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This question has been bugging me for a while? Does there exist a probability measure on the measurable space $\bigl(\mathbb{R},\mathcal{P}(\mathbb{R})\bigr)$. If so, what is it?

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    Yes, $\delta_0$ for example. (Dirac at $0$). I guess you want additional conditions.2012-09-22
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    As @DavideGiraudo said, you probably want more conditions. I am [still looking for a natural measure](http://math.stackexchange.com/questions/177282/understanding-measures-on-the-space-of-measures-via-examples) (on $\mathcal{P}([0,1])$). Though I think I now know such an example. Let me know if you're interested in that.2012-09-22
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    To read about the space $\mathcal{P}(\mathcal{P}(\mathbb{R}))$, check out Billingsley's *Convergence of Probability Measures* and Parthasarathy's *Probability Measures on Metric Spaces*.2012-09-22
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    If you drop the axiom of choice then you can have the Lebesgue measure...2012-09-22
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    @Quinn I think he means power set by ${\cal P}$, not the space of probability measures.2012-09-22
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    @Asaf Karagila: Is the weakest form of axiom of choice needed known?2012-09-22
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    @Shahab: You mean perhaps the strongest form of choice for which the Lebesgue measure is a measure on all subsets? We know that this implies that this is consistent with the principle of Dependent Choice, and it immediately negates the following: The weak ultrafilter lemma for $\mathbb N$ (i.e. every ultrafilter on $\mathbb N$ is principal), therefore ultrafilter lemma/Boolean Prime Ideal theorem; Axiom of choice for families of size $\aleph_1$, and therefore $\mathrm{DC}_{\aleph_1}$.2012-09-22

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Reference: You may be interested in the "problem of measure". There is a short treatment of this topic in Appendix C of Real Analysis and Probability by R.M. Dudley.

He proves the following result due to Banach and Kuratowski: Assuming the continuum hypothesis, there is no measure $\mu$ defined on all subsets of $I:=[0,1]$ with $\mu(I)=1$ and $\mu(\{x\})=0$ for all $x\in I$.