurable function $g_n : X \to \hointr 0 \infty$ such that $\ds \map {\nu \restriction_{\Sigma_n} } A = \int_A g_n \rd \mu \restriction_{\Sigma_n}$ for each $A \in \Sigma_n$.
We also have that $g_n$ is $\Sigma$-measurable for each $n \in \N$. 
Note that we have: 
:$g_n^{-1} \sqbrk B \in \Sigma_n$ for each Borel set $B \subseteq \R$.
Since $\Sigma_n \subseteq \Sigma$, we have: 
:$g_n^{-1} \sqbrk B \in \Sigma$ for each Borel set $B \subseteq \R$ for each $n \in \N$. 
So: 
:$g_n$ is $\Sigma$-measurable for each $n \in \N$.
From Integral with respect to Restriction of Measure to Trace Sigma-Algebra of Measurable Set:
:$\ds \int_A g_n \rd \mu \restriction_{\Sigma_n} = \int_A g_n^\ast \rd \mu$
for each $A \in \Sigma_n$, where we define the function $g_n^\ast : X \to \overline \R$ by: 
:$\ds \map {g_n^\ast} x = \begin{cases}\map {g_n} x & x \in X_n \\ 0 & x \not \in X_n\end{cases}$
Then, for each $A \in \Sigma$, we have: 
 
So the function $g : X \to \overline \R$ defined by:
:$\ds g = \sum_{n \mathop = 1}^\infty g_n^\ast$
satisfies the desired integral identity. 
We can show that $g$ is in fact a function $X \to \hointr 0 \infty$, completing the existence proof.
Clearly $g \ge 0$, since $g_n^\ast \ge 0$ for each $n \in \N$. 
Let $x \in X$. 
Then: 
:$x \in X_n$ for exactly one $n \in \N$
while: 
:$\map {g_m^\ast} x = 0$ if $x \not \in X_m$.
So:
:$\map {g_n^\ast} x = 0$ for all but possibly one $n \in \N$. 
Then: 
:$\ds \sum_{n \mathop = 1}^\infty \paren {\map {g_n^\ast} x} = \map {g_{n_x}^\ast } x$
for some $n_x \in \N$ depending on $x$. 
We have: 
:$\map {g_{n_x} } x < \infty$
We therefore have:
:$\ds \sum_{n \mathop = 1}^\infty \map {g_n^\ast} x < \infty$
for all $x \in X$. 
So: 
:$\map g x < \infty$
for all $x \in X$. 
So $g$ is a function $X \to \hointr 0 \infty$ as required.
=== Essential Uniqueness ===
Suppose that the positive $\Sigma$-measurable functions $g_1 : X \to \hointr 0 \infty$ and $g_2 : X \to \hointr 0 \infty$ are such that: 
:$\ds \map \nu A = \int_A g_1 \rd \mu = \int_A g_2 \rd \mu$
for all $A \in \Sigma$.
From Lemma 2 there exists a sequence $\sequence {X_n}_{n \mathop \in \N}$ of $\Sigma$-measurable sets with: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$
and:
:$\map \mu {X_n} < \infty$ and $\map \nu {X_n} < \infty$ for each $n$.
Let: 
:$A_1 = \set {x \in X : \map {g_1} x < \map {g_2} x}$
and:
:$A_2 = \set {x \in X : \map {g_1} x \ge \map {g_2} x}$
From Measurable Functions Determine Measurable Sets, $A_1$ and $A_2$ are $\Sigma$-measurable.
From Sigma-Algebra Closed under Countable Intersection:
:$A_1 \cap X_n$ and $A_2 \cap X_n$ are $\Sigma$-measurable for each $n$. 
Then for each $n$, we have: 
and:
Note that $A_1$ and $A_2$ are disjoint with: 
:$X = A_1 \cup A_2$
and: 
similarly:
Since the sequence $\sequence {X_n}_{n \mathop \in \N}$ is pairwise disjoint:
:the sequences $\sequence {A_1 \cap X_n}_{n \mathop \in \N}$ and $\sequence {A_2 \cap X_n}_{n \mathop \in \N}$ are pairwise disjoint.
We can then compute: 
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we then have: 
:$g_1 - g_2 = 0$ $\mu$-almost everywhere.
So:
:$g_1 = g_2$ $\mu$-almost everywhere
hence the claim.
== Sources ==
* : $4.2$
Category:Radon-Nikodym Theorem
Category:Signed Measures