Radon-Nikodym Theorem/Lemma 1

Lemma

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 $\mu$.

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 $A \in \Sigma$.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $\mathcal F$.

Then:

for each $n \in \N$, the pointwise maximum $\max \set {f_1, f_2, \ldots, f_n}$ is contained in $\mathcal F$.

Proof

Let $A \in \Sigma$.

We first prove that for each $f, g \in \mathcal F$, we have:

$\max \set {f, g} \in \mathcal F$

From Measurable Functions Determine Measurable Sets, we have:

$\set {x \in X : \map f x < \map g x}$ is $\Sigma$-measurable

and:

$\set {x \in X : \map f x \ge \map g x}$ is $\Sigma$-measurable.

Define:

$A_1 = \set {x \in X : \map f x < \map g x} \cap A$

and:

$A_2 = \set {x \in X : \map f x \ge \map g x} \cap A$

For $x \in A_1$, we have:

$\map {\max \set {f, g} } x = \map g x$

and for $x \in A_1$, we have:

$\map {\max \set {f, g} } x = \map f x$

From Sigma-Algebra Closed under Countable Intersection, we have:

$A_1$ and $A_2$ are $\Sigma$-measurable.

Clearly we also have:

$A = A_1 \cup A_2$

with $A_1$ and $A_2$ disjoint.

From Pointwise Maximum of Measurable Functions is Measurable, we have:

$\max \set {f, g}$ is $\Sigma$-measurable.

Then:

\(\ds \int_A \max \set {f, g} \rd \mu\)

\(=\)

\(\ds \int_{A_1 \cup A_2} \max \set {f, g} \rd \mu\)

\(\ds \)

\(=\)

\(\ds \int_{A_1} \max \set {f, g} \rd \mu + \int_{A_2} \max \set {f, g} \rd \mu\)

\(\ds \)

\(=\)

\(\ds \int_{A_1} g \rd \mu + \int_{A_2} f \rd \mu\)

\(\ds \)

\(\le\)

\(\ds \map \nu {A_1} + \map \nu {A_2}\)

since $f, g \in \mathcal F$

\(\ds \)

\(=\)

\(\ds \map \nu A\)

since $\nu$ is countably additive

So, we have:

$\ds \int_A \max \set {f, g} \rd \mu \le \map \nu A$

for each $A \in \Sigma$.

So:

$\max \set {f, g} \in \mathcal F$.

We now prove that:

for each $n \in \N$, the pointwise maximum $\max \set {f_1, f_2, \ldots, f_n}$ is contained in $\mathcal F$.

by induction.

For all $n \in \N$ let $\map P n$ be the proposition:

$\max \set {f_1, f_2, \ldots, f_n} \in \mathcal F$

Basis for Induction

We have:

$\max \set {f_1} = f_1$

so:

$\max \set {f_1} \in \mathcal F$

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.

Our induction hypothesis is:

$\max \set {f_1, f_2, \ldots, f_n} \in \mathcal F$

We aim to show that:

$\max \set {f_1, f_2, \ldots, f_{n + 1} } \in \mathcal F$

Induction Step

We have:

$f_{n + 1} \in \mathcal F$

and:

$\max \set {f_1, f_2, \ldots, f_n} \in \mathcal F$

so:

\(\ds \max \set {f_1, f_2, \ldots, f_{n + 1} }\)

\(=\)

\(\ds \max \set {\max \set {f_1, f_2, \ldots, f_n}, f_{n + 1} }\)

Definition of Pointwise Maximum of Extended Real-Valued Functions: General Definition

\(\ds \)

\(\in\)

\(\ds \mathcal F\)

since the pointwise maximum of two elements of $\mathcal F$ is contained in $\mathcal F$

as required.

We therefore obtain:

for each $n \in \N$, the pointwise maximum $\max \set {f_1, f_2, \ldots, f_n}$ is contained in $\mathcal F$.

$\blacksquare$