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I am learning about probability and the definition of pairwise independence is given as $P(AB) = P(A)P(B)$. My textbook motivates this definition as one to capture the intuition where the knowledge of one event occurring doesn't change the your knowledge of the likelihood of the other event occurring. I understand that this definition implies the two conditional probability statements that are closer to that above intuition, given that both events in question have non-zero probability.

But it seems to me that there are examples where this definition does not match intuition, when one or more of the events have zero probability. Consider an experiment where you pick a random number $R$ uniformly between 0 and 1. The event where that random number $R$ is equal to 0.5 has probability 0. So, by the standard definition of independence, it is trivially independent with the event that your number $R$ is in $[0.25, 0.75]$. But intuitively, the knowledge of $R = 0.5$ occurring should make you 100% confident that $R \in [0.25, 0.75]$.

I guess my question is why is independence defined for zero-probability events when the definition does not capture the intuition of independence? What is the utility of it?

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    $\begingroup$ Welcome to MO! This is a good question, though not the right forum (perhaps MSE?)—mathematically, the answer for why no exception is made is that must essentially be that it gives better theorems that way than otherwise. … $\endgroup$
    – LSpice
    2 hours ago
  • $\begingroup$ … But I think that the problem may be that you are looking at the failure of intuition in the wrong place. You explain why you don't think that two events $A$ and $B$ are independent, because $A$ implies $B$; so the real intuitive question is how these two can be compatible. They can be if $A$ has probability $0$, and you object to that; but they also can be if $B$ has probability $1$, and that may be a more tangible case on which to ground your intuition. $\endgroup$
    – LSpice
    2 hours ago
  • $\begingroup$ It’s impossible to consistently define conditional probability when the event you’re conditioning on has probability zero. This is the subject of one of the classic paradoxes of probability theory: en.m.wikipedia.org/wiki/Borel–Kolmogorov_paradox $\endgroup$
    – Andy Putman
    2 hours ago
  • $\begingroup$ (I can’t get the link to work right, so just copy it instead of clicking on it) $\endgroup$
    – Andy Putman
    2 hours ago
  • $\begingroup$ It works fine the other way around: if you are told that $0.25\le R\le 0.75$, then $R=0.5$ has just the same probability as before (namely, zero). $\endgroup$
    – Christian Remling
    2 hours ago

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A mathematical reason for this definition is that we want the independence relation to be symmetric: If $A$ is independent of $B$, then $B$ must be independent of $A$.

In particular, in your case, the chance for the event $A:=\{R=0.5\}$ to occur is $0$, whether we know that the event $B:=\{0.25\le R\le0.75\}$ occurred or not. So, by this logic, $A$ is independent of $B$. So, by the symmetry, $B$ is independent of $A$.

Other than this symmetry consideration, your question does not seem mathematical.

A "philosophical" reason here may be that an event of zero probability cannot "practically" occur. So, the "knowledge" of such an occurrence should be disregarded.

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