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I would like to know how to solve this "paradox". First of all I define some functions and syntax:

  • Let's define the Lebesgue Measure of the closed interval [x ; y] this way: LM([x;y]). So the Lebesgue Measure of the interval [0;1] is: LM([0;1])=1.

  • From the definition of a measure function, let's define L(n) as the sum of the n parts (sub-intervals) of equal length of the interval [0;1]. We have L(n) = LM([0,1]) and for any n:

$$ L(n) = \sum\limits_{i=1}^{n}LM([-1/n + \sum\limits_{j=1}^{i}1/n ; \sum\limits_{j=1}^{i}1/n]) $$

For example L(2) = LM([0;0.5])+LM([0.5;1])=LM([0;1])

Let's take the limit of L(n) when n goes to infinity with two different points of view (infinity in calculus and set theory):

Calculus

Since the limit of -1/n is zero, the first option is to consider the sub-interval a singleton [x;x]. The problem is that an infinite number of closed intervals of measure zero is zero and L(n)=0 != LM([0;1])=1;

Set Theory

The structure of the sub-intervals is [A,B], A and B being rationals. We cannot have a countable number of singleton sub-intervals like this: [A;A] because the set of reals between [0;1] is uncountable and the set of rationals is countable. From the definition of the function L(n) we are mapping an uncountable number of reals (interval [0;1]) in a countable number of parts (sub-intervals) so A must be different from B. In this case the sub-intervals are not zero in length since any real closed interval with A!=B has LM(A,B)=epsilon > 0. The problem is that an infinite number of closed interval of measure epsilon>0 is infinity and L(n)=infinity != LM([0;1])=1;

Both options make me think about the Archimedean property of reals and the infinitesimals.

Could someone please help me to understand what is wrong in both results? Thanks a lot for your comments.

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    $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n}\neq \sum_{k=1}^\infty 0$. This is why you get the wrong number in the first approach. Also a measure only satisfies countable additivity, this means when you have an uncountable sum like $\sum_{x\in[0,1]}LM\{x\}$, you should be careful about how you are going to deal with it. – 2017-01-31
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    Thanks for your comment. How would you solve the limit in the first case without saying that 1/n is 0 at infinity? $$ \lim_{n -> \infty} L(n) = \sum\limits_{i=1}^{\infty}LM([\lim_{n -> \infty}(-1/n) + \sum\limits_{j=1}^{i}(\lim_{n -> \infty} 1/n) ; \sum\limits_{j=1}^{i} (\lim_{n -> \infty} 1/n)]) $$ In the second case, why do you think that the sum is uncountable? Couldn't we have a countable but infinite number of parts? – 2017-01-31
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    Note that $\sum_{k=1}^n\frac{1}{n}=n\cdot\frac{1}{n}=1$,therefore the limit is $1$. For the second question, first I misunderstood what you wrote. So basically you are partitioning $[0,1]$ into infinitely but countablly many sub-intervals. However note that there's no way to make this partition uniform, i.e. all intervals have the same length, because as you mensioned in this case the sum of all intervals' lengths is $\infty$, and this is where the problem arises. – 2017-01-31
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    Thanks for your comment. Could you please tell me if the limit in my previous comment is correct? If this is so, my main problem with your solution to the first question is that L(n) is defined in such a way that the limit at infinity is only necessary for the difference(-1/n). This difference between A and B becomes 0 and the interval becomes a 1-point set (since n is a natural number it should be a countable collection of 1-point sets not all the interval[0;1] of reals).... – 2017-02-01
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    ...It is probably not the way to define LM(x) in math and is more oriented to computation since I want to express the "argument" (the sub-intervals [x;y]) of LM to show that it becomes a singleton at infinity. For the second question could you please tell me why we can not make a uniform partition? What is wrong with L(n) and the way it partitions the set? Where could I find information about this? Could L(n) be a sigma finite function but not a sigma infinite one? The main problem with this solution is that L(n) is simply a restatement of the definition of measure. Thanks a lot for your time. – 2017-02-01

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