I asked myself, if there is a probability space $(\Omega,\mathcal{F},P)$ such we can find pairwise disjoint sets $(A_n)_{n\in \mathbb{N}}\in\mathcal{F}$ with $P(A_n)=1$ for all $n\in \mathbb{N}$, but I couldn't find an example. My Questions is if such a space exists or not?

  • $\begingroup$ I may be missing something, but the answer seems trivial. Example: coin with both sides the same (heads). Let $A_n)$ be the outcome of the nth flip $P(An=\ heads)=1$. The $A_n$ are disjoint. $\endgroup$ – herb steinberg Aug 17 '18 at 21:34
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    $\begingroup$ On the other hand, if all these events are on the same probability space, being disjoint means that the probability of union= sum of probabilities, but the sum cannot exceed 1 on a probability space, so you can have only 1 element. $\endgroup$ – herb steinberg Aug 17 '18 at 21:56
  • $\begingroup$ @herbsteinberg - I prefer your second comment. In your first, the $A_n$ are independent in the sense that $P(A_m \cap A_n) = P(A_m)P(A_n)$, but they are not disjoint since $P(A_m \cap A_n) \not = 0$ $\endgroup$ – Henry Aug 17 '18 at 22:12

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