What actually breaks in probability when you let $\mathcal F = 2^\Omega$ for uncountable $\Omega$?

To the best of my knowledge, this is not a duplicate question. While there are similar-sounding questions, answers to them usually cite:

  1. Vitali sets. However, only translation-invariant measures prevent Vitali sets. Probability measures don’t need to be translation-invariant, so Vitali sets are irrelevant. You can totally have a probability measure and $\sigma$-algebra containing Vitali sets.

  2. Lebesgue measures. Same objection as above.

  3. Banach-Tarski. But measures must be at least rotation-invariant to prevent Banach-Tarski (or live in 2-D). Probability measures don’t need to be rotation-invariant, so Banach-Tarski is irrelevant.

AFAICT, there seems to be a logical gap in the usual explanations.

Could someone provide a concrete example of any paradoxes that emerge from uncountable sets that apply to probability theory? Any issues with letting $\mathcal F = 2^\Omega$?

Perhaps there is no such paradox. For example, $\delta_a(\cdot)$ seems like a perfectly good probability measure that works with $\mathcal F = 2^\Omega$. The answer might “There’s no paradox, but here is an extra property X we want from our probability measures, here’s why it matters, and for this restricted class of measures, they cannot be defined on $\mathcal F = 2^\Omega$.” If so, I’d love to know what this property $X$ would be.

Apologies if I’ve made an error in this post; I’m new to measure theory!

  • $\begingroup$ What makes you think something breaks? $\endgroup$ – Eric Wofsey Mar 13 '19 at 2:58
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    $\begingroup$ @EricWofsey Because people often justify measure-theoretic probability by talking about Banach-Tarski etc. However, as I mention in the question, I think the real answer might actually be that there is some other property X (e.g., translation-invariance) which we want out probability measures to satisfy, but which is not included in any definition of probability measure that I have seen. $\endgroup$ – Yatharth Agarwal Mar 13 '19 at 3:02

There's nothing that breaks. You will certainly almost never see any theorem that requires your $\sigma$-algebra to not have the form $2^\Omega$. It's just that most naturally occurring measures on uncountable sets are not defined on all subsets (because they are typically ultimately derived from Lebesgue measure, one way or another). The point of the formalism of $\sigma$-algebras is not to disallow $2^\Omega$, but to allow more general $\sigma$-algebras.

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You already listed several examples of such properties $X$. They are important because in applications our spaces do have translation and rotation invariance so if the probability measure does not have such properties then it is useless for the application.

For example, intuitively the probability that a uniformly chosen random point in $[0,1]$ lies within the interval $[a,b]$ should be equal to $b-a$. So it depends only on the length and not the position of the interval. To make this precise, we construct a translation-invariant measure on $[0,1]$.

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  • $\begingroup$ Could you elaborate on when, say, translation-invariance is important? It seems most probabilists completely ignore the actual probability space (working instead with RVs) as much as possible anyway. $\endgroup$ – Yatharth Agarwal Mar 13 '19 at 2:40
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    $\begingroup$ Just because you work with RVs doesn't mean you're ignoring the probability space. At some point you'll need to calculate the probability of something involving the random variable, and then you'll need the probability space. $\endgroup$ – Ted Mar 13 '19 at 2:43

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