My question is not research level, but I have not received any feedback on Mathstack; so I am posting it here. I am aware of the traditional proof of the Riesz Theorem that relates linear functionals on spaces of continuous functions on a locally compact space to measures. I want to do it in a different way, and would appreciate any comments.

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 $\mu (A)=\phi (\chi_A)$. Then $\mu $ is finitely additive, and in fact countably additive vacuously (because each element of $\mathcal A$ is compact). Using Carathéodory 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 $\tilde {\phi}$ 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; by 1), there is a measure $\tilde { \mu }$ such that $\tilde {\phi}(g)=\int _{2^{\mathbb N}}gd\tilde {\mu}$. Thus we compute $\phi (f)=\tilde {\phi}(f\circ p)=\int _{2^{\mathbb N}}f\circ pd\tilde {\mu}=\int _{X}fd(\tilde {\mu}\circ p^{-1})$. 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 space, and from there treat the locally compact case. I only have a very vague idea of how to proceed, but here is my idea:

Look first at the Baire sets of $X$; if $f:X\to Y$ is continuous to a compact metric space, and if $E$ is Baire (in this case it is Borel as well) in $Y$, then $f^{-1}(E)$ is Baire in $X$. 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 a 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 constitutes the Baire sets in $X$. Then, for each compact metric space $Y$ and each $f:X\to Y$, and 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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    $\begingroup$ I am not familiar with Carothers' proof. I got to thinking about this when I saw the proof of the fact that every compact metric space is the continuous image of a surjection, from $2^{\mathbb N}$- I thought that this approach might be easier than the traditional one. $\endgroup$ Jan 5, 2017 at 16:11
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    $\begingroup$ Please see math.stackexchange.com/a/1270384/17929 $\endgroup$ Jan 5, 2017 at 19:01
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    $\begingroup$ Might be good to acknowledge that the result is really Riesz-Markov-Kakutani... $\endgroup$ Jan 5, 2017 at 22:46
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    $\begingroup$ This is more or less how I do RRT when I teach Real Analysis. Carothers was my PhD student in the 18th or 19th century; probably he learned it from me. When doing the compact case, I first prove the RRT for a Boolean space $X$. This is easy since the clopen sets form an algebra that generates the Baire $\sigma$-algebra and countable additivity on the clopen sets is the same as finite additivity. The general case follows once you prove that every compact Hausdorff space $X$ is the continuous image of a Boolean space. TBC $\endgroup$ Jan 6, 2017 at 6:23
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    $\begingroup$ ... I used to do this last step the way you mention, but decided it was slightly easier to use the method mentioned in the link that Tomek Kania provided in his comment; moreover, this way gives another application of the Stone-Cech compactification. $$ $$ I have always wondered why the most used real analysis text books don't do the RRT in this way. $\endgroup$ Jan 6, 2017 at 6:24

1 Answer 1


A proof along the lines that you describe was worked out by V.S. Sunder here.

There are a few different approaches that you can take to reduce the Riesz Representation Theorem to a class of simpler spaces. For compact spaces, you can either reduce to compact metric spaces like Sunder does or directly to projective (i.e. extremally disconnected) compact spaces like Carothers.

Sunder's reduction to the compact metric case relies on the fact that every Baire set in a compact space is a continuous preimage of a Baire set in $[0, 1]^\mathbb{N}$. From here, you have two choices. You can either reduce to the case of $2^\mathbb{N}$ as you describe, or you can work directly with $[0, 1]^\mathbb{N}$. Sunder does the former, but I think you should also be able to do the latter and use an explicit proof for $[0, 1]$, e.g. in terms of Bernstein polynomials.

Carothers takes a more direct approach. Given a compact space $X$, he considers the projective cover of $X$ given by the Stone-Čech compactification of the discretization of $X$, where the clopen sets generate the Baire sets and finite additivity on clopen sets implies countable additivity. I personally feel that this better motivates advanced mathematics than the metric space proof, since the use of projective covers of topological spaces corresponds to the use of injective envelopes of $C^*$-algebras.

To generalize from compact spaces to locally compact spaces, you have a few choices. You can use the fact that locally compact spaces are compactly generated, which is roughly what Sunder does. You can also use compactifications, e.g. the one-point or Stone-Čech compactifications.

The use of the one-point compactification corresponds to the unitization of $C(X)$ as a $C^*$-algebra, where positive linear functionals have unique extensions to unitizations by taking a limit over a contractive approximate identity, i.e. again by approximating from compact subsets. If you worked it out explicitly in terms of topological spaces, it would look pretty similar to Sunder's proof.

The approach using the Stone-Čech compactification is a bit more interesting because it generalizes to arbitrary completely regular spaces. The best treatment of this I have seen is Measures on Topological Spaces by J.D. Knowles. The approach there uses both the Markov-Alexandrov approach of representing linear functionals on $C_b(X)$ in terms of finitely additive measures, as well as the Riesz-Kakutani approach on $C(\beta X)$ in terms of countably additive measures, and then studies the relationship between finite additivity, countable additivity, and inner approximations by compact sets.


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