Riesz-Markov-Kakutani Representation Theorem/Lemma 4

This article needs to be tidied. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code.

This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.

Lemma for Riesz-Markov-Kakutani Representation Theorem

Let $\struct {X, \tau}$ be a locally compact Hausdorff space.

Let $\map {C_c} X$ be the space of continuous complex functions with compact support on $X$.

Let $\Lambda$ be a positive linear functional on $\map {C_c} X$.

There exists a $\sigma$-algebra $\MM$ over $X$ which contains the Borel $\sigma$-algebra of $\struct {X, \tau}$.

There exists a unique complete Radon measure $\mu$ on $\MM$ such that:

$\ds \forall f \in \map {C_c} X: \Lambda f = \int_X f \rd \mu$

Notation

For an open set $V \in \tau$ and a mapping $f \in \map {C_c} X$:

$f \prec V \iff \supp f \subset V$

where $\supp f$ denotes the support of $f$.

The validity of the material on this page is questionable. In particular: The proof does not work with this definition. Something should be forgotten. Maybe, $f \prec V \iff 0 \le f \le {\mathbf 1}_V$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

For a compact set $K \subset X$ and a mapping $f \in \map {C_c} X$:

$K \prec f \iff \forall x \in K: \map f x = 1$

Construction of $\mu$ and $\MM$

For every $V \in \tau$, define:

$\map {\mu_1} V = \sup \set {\Lambda f: f \prec V}$

The validity of the material on this page is questionable. In particular: The definition of $\mu_1$ seems wrong, as $\map {\mu_1} V \in \set {0, +\infty}$ for all $V$. Indeed, for each $c>0$, $f \prec V \iff c f \prec V $. This means $\map {\mu_1} V = c \map {\mu_1} V$ for all $c > 0$. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Note that $\mu_1$ is monotonically increasing.

That is, for all $V, W \in \tau$ such that $V \subset W$, we have:

\(\ds \map {\mu_1} V\)

\(=\)

\(\ds \sup \set {\Lambda f: \supp f \subset V}\)

\(\ds \)

\(\le\)

\(\ds \sup \set {\Lambda f: \supp f \subset W}\)

\(\ds = \map {\mu_1} W\)

$\Box$

For every other subset $E \subset X$, define:

$\map \mu E = \inf \set {\map {\mu_1} V: V \supset E \land V \in \tau}$

Since $\mu_1$ is monotonically increasing:

$\mu_1 = \mu {\restriction_\tau}$

Define:

$\MM_F = \set {E \subset X : \map \mu E < \infty \land \map \mu E = \sup \set {\map \mu K: K \subset E \land K \text { compact} } }$

Define:

$\MM = \set {E \subset X : \forall K \subset X \text { compact}: E \cap K \in \MM_F}$

Lemma

$\mu$ is countably additive over pairwise disjoint collections of subsets of $\MM_F$.

Proof

Let $\sequence {E_i} \in \paren {\MM_F}^\N$ be pairwise disjoint with union $E$.

Let $\map \mu E = \infty$.

Then, by countable subadditivity:

$\ds \infty = \map \mu E \le \sum_{i \mathop = 1}^\infty \map \mu {E_i}$

This article, or a section of it, needs explaining. In particular: Link to evidence that $\map \mu {E_i}$ is subadditive You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

So:

$\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu E$

Suppose $\map \mu E < \infty$.

By definition of $\MM_F$, for all $\epsilon \in \R_{>0}$, for each $i$, there exists a compact $H_i \subset E_i$ such that:

$\map \mu {H_i} > \map \mu {E_i} - 2^{-i} \epsilon$

So:

\(\ds \map \mu E\)

\(\ge\)

\(\ds \map \mu {\bigcup_{i \mathop = 1}^\infty H_i}\)

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^\infty \map \mu {H_i}\)

Lemma 3

\(\ds \)

\(>\)

\(\ds -\epsilon + \sum_{i \mathop = 1}^\infty \map \mu {E_i}\)

This holds for all $n \in \N$.

So $\mu$ is countably superadditive over pairwise disjoint collections of subsets of $\MM_F$.

Therefore, by Lemma 1, $\mu$ is countably additive over pairwise disjoint collections of subsets of $\MM_F$.

$\blacksquare$