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why do we not define P on the power set of $\Omega$, which is equivalent to declaring that all sets are measurable. Even if $\Omega$ is uncountable, it's sigma-algebra and power set would also be- then how do we conclude that only the elements in sigma-algebra are measurable, while others are not

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    $\begingroup$ Closely related: stats.stackexchange.com/questions/199280/… $\endgroup$
    – Sycorax
    May 29 at 4:34
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    $\begingroup$ The answer Sycorax linked to is pretty good, and your question appears to be a duplicate. However, that answer doesn't give an explicit construction of a non-measurable set, i.e. showing that there are members of the power set that lead to a paradox for some probability measures. The Wikipedia page for Non-measurable set gives such a construction. For me, that example isn't as clear as a similar one in Avner Friedman's Foundations of Modern Analysis; I'm sure there are others. $\endgroup$
    – Mars
    May 29 at 5:34
  • $\begingroup$ @Sycorax thankyou, the question missed my attention. $\endgroup$
    – huy
    May 29 at 5:57
  • $\begingroup$ @Mars this was exactly what I was looking for. Thanks! $\endgroup$
    – huy
    May 29 at 5:57
  • $\begingroup$ Great, huy. It's really kind of an amazing thing when you understand it for the first time, and see why we need to specify a sigma algebra. At least, that's how I felt. Note that the point concerns measure in general, not only probability measures. That is, it also applies to cases in measure theory in which the measure of $\Omega$ is infinite, or is equal to some finite number other than 1. (If that doesn't make sense to you, don't worry about it--it won't matter unless you study measure theory.) $\endgroup$
    – Mars
    May 29 at 22:55