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A friend and I were having a discussion about Lebesgue measure. I attempted to be profound by making the following points:

  • Analytic geometry has been a fantastic tool, but the concept of representing a continuous "object" as a collection of points is inherently contrived (with a negative connotation). Immediately we run into the paradox that a point has no volume, and yet a collection of many points has volume.
  • The notion of Lebesgue measure attempts to resolve this essential tension by allowing only countable additivity of the measure. But it does so only by disallowing certain operations (uncountable sums) that intuitively seem reasonable. As such, it is an indispensable tool, but it remains contrived on some level.

My friend countered by saying that uncountable additivity doesn't really make sense anyway, since any uncountable sum that converges must have co-countably many terms zero.

I would say I am still on the fence about this discussion. He makes a good point, but after all, it is exactly the addition of uncountably many zeros that I am concerned with, so the notion that co-countably many of the terms must be zero may not be a decisive objection.

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  • $\begingroup$ I am especially interested in @Qiaochuyuan 's opinion on this. $\endgroup$ – Eric Auld Jun 14 '14 at 17:45
  • $\begingroup$ Please write down a proposed definition of "uncountable additivity". $\endgroup$ – Hans Engler Jun 14 '14 at 19:00
  • $\begingroup$ @HansEngler One possible notion of an uncountable sum is: Suppose $\{x_i\}_{i\in I}$ is a collection of elements of a normed vector space. Then we say $\sum_{i \in I}x_i = x$ if for all $\epsilon >0$, there exists a finite subset $S_{\epsilon} \subset I$ such that for all finite subsets $T \subset I$ such that $T \supset S_{\epsilon}$, we have $|\sum_{i \in T}x_i - x|< \epsilon$. This is just the convergence of a net defined on the directed set $\mathcal{P}_0(I)$ of finite subsets of $I$, with join being union. Then uncountable additivity means $\mu(\cup_{j \in J}S_j) = \sum_{j \in J}\mu(S_j)$ $\endgroup$ – Eric Auld Jun 14 '14 at 19:17
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    $\begingroup$ Some people feel that the right version of the additivity axiom should be that, as long as $\kappa<|\mathbb R|$, the union of $\kappa$ many disjoint measurable subsets of $\mathbb R$ should be measurable and have as measure the sum of the measures of the sets. (For any set $I$, countable or not, and for nonnegative reals $r_i$, $i\in I$, we define $\sum_{i\in I}r_i$ as the supremum of the sums of finitely many of the $r_i$.) (Cont.) $\endgroup$ – Andrés E. Caicedo Nov 18 '14 at 17:35
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    $\begingroup$ This version of the additivity axiom is actually independent of the usual axioms of set theory It follows from the continuum hypothesis (trivially), from Martin's axiom, and from many other proposed independent statements of interest, while it is false under a variety of other such assumptions. $\endgroup$ – Andrés E. Caicedo Nov 18 '14 at 17:35
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Well it may be a drawback, but, taking the pragmatic route, what's the alternative? If we want a useful measure that is translation invariant and gives the length for intervals, then there will be non-measurable Vitali sets. Lebesgue's still seems to be the best measure we can get.

Maybe you could take some sort of nonstandard analysis approach and declare the measure of a point to be an uncountably small infinitesimal, so that uncountably many of them add up to a number, but I'm not good enough with this stuff to say anything about this.

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  • $\begingroup$ I agree it is the best we can get. But do you agree that it resolves the paradox by defining it away? $\endgroup$ – Eric Auld Jun 14 '14 at 18:25
  • $\begingroup$ Yes, I agree. Essentially my point is that choosing the definitions such that we never get to the paradox seems to be the only way out. I mean in a way it is the same as with Russell's paradox in set theory, the only way to avoid the paradox is never to get there in the first place. $\endgroup$ – mlk Jun 14 '14 at 18:33
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We have $[0,1] = \bigcup_{x \in [0,1]} \{ x \}$. How could you make a definition of uncounteable additivity which would produce $1 = \sum_{x \in [0,1]} 0$?

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  • $\begingroup$ I agree that that is the central tension that Lebesgue measure tries to resolve. I'm arguing that it does so by saying, somewhat arbitrarily, that uncountable sums are not measure-additive. In any notion of uncountable sums (I know of at least one, the "net sum", see link below), the sum of any number of zeros should be zero. I think what this means is that the Cartesian notion of treating a continuum as a collection of points is somewhat contrived, and so the definition of Lebesgue measure reflects this. $\endgroup$ – Eric Auld Nov 18 '14 at 17:33
  • $\begingroup$ math.stackexchange.com/questions/20661/… $\endgroup$ – Eric Auld Nov 18 '14 at 17:34
  • $\begingroup$ But I admit, this whole line of reasoning is looking less impressive to me now than when I posted it. $\endgroup$ – Eric Auld Nov 18 '14 at 17:35

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