Riesz-Markov theorem: Let $X$ be a locally compact Hausdorff space. For any continuous linear functional $\Psi$ on $C_0(X)$, there is a unique regular countably additive complex Borel measure $\mu$ on $X$ that $$ \forall f \in C_0(X): \Psi(f)=\int_X f(x)d\mu(x). $$
The norm of $\Psi$ as a linear functional is $||\Psi||=|\mu|(X)$. Finally, $\Psi$ is positive if and only if the measure $\mu$ is non-negative.
The different versions I have found of the Riesz-Markov theorem do not specify if the set of continuous functions on X which vanish at infinity are real or complex valued functions.
I want to say that if I have a positive linear functional defined on $C^\mathbb{R}_0(X)$ where $X$ is a compact Hausdorff space, and $||\Psi||=1$, then the measure given by the theorem is a probability measure. Any help on this topic? Thanks.
Best Answer
Given what you're trying to prove, it's probably most convenient to use the version of the Riesz-Markov theorem for positive functionals. This is the content of Theorems 2.14 and 2.17 in Rudin's Real and Complex Analysis.
Below, I have summarised the relevant conclusions from these theorems in the book. (I have simplified the hypotheses somewhat to make the statement easier to digest. Also, the statement in the book is written for $C^{\mathbb C}(X)$ rather than $C^{\mathbb R}(X)$, but the statement does hold for $C^{\mathbb R}(X)$; in fact, the book proves the statement first for $C^{\mathbb R}(X)$, and then deduces the statement for $C^{\mathbb R}(X)$ from this.)
Now let's address your question. $X$ is compact Hausdorff, and we have a positive linear functional $\Psi : C_0^{\mathbb R}(X) \to \mathbb R$ such that $\left\| \Psi \right\| = 1$.
First, a quick observation. Since $X$ is compact, $C_0^{\mathbb R}(X)$ is in fact the same thing as $C^{\mathbb R}(X)$.
Now let $\mu$ be the (positive real) regular Borel measure provided by the Riesz-Markov theorem, which satisfies $\Psi(f) = \int_X f \ d\mu$ for all $f \in C^{\mathbb R}(X)$.
If I understood your question correctly, you want to show that this $\mu$ is a probability measure. By construction, $\mu$ is a positive real measure. It remains to check that $\mu(X) = 1$.
To show that $\mu(X) = 1$, observe that the unit function $\mathbf 1_X$ is in $C^{\mathbb R}(X)$, and $$ \Psi(\mathbf 1_X) = \int_X \mathbf 1_X \ d\mu = \mu(X).$$
Meanwhile, $$ \left\| \Psi \right\| := \sup_{f \in C^{\mathbb R}(X), \ |f| \leq 1} |\Psi(f) |.$$ But $\Psi$ is a positive linear functional, so $f_1 \geq f_2 \implies \Psi(f_1) \geq \Psi(f_2)$. Thus $$ \Psi(-\mathbf 1_X) \leq |\Psi(f) | \leq \Psi(\mathbf 1_X) $$ for all $f \in C^{\mathbb R}(X)$ such that $|f| \leq 1$. So $$ \sup_{f \in C^{\mathbb R}(X), \ |f| \leq 1} |\Psi(f) | = \Psi(\mathbf 1_X).$$
Hence $$ \mu(X) = \Psi(\mathbf 1_X) = \left\| \Psi \right\| = 1,$$ which shows that $\mu$ is a probability measure.