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The following theorem seems to have folk status:

The topological dual of the space $C_b(X)$ of bounded continuous functions on a topological space $X$ is isomorphic to the space $rba(X)$ of finite, regular, finitely additive Borel set functions.

This fact is often mentioned (for instance in the answer to Dual of bounded uniformly continuous functions) but I'm having great difficulty actually finding a reference. Often Dunford & Schwartz is mentioned as a reference; D&S defines $rba$, but doesn't prove the connection to the dual of $C_b$. Hildebrandt 1934 proves a characterization in terms of limits of Stieltjes integrals, but that is still some steps away from the characterization above. I haven't been able to find anything coming closer than this.

Does anyone know of a real proof of this statement? Am I maybe overlooking a very simple proof?

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    $\begingroup$ What's wrong with the discussion in IV.6, p.261ff in DS? They prove the duality at least for $X$ normal. $\endgroup$
    – Theo Buehler
    Commented Jul 18, 2011 at 11:51
  • $\begingroup$ What hypotheses (if any) are you putting on your space $X$? $\endgroup$
    – Yemon Choi
    Commented Jul 18, 2011 at 12:56
  • $\begingroup$ Perhaps confusion: Dunford & Schwartz notation $C(X)$ is the space of bounded continuous functions on $X$. The dual (for $X$ normal, but probably completely regular will also work) is Theorem IV.6.2, as Theo says. $\endgroup$
    – Gerald Edgar
    Commented Jul 18, 2011 at 13:47
  • $\begingroup$ Thanks @Theo and @Gerard - I obviously overlooked Theorem IV.6.2, possibly because of the different notation. Thanks for pointing that out! $\endgroup$
    – Mark Peletier
    Commented Aug 14, 2011 at 18:14

2 Answers 2

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The topological dual of the space of bounded continuous functions on a topological space X is isomorphic to the space of finite, zero set regular, finitely additive Baire set functions; see: R. F. Wheeler, A survey of Baire measures and strict topologies, Exposition. Math. 2 (1983), 97–190 (a proof is on pp. 115-117).

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    $\begingroup$ The point being: in exotic spaces, if you use Baire sets (not Borel sets) and zero sets (not closed sets) you get what you want. A reference for more discussion: Gillman & Jerison, Rings of Continuous Functions. $\endgroup$
    – Gerald Edgar
    Commented Aug 11, 2011 at 15:36
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In the answer you mentioned, the space $X$ is metrizable, hence normal, so the proof from Dunford & Schwartz that appeared in the comments is aplicable in that case.

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