4 ms·

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)

by rsj_hn 5y ago

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 5y 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_hn 5y 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.
  - danbruc 5y 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%29%28+sum_%28i%3D0%29%5E%28n+-+1%29+1%2Fn%29 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%29%28lim_%28b-%3E%2B%E2%88%9E%29+integral_a%5Eb+x+dx%29 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%29+integral_%28-a%29%5E%28%2Ba%29+x+dx https://www.wolframalpha.com/input/?i=lim_%28a-%3E%E2%88%9E%... [3] http://cetus.stat.cmu.edu/tr/tr814/tr814.pdf http://cetus.stat.cmu.edu/tr/tr814/tr814.pdf

by rsj_hn 5y ago

danbruc 5y 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_hn 5y 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.
- danbruc 5y 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%29%28+sum_%28i%3D0%29%5E%28n+-+1%29+1%2Fn%29 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%29%28lim_%28b-%3E%2B%E2%88%9E%29+integral_a%5Eb+x+dx%29 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%29+integral_%28-a%29%5E%28%2Ba%29+x+dx https://www.wolframalpha.com/input/?i=lim_%28a-%3E%E2%88%9E%... [3] http://cetus.stat.cmu.edu/tr/tr814/tr814.pdf http://cetus.stat.cmu.edu/tr/tr814/tr814.pdf