Riesz-Markov-Kakutani Representation Theorem

== 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 ===
=== Construction of $\mu$ and $\MM$ ===
== Proof ==
=== Lemma $1$ ===
=== Lemma $2$ ===
=== Lemma $3$ ===
=== Lemma $4$ ===
=== Lemma $5$ ===
=== Lemma $6$ ===
=== Lemma $7$ ===
=== Lemma $8$ ===
=== Lemma $9$ ===
Let $K\subset X$ be compact. 
Suppose $A \in \MM$. 
Then, by Lemma $8$: 
:$A^C \cap K = K \setminus \paren {A \cap K} \in \MM_F$
So, $\MM$ is closed under complement. 
Let $\sequence {A_n} \in \MM^\N$ and $A = \ds \bigcup_{i \mathop = 1}^\infty$.
Let $B_1 = A_1 \cap K$ and:
:for $n \ge 2: B_n = \paren {A_n \cap K} \setminus \ds \bigcup_{i \mathop = 1}^{n - 1} B_i$
$\sequence {B_n}$ is disjoint and, by Lemma $8$, a sequence of $\MM_F$.
Then, by Lemma $7$:
:$A \cap K = \ds \bigcup_{i \mathop = 1}^\infty \in \MM_F$
Therefore, $M$ is closed under countable union. 
Now:
:$C \text{ closed} \implies C \cap K \text{ compact} \implies C \cap K \in \MM_F \implies C \in \MM$.
Therefore, by Borel $\sigma$-algebra generated by closed sets, $\MM$ is a $\sigma$-algebra which contains the Borel $\sigma$-algebra of $\struct {X, \tau}$.
Now, by Lemma $2$, the definition of $\mu$, Lemma $9$ and Lemma $4$, $\mu$ is a Radon measure on $\MM$.
Let $f \in \map {C_c} X $ be real. 
Let $K = \supp f$ and $\epsilon \in \R_{>0}$.
Let $\tuple {a, b} \in \R^2$ be such that $\closedint a b \supset K$.
Choose $\sequence {y_i} \in \R^n$ such that for all $i$, $y_i - y_{i - 1} < \epsilon$ and:
:$y_0 < a < y_1 < \cdots < y_n = b$
Define:
:$E_i = \map {f^{-1} } {\hointl {y_{i - 1} } {y_1} } \cap K$
By definition of $\mu$, there exist open sets $W_i \supset E_i$ such that:
:$\map \mu {W_i} < \map \mu {E_i} + \dfrac \epsilon n$
Define:
:$V_i = W_i \cap \map {f^{-1} } {\openint \gets {y_i + \epsilon} }$
Let $\set {h_i}$ be a partition of unity on $K$ subordinate to $\set {V_i}$. 
Then $\ds f = \sum h_i f$ on $K$ and, by Lemma 2:
:$\ds \map \mu K \le \Lambda \sum_{i \mathop = 1}^n h_i = \sum_{i \mathop = 1}^n \Lambda h_i$
Since, for all $i$, for all $x \in E_i$:
:$h_i f \le \paren {y_i + \epsilon} h_i$
and:
:$y_i - \epsilon < \map f x$
we have:
Since $\epsilon$ was arbitrary, for all real $f \in \map{C_c} X$:
:$\Lambda f \le \int_X f \rd \mu$
Now, by linearity of $\Lambda$:
:$-\Lambda f = \map \Lambda {-f} \le \int_X \paren {-f} \rd \mu = -\int_X f \rd \mu$
Therefore, for all real $f \in \map {C_c} X$:
:$\Lambda f = \int_X f \rd \mu$
Suppose $u, v \in \map {C_c} X$ are real.
Then by linearity of $\Lambda$:
:$\map \Lambda {u + i v} = \Lambda u + i \Lambda v = \int_X u \rd \mu + i \int_X v \rd \mu = \int_X \paren {u + i v} \rd \mu$
That is, equality holds for all functions in $\map {C_c} X$.
Suppose $\mu_1$ and $\mu_2$ are two Radon measures on $\MM$. 
For all compact $K \subset X$ and $\epsilon>0$, there exists an open $V \supset K$ such that:
:$\map {\mu_2} V < \map {\mu_2} K + \epsilon$
By Urysohn's Lemma, there exists:
:$f \in \map{C_c} X: K \prec f \prec V$
Then:
Thus:
:$\map {\mu_1} K \le \map {\mu_2} K$
Interchanging $\mu_1$ and $\mu_2$ yields the opposite inequality.
Therefore, $\mu_1$ and $\mu_2$ coincide on all compact sets. 
By the definition of a Radon measure, the measure of all measurable sets is uniquely determined by the measures of the compact sets. 
So:
:$\mu_1 = \mu_2$
and the measure is unique, completing the proof.
Category:Measure Theory
Category:Riesz-Markov-Kakutani Representation Theorem