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Suppose $P$ is a finitely additive probability measure on a space $\Omega$, and $X_1,X_2,X_3,\dots$ are i.i.d. random variables with mean 0 and variance 1. Is it true that for all $r \in \mathbb{R}$, $$\lim_{n\to\infty} P \left( \sqrt{n}\sum_{i = 1}^n X_i \leq r \right) = \frac{e^{-r^2/2}}{\sqrt{2\pi}}?$$

Basically I am asking whether countable additivity plays an essential role in the proof of the classical CLT.

See this question for a brief discussion of random variables in the finitely additive setting.

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  • $\begingroup$ Is this question not solved in doi.org/10.1016/0047-259X(88)90035-8 , Theorem 2.1? In particular use ch.3, Consequences (i) of this paper. $\endgroup$ Commented Aug 27 at 9:24
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    $\begingroup$ @DieterKadelka Thanks for pointing that out. Not sure but I think the statement of the CLT implicitly assumes that the integral of $X_1$ (the mean) exists according to the definition in that paper, which means that there is a kind of "Cauchy sequence" of simple functions converging to $X_1$ in measure. In the finitely additive setting, this is not, apriori, necessarily the broadest class of measurable functions for which a well-defined integral exists. So if my understanding is correct, he solves the question affirmatively for his $L_1$ random variables. Good enough for most purposes. $\endgroup$ Commented Aug 27 at 11:32
  • $\begingroup$ A good place to check would be "How to gamble if you must" by Dubins and Savage. I don't have a copy on hand, so I cannot look at it myself. $\endgroup$ Commented Aug 27 at 15:07

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