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Let X be a complete separable metric space.
Denote by PX the set of Borel probability measures.
We can construct a priori two sigma-algebras on PX, both mentioned by Giry:
1) take the weak topology on PX (the coarsest topology making integration of bounded continuous functions itself continuous), and form its Borel sigma-algebra;
2) take the coarsest sigma-algebra making integration of bounded measurable functions itself measurable.
Are the two sigma-algebras actually equal? It's fairly easy to see that 1) implies 2), but what about the converse?
Is 2) not the standard sigma-algebra on a standard Borel space? If so, then we know that PX is a standard Borel space in both sigma-algebras, we have one measurable bijection from one to the other, and there is the result about inverses of bijective measurable maps between standard Borel spaces being measurable. But maybe that’s using an A-bomb to kill a mosquito (or whatever the correct idiom is)?
Of course! Very good point. Thank you!