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

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.