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Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and $(A_n)_n$ a sequence in $\mathcal{A}$.

If $\mathbb{P}(\bigcup_n A_n) = \sum\limits_{n} \mathbb{P}(A_n)$, then $\mathbb{P}(A_i \cap A_j) = 0$ for all $i,j \in \mathbb{N}, i \not= j$.

I have to prove that the $\sigma-$additivity of the probability measure implies that the events are pairwise disjoint.

It is clear that I could construct a disjoint sequence $B_n := A_n \setminus \bigcup_{k = 1}^{n-1} A_k$ that would satisfy the condition but that doesn't really help with this excercise.

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    $\begingroup$ Have you tried contradiction? $\endgroup$ – Camille Nov 26 '16 at 12:02
  • $\begingroup$ @Camille I'll try. Let's suppose that $\mathbb{P}(A_i \cap A_j) > 0$ for all $i,j \in \mathbb{N}, i \not= j$. Then obviously $A_i \cap A_j \not= \emptyset$. Therefore $\mathbb{P}(\bigcup_n A_n) < \sum_{n}\mathbb{P}(A_n)$, delivering the contradiction. $\endgroup$ – PeterMcCoy Nov 26 '16 at 12:08
  • $\begingroup$ If you're looking for a brief answer that's alright I think, but your assumption for contradiction should read for some $i,j\in\mathbb{N},i\neq j$, not for all. $\endgroup$ – Camille Nov 26 '16 at 12:10
  • $\begingroup$ @Camille Surely, you're right. I just used copy/paste at the wrong place, "for all" negates into "exist". $\endgroup$ – PeterMcCoy Nov 26 '16 at 12:12
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    $\begingroup$ Very good! Now can you show that $P(\bigcup A_n)\le \sum_n P(A_n) - P(A_i\cap A_j)$ for any $i\neq j$? $\endgroup$ – zhoraster Nov 26 '16 at 15:10
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Define $f:=\sum_n \mathbf 1_{A_n}-\mathbf 1_{\bigcup_n A_n}$. This function is non-negative and its integral is zero, so $f=0$ almost surely. Therefore, the function $\sum_n \mathbf 1_{A_n}$ can take only the values $0$ and $1$.

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