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Is it possible to construct a probability space $(\Omega,\mathcal{A},\mathbb{P})$ such that $\mathcal{A}$ is uncountable, there are uncountable events with probability $>0$ and there are also countably many measurable singleton ${x}$ that have probability $>0$?

This seems like a really weird probability measure to have - and I wasn't able to construct one. All non-toy example with uncountable $\mathcal{A}$ that I know assign measure $0$ to singletons.

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    $\begingroup$ Hint: take a probability measure $\mu$ with the first property, a probability measure $\nu$ with the second one, and consider $1/2(\mu+\nu)$ $\endgroup$ – Lucio Feb 16 at 20:19
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Why be fancy? A countable set can have uncountably many subsets. For example, take $\Omega=\mathbb N$ and $\mathcal A=2^\Omega$ and $P(X=n)=2^{-n}$.

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Let $\mu=\frac 1 2 \nu+\frac 1 2\sum \frac 1 {2^{n}} \delta_n$ where, for example, $\nu$ is the $N(0,1)$ measure. [$\mu(x,x+1)>0$ for all $x$ and $\mu\{n\}>0$ for all $n$].

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Take a right continuous function $F: \mathbb{R} \to \mathbb{R}$ with $F(1)- F(0) = 1$ that has countably many discontinuities in $]0,1]$ and consider the measure space $(]0,1], \mathcal{B}(]0,1]), \mu_F)$ where $\mu_F$ is the Lebesgue-Stieltjes measure restricted to the Borel sets on $]0,1]$.

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