Can a Bounded Function on a Rectangle be Integrable over Q?

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Homework Help Overview

The discussion revolves around finding a bounded function defined on the rectangle Q = I × I (where I = [0,1]) that has existing iterated integrals but is not integrable over Q. The original poster seeks to explore examples and properties of such functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the idea of constructing a dense subset S of Q and its implications for the function's integrability. There are suggestions to consider space-filling curves and the behavior of functions near the origin. Questions arise about the requirements for integration measures and the nature of the function's boundedness.

Discussion Status

The discussion is ongoing, with various ideas being explored regarding the construction of the function and the properties of integrability. Participants are sharing insights and questioning assumptions, particularly about the nature of the measure used for integration.

Contextual Notes

There is a focus on the definition of integrability, particularly in the context of Riemann integrability, and the constraints of using bounded functions. The original poster emphasizes that the function must be bounded, which influences the direction of the discussion.

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Homework Statement


Let Q=I\times I (I=[0,1]) be a rectangle in R^2. Find a real function f:Q\to R such that the iterated integrals

\int_{x\in I} \int_{y\in I} f(x,y) \; and \int_{y\in I} \int_{x\in I} f(x,y)
exists, but f is not integrable over Q.

Edit: f is bounded

Homework Equations





The Attempt at a Solution


I am suggested to find a dense subset S of Q such that the intersection of S and each vertical/horizontal line contains at most one point. The most related idea once in my head was some space-filling curve, but it's probably not the right idea, since the function of the curve is, a little complicated...
Any hint? Thanks!
 
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An idea:

An ordinary multivariable calculus textbook might have an example (I found one). It is in the same spirit as the standard examples of functions that don't have limit at the origin even though the limits along paths do exist.

I can't yet figure out how the example was created, even though it works. Intuitively, I think the absolute value of the function behaves like 1/r^2, so that when you integrate r dr d\theta, you get 1/r near the origin. However, the function has enough "positive vs negative compensation" to get different finite iterated integrals.
 
thank you. I don't find an example in textbooks yet but your idea sounds workable, in which f seems to be unbounded, and I'll try to figure it out... well, I forgot to tell that f is supposed to be bounded in my original question. Anyway it's helpful and thanks a lot.
 
Are you required to use ordinary Lebesgue measure for integration with respect to x and with respect to y? Or can you use, say, counting measure for the integration with respect to y?
 
Oh, this is what I know about measure: a set A has measure zero means there is a countable collection of open/or closed rectangle that covers A that the sum of the volumn of each rectangle is less than an arbitrary given epsilon.
my original goal is to construct the subset S of Q, then \chi_{S} will be the desired function.
I'm wondering what the example is in your textbook...I haven't figured it out yet.
Thanks!
 
OK, I think I understand. I assume "integrable" means "Riemann integrable."

Some incomplete ideas:

Similar to your space filling curve idea, how about a line of irrational slope through the unit square, and whenever it hits the top or the right, it wraps around to the bottom or left, respectively.

Would it work to take two of those lines, and take their intersections.

Or, let {x} denote the fractional part of x. A sequence like {n*sqrt(2)} is dense in the unit interval. Can you use this idea in two dimensions.

Or, instead of only allowing one point per horizontal (or vertical) line, allow a finite number of points. Then use a set like (1/2,1/2), (1/3,1/3), (1/3,2/3), (2/3,1/3), (2/3,2/3), (1/5,1/5),... using prime denominators.
 
Thanks! the ideas are great, especially the last one, clear and efficient... well my space-filling-curve thought was...nasty... Thanks a lot!
 

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