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The third axiom of Probablity states that,

If $A_1,A_2,\ldots$ be a finite or infinite sequence of pairwise mutually exclusive events, then

$$P\left(\bigcup A_i\right)=\sum P(A_i)$$

We know that, $P\left(\bigcup A_i\right)\le 1$ but what guarantees that, $\sum P(A_i)\le 1$ for infinite case.

I am an beginner in this topic,so pardon me, if this question is a trivial one

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First of all the definition of countable additivity of a probability measure says that $\sum\limits_{i=1}^{\infty} P(A_i)$ converges and the sum is equal to $P(A)$ (if $A_i$'s are mutually exlcusive events with union $A$). Convergence is part of the definition .

The second point is if $a_i \geq 0$ for all $i$ and $\sum\limits_{i=1}^{n} a_i \leq a$ for al $n$ then automatically $\sum\limits_{i=1}^{\infty} a_i$ converges and $\sum\limits_{i=1}^{\infty} a_i \leq a$.

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  • $\begingroup$ Will you elaborate more?I am not familiar with the notion of measure. $\endgroup$ – who Dec 14 '19 at 6:43
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The fact that it (by assumption) equals the LHS does!

The intuition for why we should assume these add up to the LHS in our axioms for probability, is because the events are mutually exclusive. If we think in terms of Venn-diagrams, the regions corresponding to the events do not overlap so the area of the union is just the sum of the areas.

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  • $\begingroup$ Can we extend this to any collection (uncountable) of mutually exclusive events? $\endgroup$ – who Dec 14 '19 at 12:13
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    $\begingroup$ @who No, the third axiom only applies to countable collections. We want situations like distributions on the real line where the probability $X=a$ is zero for any $a\in \mathbb R,$ so we don't require uncountable additivity... it's too strong and doesn't let us do what we want to do. Why we should demand countably infinite instead of just finite additivity is more subtle, but keep in mind the way to look at it is that not requiring countable additivity (and only requiring finite additivity) would be a generalization of the usual probability theory. $\endgroup$ – spaceisdarkgreen Dec 14 '19 at 17:06
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    $\begingroup$ @who One way to look at it is that additivity requirements accord with our intuition and we want to require as much as we can without going overboard and ruining our ability to work with continuous distributions. Another way is that the additional 'distributions' we get when we weaken to finite additivity are very abstract and it's unclear they do anything useful for us, so why not exclude them? Another reason is more mathematical: there are very nice properties associated with countable additivity. Not everyone agrees: see bof's answer here $\endgroup$ – spaceisdarkgreen Dec 14 '19 at 17:16

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