Lifting a probability measure to the power set

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 $$\bar{\mu}({\cal P}(S)) =\mu(S)$$.

• Still wondering whether it is doable for $X=\mathbb{N}$? – Dominic van der Zypen Jun 25 at 16:26
• 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. – Anthony Quas Jun 25 at 17:12

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 $$\bar \mu (\mathcal A) := \mu(S_{\mathcal A})$$.

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 $$\overline{\mu}(S)=0$$.

• 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.) – Philipp Lampe Jun 25 at 16:24
• Grateful for the nice answer in the finite case, what about $X$ infinite? (I am especially curious for $X=\omega$.) – Dominic van der Zypen Jun 25 at 16:24
• I will edit my answer: I got it slightly off... – Bugs Bunny Jun 25 at 19:07