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
We first prove the case of $\mu$ and $\nu$ finite.

Define $\mathcal F$ 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 $\mathcal F$ is non-empty.

From Integrable 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 $\mathcal F$.

We will now show that there exists $g \in \mathcal F$ such that:


 * $\ds \int f \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

and that this $g$ has:


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

for each $A \in \Sigma$.

Lemma 1
We now construct a increasing sequence of functions $\sequence {g_n}_{n \mathop \in \N}$ such that:


 * $\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

with a view to then apply the monotone convergence theorem.

Lemma 2
In the case:


 * $\ds \sup \set {\int f \rd \mu : f \in \mathcal F} = \infty$

we pick $f_n \in \mathcal F$ such that:


 * $\ds \int f_n \rd \mu \ge n$

for each $n \in \N$.

Then:


 * $\ds \lim_{n \mathop \to \infty} \int f_n \rd \mu = \infty$

Now consider the case:


 * $\ds \sup \set {\int f \rd \mu : f \in \mathcal F} < \infty$

we pick $f_n \in \mathcal F$ such that:


 * $\ds \sup \set {\int f \rd \mu : f \in \mathcal F} - \frac 1 n < \int f_n \rd \mu \le \sup \set {\int f \rd \mu : f \in \mathcal F}$

From the Squeeze Theorem, we have:


 * $\ds \lim_{n \mathop \to \infty} \int f_n \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

In either case, we have a sequence $\sequence {f_n}_{n \mathop \in \N}$ of functions in $\mathcal F$ such that:


 * $\ds \lim_{n \mathop \to \infty} \int f_n \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

So by Lemma 2, there exists an increasing sequence $\sequence {g_n}_{n \mathop \in \N}$ in $\mathcal F$ such that:


 * $\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

Now let:


 * $\ds g = \lim_{n \mathop \to \infty} g_n$

From the Monotone Convergence Theorem, we have:


 * $\ds \int g \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

Since $f_n \in \mathcal F$ 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 $g \in \mathcal F$.

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:


 * $g + \epsilon \chi_P \in \mathcal F$.

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:

Since we also have:


 * $\ds \int g \rd \mu = \sup \set {\int f \rd \mu : f \in \mathcal F}$

we have a contradiction.

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:


 * $\map g x < \infty$ for $\mu$-almost all $x \in X$

from Integrable Function is A.E. Real-Valued.

That is:


 * there exists a $\mu$-null set $N \subseteq X$ such that whenever $\map g x = \infty$ we have $x \in N$.

Define the function $h : X \to \hointr 0 \infty$ by:


 * $\map h x = \begin{cases}\map g x & \map g x < \infty \\ 0 & \text{otherwise}\end{cases}$

for each $x \in X$.

From Piecewise Combination of Measurable Mappings is Measurable, we have:


 * $h$ is $\Sigma$-measurable.

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


 * $h \times \chi_A$ is $\Sigma$-measurable

from Pointwise Product of Measurable Functions is Measurable.

From A.E. Equal Positive Measurable Functions have Equal Integrals, we then have:


 * $\ds \int \paren {h \times \chi_A} \rd \mu = \int \paren {g \times \chi_A} \rd \mu < \infty$

From the definition of the integral of a $\mu$-integrable function over a $\Sigma$-measurable set, we have:


 * $\ds \int_A h \rd \mu = \int_A g \rd \mu$

for each $A \in \Sigma$.

In particular:


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

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 3
For each $n \in \N$, define a function $\nu_n : \Sigma \to \overline \R$:


 * $\map {\nu_n} {A} = \map \nu {A \cap X_n}$

for each $A \in \Sigma$.

By Intersection Measure is Measure, we have:


 * $\nu_n$ is a measure.

We show that for each $n \in \N$:


 * $\nu_n$ is finite and absolutely continuous with respect to $\mu$.

We have:

so $\nu_n$ is finite 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 $\map \nu A = 0$.

From Intersection is Subset, we have:


 * $A \cap X_n \subseteq A$

for each $n \in \N$.

From Measure is Monotone, we have:


 * $\map \nu {A \cap X_n} \le \map \nu A$

So, if $\map \nu A = 0$, we have:


 * $\map \nu {A \cap X_n} \le 0$

giving:


 * $\map \nu {A \cap X_n} = \map {\nu_n} A \le 0$

so:


 * $\map {\nu_n} A = 0$

So:


 * for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map {\nu_n} A = 0$.

for each $n \in \N$.

So:


 * $\nu_n$ is absolutely continuous with respect to $\mu$

for each $n \in \N$.

Applying the finite case to each $\nu_n$:


 * for each $n \in \N$ there exists a $\Sigma$-measurable function $g_n : X \to \hointr 0 \infty$ such that $\ds \map {\nu_n} A = \int_A g_n \rd \mu$ for each $A \in \Sigma$.

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

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