# Riesz's representation theorem for non-locally compact spaces

Every version of Riesz's representation theorem (the one expressing linear functionals as integrals) that I have found so far assumes that the underlying topological space is locally-compact. (For instance, Rudin in "Real and Complex Analysis" uses this assumption in the proof of Urysohn's lemma, upon which he bases the proof of Riesz's theorem.) Nevertheless, in an answer on MO, @jbc (inactive on MO since 2013 and with no real name or contact information available) claims that the theorem can be stated in much more generality, for Hausdorff completely regular spaces (which suggests that Urysohn's lemma is indeed the essential ingredient here). Unfortunately, no detail or bibliography is given in that answer. Currently, I have only found a practically useless theorem in Bourbaki's "Elements of Mathematics - Integration II", chapter IX, paragraph 5, page IX.59, and a very difficult to use version in "Bounded Continuous Functions On A Completely Regular Space" by Dennis F. Sentilles ("Transactions of the AMS", vol.168, June 1972, page 311, theorems 9.1.c and 9.2).

My question is: does anybody know of a "definitive", recent statement concerning the dual of the algebra of bounded continuous functions on non-locally-compact spaces? What is lost when one gives up local-compactness? (Please notice that I am not interested in the algebra of functions with compact support or vanishing at infinity.)

• You certainly do need continuity in the locally compact case. A discontinuous linear functional is not a measure, and Zorn's lemma shows that discontinuous linear functionals exist. You may be thinking of the fact that a positive linear functional is automatically continuous. – Nate Eldredge Mar 9 '14 at 14:53
• By the way, I don't know the answer offhand, but the first place I would look is in Dunford and Schwartz. They have lots of results classifying the duals of various spaces. – Nate Eldredge Mar 9 '14 at 14:54
• Thank you @NateEldredge, I have corrected my question following your comment. Indeed, I was ignoring the fact that positivity implies continuity. – Alex M. Mar 9 '14 at 16:29
• @NateEldredge: I've just looked in Dunford & Schwartz: if X is normal then $C(X)^*$ is isometrically and order-preserving isomorphic to $rba(X)$ (the space of finitely-additive regular complex measures); if X is compact and Hausdorff then $C(X)^*$ is isometrically and order-preserving isomorphic to $rca(X)$ (as above, but countably-additive). These results are weaker than what I'm looking for, but this is natural given that the edition that I have dates back to 1957, when the results that I'm looking for were still in their infancy. – Alex M. Mar 9 '14 at 16:54
• The theorem of Bourbaki mentioned by the OP also appears in Bogachev, Measure Theore, Volume II, as Theorem 7.10.6. See also p.446-447 for some discussion and other references. – MichaelGaudreau Sep 24 '18 at 6:50

The basic idea is that of the strict topology on $C^b(X)$. This locally convex topology was introduced in the case of a locally compact space by R.C. Buck in the fifties using weighted seminorms and generalised to the completely regular case by many authors in the sixties and seventies. It can be succinctly described as the finest locally convex topology which agrees with that of compact convergence on the unit ball for the supremum norm and the dual is the space of bounded Radon measures. One of many references: "Bounded measures in topological spaces" by Fremlin, Garling and Haydon (Proc. Lond. Math. Soc. 25 (1972) 115-135). The role of complete regularity is to ensure that the space of continuous functions is large enough for the purposes of this result.

Historical notes of chapter 9 of Bourbaki's integration give the following as original reference for the case of completely regular spaces:

A. D. Alexandroff, Additive set functions in abstract spaces, Mat. Sbornik, I (chap. 1), t. VIII (1940), p. 307-348; II (chap. 2 et 3), t. IX (1941), p. 563-628; III (chap. 4 6), t. XIII (1943), p. 169-238.