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0 pointsrsj_hn4y ago0 comments

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)

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

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danbruc4y ago

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_hnOP4y ago

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.

danbruc4y ago

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

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