Prove that a double integral exists

In summary, the function f(x,y) = 1 if x=1/3 and y is rational, and 0 otherwise, is not integrable when integrated from 0 to 1, as irrational numbers are infinitely dense in any interval. However, the double integral of f over the region Q = [0,1]x[0,1] in R2 exists, as the difference between the upper and lower Riemann sums can be made arbitrarily small through refining the partitions and reducing the area containing points where f=1.
  • #1
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Homework Statement


Let f(x,y) = 1 if x = 1/3 and y is rational, and let f(x,y) = 0 otherwise. Show that the double integral of f over the region Q = [0,1]x[0,1] in R2 exists (SSQ f dA exists) yet the integral from 0 to 1 of f(1/3, y) does not exist.

(sorry for the weird way of writing, I'm not sure how to do integral signs)


Homework Equations





The Attempt at a Solution


For the second part - a function is integrable if its set of discontinuities is finite or has zero content. However, irrational numbers are infinitely dense in any interval so the set of discontinuities are infinite and they cnanot be covered by a finite amoutn of intervals with arbitrarily small lengths, so the function f(1/3, y) is not integrable

But the second part is kind of counter-intuitive to me. No matter how i partition that square, as long as the partitions have real lengths, then the lower riemann summ will always be 0 (since every square that contains a point where x= 1/3 and y is rational, will also contain points where x is not 1/3 and/or y is not rational) and the upper riemann summ will always be non-zero (since, any partitioning into squares of real lengths, will result in at least some squares containing points where x = 1/3 and y is rational). So, the lower integral and upper integral will not agree (one is zero, one non-zero), so how can the double integal over the region exist?
 
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  • #2
Not every 'square' (really the partition can consist of rectangles as well) contains a point where x=1/3. How about [1/2,1]x[1/2,1]? How about [0,1/4]x[0,1]? Etc etc. If you draw small enough rectangles, very few of them will overlap x=1/3 and they can have very small total area.
 
  • #3
Ok, I think I know what you're saying
The integral exists if the difference between the upper and lower riemann sums can be made arbitrarily small (There exist partitions p of Q such that Spf - spf < epsilon for any epsilon >0). I know spf is always 0, and, while Spf will not be zero for any partition into rectangles of real lengths, it can be made to approach zero by refining the partitions and making the total area of all the rectangless that contain x = 1/3 approach zero (since the area containing points where f=1 is reduced, the upper riemann sum is also reduced).

So, for any epsilon we choose, there exists a particion p of Q such that Spf - spf < e

is this what you meant?
 
  • #4
That's exactly what I mean. If all of the rectangles have side length less than e/2, then all of the rectangles that touch x=1/3 will lie between x=1/3-e/2 and x=1/3+e/2. Hence the total area is less than e.
 
  • #5
It makes a lot more sense now.. :) Gotta watch out for my faulty intuition haha. Thank you so much for your help!
 

1. What is a double integral?

A double integral is a type of mathematical calculation that involves finding the volume under a surface by dividing it into smaller rectangles and summing their areas. It is often used in calculus and physics to solve problems related to area, volume, and mass.

2. How do you know if a double integral exists?

A double integral exists if the function being integrated is continuous over the given region and the region itself is bounded. If these conditions are met, then the double integral will have a finite value.

3. Can you give an example of a double integral?

Sure, for example, if we want to find the volume of a cone with a circular base of radius 3 and height 4, we can set up a double integral by integrating the function f(x,y) = 4 over the region bounded by the circle x^2 + y^2 = 9 and the line y = 4x/3. This will give us the volume of the cone, which is 12π.

4. Why is it important to prove that a double integral exists?

Proving that a double integral exists ensures that our calculation is valid and accurate. It also allows us to use the integral to solve real-world problems and make predictions in fields such as physics, engineering, and economics.

5. Are there any techniques for proving that a double integral exists?

Yes, there are several techniques for proving the existence of a double integral, such as the Riemann sum method, the Fubini's theorem, and the Cauchy criterion. These methods involve breaking down the region of integration into smaller parts and evaluating the integral over each part to show that it converges to a finite value.

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