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Let $X\neq\emptyset$ be a set and let $\mu:{\cal P}(X)\to [0,1]$ be a probability measure. Is there a probability measure $$\bar{\mu}:{\cal P}({\cal P}(X))\to [0,1]$$ with the following property?

For all $S\subseteq X$ we have $\bar{\mu}({\cal P}(S)) =\mu(S)$.

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  • $\begingroup$ Still wondering whether it is doable for $X=\mathbb{N}$? $\endgroup$ – Dominic van der Zypen Jun 25 at 16:26
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    $\begingroup$ If the measure is atomic (as it necessarily is in the case of $\mathbb N$), just define $\bar\mu({{x}})=\mu({x})$ and set all other subsets to have measure 0. $\endgroup$ – Anthony Quas Jun 25 at 17:12
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For notational and conceptual simplicity, assume that $X$ and $P(X)$ are disjoint.

As Bugs Bunny implicitly suggested: Let $\mathcal E = \{\{n\}: n \in X \}$ be the set of singletons. This is a subset of $P(X)$. The measure $\bar\mu$ will concentrate on this set.

For any $\mathcal A \subseteq P(X)$, we will have $\bar \mu(\mathcal A)=\bar \mu(\mathcal A\cap\mathcal E)$; note that $\mathcal A\cap\mathcal E$ is morally the same as a subset of $X$, so $\mu$ will measure it.

Formally, let $S_{\mathcal A} = \{n\in X: \{n\}\in \mathcal A\}$ be the ``singleton support'' of $\mathcal A$. Define $\bar \mu (\mathcal A) := \mu(S_{\mathcal A})$.

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If $|X|>2$, then there is a single solution: $\overline{\mu}(\{x\})=\mu(x)$ and $\overline{\mu}(S)=0$ for any other set $S$.

Indeed, let $a=\overline{\mu}(\emptyset)$. By the condition for $S=\{x\}$, $\overline{\mu}(\{x\}) =\mu (x)-a$. By the condition for $S=\{x,y\}$ with $x\neq y$, $\overline{\mu}(\{x,y\}) =a$.

Now if $|X|>2$, there are at least $1+|X|$ two-element subsets. Thus, $\sum_{|S|\leq 2} \overline{\mu}(S) \geq a+ \sum_{x\in X}\mu (x) =1+a$. This forces $a=0$.

Now it is easy to show that for all larger subsets $\overline{\mu}(S)=0$.

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    $\begingroup$ I am afraid I cannot follow the argument. What is $x$? Is it an element of $X$? Then I do not see how to apply the map $\mu$ to it. Or is it a set with $1$ element? Then $\mathcal{P}(x)=\{\varnothing, x\}\neq \{x\}$. Moreover, it would be helpful if you could explain what you mean by the phrase "no measure [...] left". (Some probabilities can be zero I guess.) $\endgroup$ – Philipp Lampe Jun 25 at 16:24
  • $\begingroup$ Grateful for the nice answer in the finite case, what about $X$ infinite? (I am especially curious for $X=\omega$.) $\endgroup$ – Dominic van der Zypen Jun 25 at 16:24
  • $\begingroup$ I will edit my answer: I got it slightly off... $\endgroup$ – Bugs Bunny Jun 25 at 19:07

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