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The space $\mathcal{M}$ of signed measures on a measurable space $(\Omega, \mathcal{F})$ is a Banach space when equipped with the total variation norm. So $(\mathcal{M},d)$ is a complete metric space, where $d$ is the total variation metric.

I thought there should be a very quick argument to show that the space of probability measures on $(\Omega, \mathcal{F})$ is of second category in the total variation topology using the Baire category theorem, but somehow I don't see how it follows. The set of probabilities isn't open, so I can't conclude immediately that it's second category. Is there some simple way to show this that I'm just missing, or does it actually require some work?

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  • $\begingroup$ Are you trying to show the space of probability measures has second category in itself, or as a subset of $\mathcal{M}$? $\endgroup$ – Eric Wofsey May 28 '17 at 16:59
  • $\begingroup$ If the space of probability measures on $(\Omega, \mathcal{F})$ is a $G_\delta$ subset (in particular, closed) of $(\mathcal{M},d)$ then it is completely metrizable (for instance, by Theorem 4.3.23 from Engelking's “General Topology”) and therefore Baire and second category in itself. $\endgroup$ – Alex Ravsky May 28 '17 at 17:00
  • $\begingroup$ @EricWofsey In itself. Ah I see, I guess my remark about the set not being open is beside the point. Though I guess I'd be interested in the answer to both questions. $\endgroup$ – grndl May 28 '17 at 17:06
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The set of probability measures is closed in $\mathcal{M}$. Indeed, it is just the set of nonnegative measures of norm $1$. The set of nonnegative measures is closed (since evaluation of a measure on any measurable set is continuous), as is the set of measures of norm $1$. So the space of probability measures is complete with respect to the metric $d$, and hence has second category in itself.

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  • $\begingroup$ Great thanks. For some reason I was thinking the set of probabilities is not closed because I thought a sequence of countably additive probabilities could converge to a set function that's merely finitely additive. Maybe that example is under a different mode of convergence than total variation, though. I'll need to check. $\endgroup$ – grndl May 28 '17 at 17:14
  • $\begingroup$ Yes, I think the example I was thinking of was for weak* convergence. Thanks again! $\endgroup$ – grndl May 28 '17 at 17:20

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