2
$\begingroup$

I was reading about Riesz Representation Theorem for compact Hausdorff spaces $X$ on Reed & Simon's book. As far as I understood, one starts with the $\sigma$-algebra generated by $G_{\delta}$ sets (countable intersections of open sets), which is called the Baire $\sigma$-algebra and a Baire measure $\mu$ which is positive and finite measure on this $\sigma$-algebra.

We want to study the dual of $C(X)$, the space of all continuous functions $f: X \to \mathbb{C}$, which is a Banach space when equipped with the sup norm. If $f \in C(X)$ is real valued, it is automatically measurable with respect to the Baire $\sigma$-algebra; however, this is not true for general complex valued functions on $C(X)$. Thus, we ultimately we want to consider the Borel $\sigma$-algebra on $X$; it turns out that to every Baire measure on the Baire $\sigma$-algebra there exists a unique regular Borel measure on the Borel $\sigma$-algebra, so there is a natural identification between Baire measures and Borel measures.

Given a regular Borel measure $\mu$, the mapping $C(X) \ni f \mapsto \mu(f) := \int f \mu$ is a positive linear functional. Riesz Representation Theorem states that every positive linear functional is precisely of that form, so that $C(X)^{*}$ is identified with the set of regular Borel measures on $X$.

Question 1: Is this all correct?

Question 2: Can someone please point out a reference which provides the proofs of the above facts? I know this is quite standard material, but it is really hard to find more focused material. Most of the references I know prove the results for locally compact spaces, or compact metric spaces and I don't want to lose myself in all these different cases.

$\endgroup$
2
  • $\begingroup$ I haven't read too carefully but Royden's real analysis covers various different Riesz representation theorems $\endgroup$ Commented Jan 23, 2023 at 21:06
  • $\begingroup$ (fourth edition, at least) $\endgroup$ Commented Jan 23, 2023 at 21:15

2 Answers 2

Reset to default
3
+50
$\begingroup$

No: complex-valued continuous functions are Baire measurable. The real and imaginary parts of a $\mathbb C$-valued continuous function are $\mathbb R$-valued continuous functions.
Note: even for compact $X$, you do not use merely $G_\delta$ sets to generate the Baire sets, but closed $G_\delta$ sets.

When the compact space $X$ is metrizable, Borel=Baire and all measures are regular. When $X$ is not metrizable, it could happen that two different Borel measures induce the same linear functional on $C(X)$. So, to represent the dual space $C(X)^*$, you can either (i) use Baire measures, or (ii) require regularity.

A reference:

Varadarajan, V. S., Measures on topological spaces, Am. Math. Soc., Transl., II. Ser. 48, 161-228 (1965); translation from Mat. Sb., n. Ser. 55(97), 35-100 (1961). ZBL0152.04202.

Of course it considers not only compact $X$. For completely regular Hausdorff $X$ we can define the Baire sets as the sigma-algebra generated by the zero-sets. [Here, "zero-sets" are sets of the form $\{x \in X : f(x) = 0\}$ where $f : X \to \mathbb R$ is continuous.] If $X$ is compact Hausdorff, these are the same Baire sets. But in general $$ \{\text{compact }G_\delta\} \subsetneq \{\text{zero-sets}\} \subsetneq \{\text{closed }G_\delta\} $$

$\endgroup$
1
$\begingroup$

This Answer is on helpful comments purely on Representation theorem part:

If you are applying Riesz representation theorem derived in Hilbert space, note that $C(X)$ is not complete w.r.t inner product $<f,g>= \int fg d\mu $. So we have to go to a completion of $C(X)$ w.r.t the norm induced by this product and apply Riesz representation in the resulting completed space (a Hilbert space). So when we apply Riesz representation theorem and write $L(f) = <f,g_L> = \int f g_L d\mu $ where $\mu$ is a borel measure, we have that $g_L$ need not be in $C(X)$ but actually in $L_2(X)$. Further by Hahn banach theorem we can extend $L$ to apply also in the completion $L_2(X)$. So you can also identify the dual space $C^*(X)$ with positive functions in $L_2(X)$. But for above to apply we need $L$ to be of bounded norm when applied on $C(X)$ w.r.t norm on $C(X)$ induced by this inner product.

EDIT: There is another version of Riesz theorem w.r.t measure theory which applies to this case. See: https://en.m.wikipedia.org/wiki/Riesz%E2%80%93Markov%E2%80%93Kakutani_representation_theorem#:~:text=In%20mathematics%2C%20the%20Riesz%E2%80%93Markov,to%20measures%20in%20measure%20theory.

This theorem shows that if $L(f) \leq L(g)$ for any $f,g$ such that $f(x) \leq g(x)$, $\forall x \in X$ then such an $L$ can be represented as $L(f) = \int f d\mu_R$ for a unique Radon Measure $\mu_R$. Hence this applies for any such monotonic Linear functional $L$. If we have a decomposition that $L=L_1 - L_2$ with $L_1,L_2$ positive Linear functional then a form of Riesz Representation theorem holds for this $L$ on $C(X)$ with $X$ being compact Hausdorff space i.e., $L = \int f d\mu_{R}-\int f d\mu'_{R}$. I find such a decomposition interesting.

This extends to Locally Compact Hausdorff space $X$ but now we can represent only linear functionals on $C_c(X)$. https://math.berkeley.edu/~arveson/Dvi/rieszMarkov.pdf Can we extend $C_c(X)$ to $C(X)$ for $X$ being Locally compact Hausdorff space ?

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.