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.

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 Restricted Measure on Trace Sigma-Algebra:


 * $\ds \int_A g_n \rd \mu \restriction_{\Sigma_n} = \int_A g_n^\ast \rd \mu$

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.