As a quick revision, let me recollect the way we define the addition of countably many numbers:

Suppose are non-negative real numbers. Then the series is said to converge to if the sequence defined by of partial sums of converges to .

In other words, the tail of the sequence must add up to atmost as small a number as one desires.

Now, in the quest of adding up uncountably many non-negative numbers, let us index the set of these numbers by . Suppose is a function that associates every element of the index set to a non-negative number. Our mission is to define .

Turn into a discrete measure space by defining the measure of each point in to be 1. Clearly, now , with the discrete measure is a measure on the measure space .

Now define

In other words, the sum of all elements indexed by , is the integral of f with respect to the discrete measure taken over the entire index set .

I then wondered if the sum of uncountably many positive numbers could be finite. (Note that a countable sum of positive numbers can be finite; the reciprocals of positive powers of two add up to unity.) The following proposition helps (cf. [Rud]) :

**Proposition: **Let be a measure space. If is such that , then except perhaps on a countable subset of I.

**Proof:**

Decompose as:

Call the right set as and the further right ones as . Suppose that some of the is infinite. Then

contrary to our assumption that the integral is finite. Hence, every is finite and so its union is countable and the proposition is proved.

Reference:

[Rud] Walter Rudin, “Real and Complex Analysis”, 3rd ed. TMH

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August 30, 2009 at 13:18

AnilHi,

I was looking up for some TIFR interview experiences and landed up on your blog.

Didn’t get much of your proof, but isn’t Integral of sine over [0,PI] a sum of uncountably many positive numbers – and it is finite.

Regards,

Anil

August 30, 2009 at 15:54

abhishekparab@Anil We need some measure theory here; I am not talking of the classical Riemann integral but the Lebesgue integral. On the measure space [0,1], there are two measures under consideration; the usual Borel measure, which assigns to each interval, its length AND the discrete measure (which I used in the post above) inducing to each point, a unit measure. Roughly speaking, this means that to integrate the function 1 from [0,1], the standard integral would be 1 whereas the integral with respect to the point measure would be infinity. The two integrals are not the same.

August 30, 2009 at 21:09

AnilSo from what I understood from your post and wikipedia is… you are using mu(x) = 1 as the measure and asserting that any function which has finite integral over such measure can acquire non-zero values only at a countable number of points.

That’s insightful! Thanks…

August 30, 2009 at 22:20

abhishekparabIn fact, we have the stronger result which I state as follows:

If is a measure space and is a measurable and integrable function, then support of is -finite.

September 4, 2009 at 21:42

SunderDo you know the proof for “The set of all real numbers being uncountable”?

September 5, 2009 at 15:16

abhishekparabYes.

One can use the Cantor’s diagnalization proof applied to the binary expansion of points in . But one has to be careful; the expansion might not be unique!

There is also a topological proof, but I am not sure if I am not erring in the hypothesis: A second-countable Hausdorff space with no isolated points is uncountable.