[danbruc]

Just out of curiosity, is there are short answer why it requires uncountability? Naively something like pick a random natural number would also seem to lead to probability zero. I can see that pick a random natural number might be problematic, how would you do this? Pick on digit and then with some probability either stop or continue and pick another digit, but it is at the very least not obvious that one could make this work without larger numbers just having smaller and smaller probabilities and there might also be issues with termination. On the other hand it is not obvious to me why one could not work with uniform distributions over the set [0, n) and then look at the limit as n goes to infinity.

[rsj_hn]

If you have a measure on a countable set, lets number it 0, 1, 2, .. then you must have: m(i) >= 0 (since it's a measure).

And must also have

1 = m(0) + m(1) + ... (because it's a measure)

so

1 = Lim S(i)

Where S(i) is the partial sum going from 0 to i.

But if each m(i) = 0, then each partial sum is zero.

So 1 = Lim 0 = 0

[danbruc]

There is probably something wrong with this, but I was thinking something like the following.

Take the sequence of sets M(n) = { 0, 1, 2, ... n - 1 } with measure m(n, i) = 1 / n. The m(n, i) are non-negative and the sum over all m(n, i) for a fixed n is 1. Then take the limit. The set M(n) will seemingly approach the natural numbers but I am not sure that this is valid. The m(n, i) will approach 0, I think that is uncontroversial. But I guess it might not be valid to argue that the sum remains 1 even though it seemingly equals n * 1 / n.

[rsj_hn]

No, you can't interchange limits like that.

In your case, when you say "sum" of the n identical things, you just mean multiplying n by the integer 1/n.

So you have 1 = 1/n *n != lim(1/n)lim(n). The last is an indeterminate form of 0*infinity and so you don't get to conclude that it's one.

[danbruc]

I would actually want to use only one limit, not two independent limits. And if I throw this [1] at Wolfram Alpha it actually says 1. I have a set of n elements with measure 1/n and then grow the set towards infinity while simultaneously shrinking the measure.

I agree that this does not work if growing the set and shrinking the measures are two independent limiting processes very similar to how integrating x dx from minus to plus infinity yields infinity if you have independent integration limits [2] but yields 0 if the integration limits are not independent [3].

I am still happy to accept that it requires uncountable sets but I am not convinced by the argument you provided, that the limit does not work out. I think there must be a different issue, some other property of probability measures that fails.

EDIT: I also finally did a little bit of searching and while I did not read much yet, it seems that the problems indeed arise from additivity as you hinted at with the partial sums. But I also found that there are actually ways to have uniform distributions on the natural number [4] if one uses non-standard axioms, but I only skimmed the paper for the moment.

[1] https://www.wolframalpha.com/input/?i=lim_%28n-%3E%E2%88%9E%...

[2] https://www.wolframalpha.com/input/?i=lim_%28a-%3E-%E2%88%9E...

[3] https://www.wolframalpha.com/input/?i=lim_%28a-%3E%E2%88%9E%...

[3] http://cetus.stat.cmu.edu/tr/tr814/tr814.pdf