Timeline for Existence of dominating measure for weak*-compact set of measures
Current License: CC BY-SA 3.0
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| May 20, 2013 at 10:42 | comment | added | jbc | By "compatibility" above I of course meant "non-compatibility". It is perhaps worth mentioning that in the topological situation, the universal property works in the other direction. $S$ embeds into $M^t(S)$ in such a way that every continuous, bounded function on $S$ with values in a Banach space lifts in a unique fashion to a continuous linear mapping with the appropirate (which, again, is not the norm). | |
| May 20, 2013 at 10:06 | comment | added | jbc | Compatibility in the first part means that the corresonding dual spaces are too large. Thus the Banach space duals of the function spaces consist in both cases of the finitely additive measures, not the countably additive or Radon ones. The unversal property is that every countably additive meaure on the $\sigma$-algebra with values in a Banach space (for which see Diestel and Uhl "Vector measures") lifts to a unique continuous linear mapping on $L^\infty$ with the topology mentioned in my answer. | |
| May 20, 2013 at 9:31 | comment | added | andy teich | What do you mean by universal property?Can you specify a bit more what you mean by compatibility in the first part? | |
| May 20, 2013 at 6:42 | history | edited | jbc | CC BY-SA 3.0 |
corrected some irritating typos.
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| May 20, 2013 at 6:29 | history | answered | jbc | CC BY-SA 3.0 |