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Nov 18, 2018 at 17:39 comment added Dirk Werner The notion of maximal measure seems to be the right substitute for a measure supported on the extreme boundary in the narrow sense, which is not always available as shown by the "porcupine topology'' in the Bishop/DeLeeuw paper. That this is a meaningful generalisation is indicated e.g. by the fact that it allows to prove Rainwater's theorem in the nonseparable case; see R. Phelps's ``Lectures on Choquet's Theorem''. It is also important for uniform algebras.
Nov 18, 2018 at 11:00 comment added Tanmoy Paul Prof. Werner, thank you for the explanation. One small doubt; If $D$ is not metrizable then what is the significance of Choquet-Bishop-De Leeuw Theorem? It only guarantees that $|\mu|(C)=0$ for any boundary measure $\mu$ and for any Baire set $C\subset D\setminus ext(D)$. Although it is not possible to conclude that $Supp(\mu)\subseteq ext(D)$ or even $\int_Df(t)d\mu(t)=\int_{ext(D)}f(t)d\mu(t)$, for ant $f\in C(D)$.
Nov 2, 2018 at 0:38 comment added Tanmoy Paul Yes, of course, $D$ is compact.
Nov 1, 2018 at 20:04 comment added Dirk Werner I assume that you want $D$ to be compact. If $L$ is in the dual of $A(D)$, extend it to $L'$ in the dual of $C(D)$, represent the latter by a signed measure, and decompose this into positive measures, $\mu=\mu^+-\mu^-$. By Choquet theory, each probabilty measure on $D$ has a barycentre in $D$, which can be represented by a maximal measure. Thus, $\mu^{\pm}\in C(D)^*$, restricted to $A(D)$, can be represented by positive boundary measures -- this is the Choquet resp. Bishop-de Leeuw theorem. In the case of a simplex, the representing boundary measure is unique (Choquet-Meyer's theorem).
Nov 1, 2018 at 18:52 comment added Tanmoy Paul Thanks for pointing out. Is it okay now?
Nov 1, 2018 at 18:39 history edited Tanmoy Paul CC BY-SA 4.0
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Nov 1, 2018 at 16:47 history asked Tanmoy Paul CC BY-SA 4.0