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Let $A$ and $B$ be compact subsets of a metric space. Denote by $\mathcal{P}(A)$ the set of (Borel) probability measures in $A$, and similarly define $\mathcal{P}(A \cup B)$. They are compact metric. Define $C$ as the set of all elements of $\mathcal{P}(A \cup B)$ which assign probability 1 to $A$.

Question: Is there an embedding between $C$ and $\mathcal{P}(A)$?

As far as I can see, when $A$ and $B$ are finite (and assuming without loss they are also disjoint), the answer is yes because one can identify $\mathcal{P}(A)$ with a closed subset of $\mathcal{P}(A \cup B)$. But I am a little lost on how to (try to) show it in general.

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    I suspect I'm missing something. Isn't $C$ isometrically identified with $\mathcal{P}(A)$ by sending $\mu\in C$ to the measure $\mu'$ on $A$ defined by $\mu'(E)=\mu(E)$? The inverse map sends $\nu\in\mathcal{P}(A)$ to the measure $\tilde{\nu}$ on $A\cup B$ defined by $\tilde{\nu}(E)=\nu(E\cap A$. Part of the reason I suspect I'm missing something is because of the claim that $\mathcal{P}(A)$ is compact, which for the metric I'm aware of would be true only if $A$ is finite. What metric are you considering on spaces of probability measures?2010-10-10
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    I believe you are right in your first question. As for the second, I am aware that $A$ is metrizable and compact iff $\mathcal{P}(A)$ is metrizable and compact (for example, we can use the Prokhorov metric).2010-10-10
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    Thank you. I was correct that I was missing something on my second question. I was naively thinking of the total variation distance.2010-10-10

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Disclaimer: I am new to this metric on spaces of probability measures, so hopefully there are not silly errors in what follows.

For $\mu\in C$, define $\mu'\in\mathcal{P}(A)$ by $\mu'(E)=\mu(E)$. The map $\mu\mapsto\mu'$ is surjective, because if $\nu$ is in $\mathcal{P}(A)$, then $\nu=\tilde{\nu}'$, where $\tilde{\nu}\in C$ is defined by $\tilde{\nu}(E)=\nu(E\cap A)$. The map is also isometric with respect to the Lévy–Prokhorov metric. To see this, note that although the $\epsilon$-neighborhoods of a Borel subset $E$ of $A$ will be different in $A$ versus in $A\cup B$, the $\mu$-measures of these neighborhoods are the same for all $\mu\in C$, due to the fact that $\mu(B\setminus A)=0$ for such $\mu$. Hence the sets appearing in the definition of the metric (and therefore also their infima) will be identical whether considering the distance between $\mu$ and $\nu$ in $C$ or between $\mu'$ and $\nu'$ in $\mathcal{P}(A)$.