Let $X$ be a compact Hausdorff space, $C(X)$ the set of all real continuous functions on $X$, and $\mathcal{B}$ be the Baire $\sigma$-algebra of $X$, which is the $\sigma$-algebra generated by the functions in $C(X)$. Furthermore $M(X)$ is the set of all finite signed measures on $(X,\mathcal{B})$ with the norm $||\mu||=|\mu|(X)$, $|\mu|$ being the total variation of $\mu$. Finally, for all $\mu \in M(X)$ define $\phi_{\mu}:C(X)\rightarrow \mathbb{R}$ by $\phi_{\mu}(f)=\int f d\mu$.

It is possible to formulate the Riesz-Markov-Kakutani theorem as follows:

The application $\mu\mapsto \phi_{\mu}$ is a surjective isometry from $M(X),||.||$ onto $C(X)^{*},||.||_{*}$ the dual space of $C(X),||.||_{\infty}$.

In other words the topological dual of the Banach space $(C(X),||.||_{\infty})$ (with as usual $||f||_{\infty}=\sup_{x\in X} f(x))$) can be identified with $M(X)$.

My question is: since there is a functional analysis formulation of the theorem, is there a functional-analysis-flavoured proof of it ?

Note that this version of the Riesz-Markov-Kakutani theorem is much stronger than the usually stated one, which is concerned positive functionals on $\mathbb{R}$. The fact that the dual norm is the total variation one is equivalent to the fact that Baire measures are necessarily regular, a not so trivial fact proved in Halmos's Measure Theory.

I find proofs like the one in Rudin's Real and complex analysis disturbingly artificial and complex for such what seems such a natural and important result: an integral is nothing more or less than a bounded linear functional on continuous functions (when integrating over a compact space).

I had a professor who used Daniell's integral with which the Riesz-Markov-Kakutani theorem follows almost instantly. That is a very good proof, the best I've seen so far, but I'm still under the impression that there might be other ways to look at the problem.


1 Answer 1


My favourite proof is due to Hartig:

D. G. Hartig, The Riesz representation theorem revisited, American Mathematical Monthly, 90(4), 277–280.

Hartig claims that his proof is ...category-theoretic but having unwrapped the details I must say it is really functional-analytic. The idea is as follows.

  • Step 1. We can easily prove this theorem for extremely disconnected compact Hausdorff spaces. Indeed, everything can be reduced there to messing around with indicator functions of clopen subsets.

  • Step 2. We employ the Stone–Čech compactification of a discrete space. One can use the Banach–Alaoglu theorem to show that every completely regular space admits the Stone–Čech compactification. The Stone–Čech compactification of a discrete space is extremely disconnected.

  • Step 3. Now, we prove that every compact space $X$ is a continuous image of an extremely disconnected space. Indeed, give $X$ the discrete topology. Call this space $X_d$ and extend the identity map $\iota \colon X_d \to X$ to a continuous map from $\beta X_d$ to $X$. Then, transfer the Riesz theorem to $X$ via the adjoint map to $f\mapsto f\circ (\beta\iota)$ ($f\in C(X)$).

  • $\begingroup$ The proof is due to Garling, Hartig is only giving an exposition. $\endgroup$ Jan 7, 2017 at 0:32

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