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Lemma: Let $A_1,\ldots,A_n$ are events $n\in\mathbb{N}$ then $$ \sum_{i=1}^n \mathbb{P}(A_i) = \mathbb{P}(\cup_{i=1}^n A_i) $$ if and only if $A_1,\ldots,A_n$ are mutually exclusive. Both ways are shown by an easy induction.
However, I think that we are assuming that the probability spaces are finite. Does this lemma still hold if we have a probability space $(\Omega,\mathcal{F},\mathbb{P})$ where $\mathcal{F}$ is countable or uncountably infinite. Thanks in advance.

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This question is too easy for MO. Is it a homework problem? –  Bill Johnson Jan 29 '10 at 18:39
    
I am asking about when we are dealing with an infinite probability space and not the easy finite probability space case. I am wondering if the easy proof extends. –  alext87 Jan 29 '10 at 19:04
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Alex, part of the definition of a measure is countable additivity. See en.wikipedia.org/wiki/Measure_(mathematics)#Definition –  Tom LaGatta Jan 29 '10 at 19:31
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This has nothing to do with the size of the measure (probability) space, either. –  Tom LaGatta Jan 29 '10 at 19:32
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Consider [0,1] with uniform probability density. Then 1 = Pr(x is in [0,1/2] OR x is in [1/2,1]) = Pr(x is in [0,1/2]) + Pr (x is in [1/2,1]), but the two are NOT mutually exclusive as x = 1/2 fulfills both. In general, the statement is "up to probability 0" pieces, i.e., that Pr(A_i intersect A_j) = 0 for all i different from j. –  Jason DeVito Jan 29 '10 at 21:07

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Your "lemma" is false for finite probability spaces, e.g., $\Omega = \{a,b\}, \mathbb P(\{a\})=0,\mathbb P(\{a,b\})=1, \mathbb P(\{a\} \cup \{a,b\})=1.$

After you fix it, a cannon to swat the fly is inclusion-exclusion, or more specifically, the Bonferroni inequalities.

I think people are confusing your question as stated with the natural and very elementary question of whether countably additive probabilities must be uncountably additive, and the example of Lebesgue measure on $[0,1]$ shows this this is not the case.

You should very rarely do anything with the sample space itself in any intrinsic probability question. See the answer gowers gave to this question and this Tao blog entry. It's ok to have a sample space when you apply probability to something like an analysis question (e.g., proving the Weierstrauss approximation theorem using probability) or use the probabilistic method in combinatorics.

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Is it too charitable to interpret "A and B are mutually exclusive" as $\mathbb{P}(A \cap B) = 0$? This seems in keeping with the "ignore the sample space" philosophy. –  Pete L. Clark Jan 30 '10 at 0:15
    
Yes, I agree that is the most natural way to fix it, to change the definition of mutually exclusive or replace that in the statement. Since you sometimes deal with different measures on the same space, I'd be uncomfortable redefining mutually exclusive in all circumstances. –  Douglas Zare Jan 30 '10 at 0:39
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Well, one could reserve the term disjoint for when the set-theoretic meaning is truly needed... –  Pete L. Clark Jan 30 '10 at 1:02
    
I'd qualify the last sentence with a "sometimes", namely when the sample space itself is part of the problem of interest. In many (probably most) applications of the probabilistic method the sample space will still be a distraction. –  Mark Meckes Jan 30 '10 at 1:26
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I understand. The definition of mutually exclusive is not exactly how I was using it. I assume that if two events can't occur together then they are mutually exclusive. Since events with probability $0$ can't occur I assumed they were mutually exclusive with all events. The wikipedia definition is not exactly precise on this. Thanks for all your help. –  alext87 Jan 30 '10 at 17:59

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