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Let us assume, as in here Measurable structure on the space of probability measures that $X$ is a locally compact Polish space. Then can the same thing be said of $P(X)$, its probability measures with weak * convergence? I am only missing local compactness and completeness relative to one metrization now, as I realized that appeal to the 1 point compactification of X helps. (This process gives a bigger space that is actually a metric space.)

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@EnjoysMath I'm studying from Aldous' notes on exchangeability in preparation to work on one of his research problems. My advisor and him have different ways of doing things (See the question I linked for how it differs but how I hope they are equivalent.) –  Jeff Sep 27 '13 at 5:39
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That $P(X)$ is Polish is true and can be found in many books; see e.g. Chapter 17 of Kechris' Classical descriptive set theory. One way to prove it is to observe that, denoting by $\widehat X=X\cup\{\infty\}$ the one-point compactification of $X$, the space $P(X)$ can be identified with $G=\{ \mu\in P(\widehat X);\; \mu(\{\infty\})=0\}$, which is $G_\delta$ in the compact metrizable space $P(\widehat X)$, and hence Polish.

On the other hand, $P(X)$ is not locally compact unless $X$ is compact.

For simplicity, consider $X=[0,\infty)$. Take any neighbourhood $\mathcal U$ of $\delta_0$ in $P(X)$. Then you will have no problem to show that one can find $\varepsilon >0$ such that $(1-\varepsilon)\delta_0+ \varepsilon \delta_a\in\mathcal U$ for every $a\in\mathbb R$.

Now take a sequence $(a_n)$ tending to $\infty$ and put $\mu_n:=(1-\varepsilon)\delta_0+\varepsilon\delta_{a_n}$. Then $(\mu_n)$ is a sequence in $\mathcal U$ which has no subsequence converging in $P(X)$. (The sequence $(\mu_n)$ is of course convergent in $P([0,\infty])$, but the limit $(1-\varepsilon)\delta_0+\varepsilon\delta_\infty$ is not in $P(X)$). So $\mathcal U$ is certainly not compact.

Exactly the same argument gives the result for a general locally compact and non-compact $X$.

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Local compactness is required before one can talk about iterating this procedure, because the Riesz representation theorem is used. I suppose then you are saying that I misunderstood my advisor because it is simply not possible to iterate the procedure unless $X$ is compact. Thanks. –  Jeff Sep 28 '13 at 7:41
    
By "iterate the procedure" you mean consider $P(X)$, $P(P(X))$ and so on? This you can still do with $X$ Polish. –  Etienne Sep 28 '13 at 10:11
    
how can one consider $PP(X)$ when $X$ is only Polish? There is no Riesz correspondence of measures with functionals. I suppose the map of measures into functionals is still injective, but just not onto. So then you inherit the topology this way I guess. –  Jeff Sep 28 '13 at 15:33
    
Wel, $Z=P(X)$ is Polish. $P(Z)$ is perfectly well-defined (Borel probability measures on $Z$), and the topology as well (it is the topology on $P(Z)$ generated by the evaluation maps $\mu\mapsto \int fd\mu$, where $f$ ranges over the bounded continuous functions on $Z$; usually, this topology is called the Prokhorov topology). Then $P(Z)$ is Polish whenever $Z$ is, so you can indeed iterate. –  Etienne Sep 28 '13 at 17:36
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