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Let$X$ be a compact Hausdorff space, $C(X)$ the space of continuous functions on $X$, and $\mathcal B_X$ the Borel sets on $X$. Then, if $\phi:C(X)\to \mathbb C$ is a positive linear functional, there is a measure $\mu:\mathcal B_X \to \mathbb R$. such that $\phi (f)=\int_X fd\mu$ for each $f\in C(X)$.

I want to prove this using the fact that every compact metric space $X$ is the continuous image of a surjection, $p$, from $2^{\mathbb N}$. Of course, $2^{\mathbb N}$ has the product topology which is also induced by the metric $d(x,y)=1/n$ where $n$ is the least integer such that $x_n\neq y_n$. Here is an outline, and then questions at the end:

1). First, suppose $X=2^{\mathbb N}$. Then, $\mathcal A=\left \{ \pi^{-1}_n(E):E\in 2^{n} \right \}$ is a subbase for the topology of $2^{\mathbb N}$, and so is an algebra that generates $\mathcal B_{2^{\mathbb N}}$. We have that $\chi_A$ is continuous for each $A\in \mathcal A$ so we set $\mu (A)=\phi (\chi_A)$. $\mu $ is finitely additive, and in fact countably additive vacuously because each element of $\mathcal A$ is compact. Using Carotheodory and a density argument now gives the result if $X=2^{\mathbb N}$.

2). Let $X$ be any compact metric space. We may assume that $\left \| \phi \right \|=1$. Moreover, it is not hard to show that in general $\phi$ is a positive linear functional if and only if $\left \| \phi \right \|=\phi (1)$. Now, the map $f\mapsto f\circ p$ is an isometry and $M=\left \{ f\circ p:f\in C(X) \right \}$ is a subspace of $C(2^{\mathbb N})$. Then, $\tilde{\phi}_M(f\circ p):=\phi (f)$ is a functional on $M$ which extends via Hahn-Banach, to a $\tilde {\phi}$ on all of $C(2^{\mathbb N})$. Since $\left \| \tilde { \phi }\right \|=\left \| \tilde { \phi_M }\right \|=\left \| \phi \right \|=\phi (1)=\tilde { \phi }(1\circ p)=\tilde { \phi (1) }$, $\tilde { \phi }$ is a positive linear functional, so by 1)., there is a measure $\tilde { \mu }$ such that $\tilde {\phi}(g)=\int _{2^{\mathbb N}}gd\tilde {\mu}$. Thus we compute that $\phi (f)=\tilde {\phi}(f\circ p)=\int _{2^{\mathbb N}}f\circ pd\tilde {\mu}=\int _{X}fd(\tilde {\mu}\circ p^{-1})$. All that remains is uniqueness of the measure $\nu =\tilde {\mu}\circ p^{-1}$, but this follows exactly as in the traditional proof.

Now I want to extend this to an arbitrary compact Hausdorff spaceand from there treat the locally compact case. I only have a very vague idea of how to proceed, but here is my idea:

Now look first at the Baire sets of $X$, then if $f:X\to Y$ is continuous into a compact metric space, then if $E$ is Baire (in this case it is Borel as well) in $Y$, then $f^{-1}(E)$ is Baire in $X$. Then, consider the collection $\mathcal C$ of all pairs $(f,Y)$ such that $f:X\to Y$ is a continuous map into a compact metric space. Now consider all sets of the form $f_Y^{-1}(E)$ where $Y$ is any compact metric space, $f:X\to Y$ is continuous and $E$ is Baire in Y. Next I would show that the union of all these is exactly the Baire sets in $X$. Then, for each compact metric space $Y$, and each $f:X\to Y$, using the trick in 2), I find a measure $\mu_Y$ such that $\int_Y gd\mu_Y=\tilde {\phi}(g):=\phi (g\circ f)$. Finally, I patch these measures together to get the measure on the Baire sets of $X$ and from there, to the Borel sets.

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