Double integral of piecewise function

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The discussion focuses on computing the double integral of the piecewise function f(x,y), defined as 1 for rational x and 2y for irrational x, over the unit square [0,1]x[0,1]. One participant suggests that integrating with respect to y yields areas of 1 for both cases, leading to the conclusion that both integrals should equal 1. However, another contributor clarifies that the reasoning is flawed, emphasizing that any region contains both rational and irrational points, which complicates the integration. The conversation also touches on the relevance of Lebesgue measure, noting that the set of points where x is irrational has a two-dimensional Lebesgue measure of zero. Overall, the integral's evaluation hinges on understanding the behavior of the function across rational and irrational inputs.
Dr. Lady
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Homework Statement


Let f(x,y)= 1 if x is rational, 2*y if x is irrational
Compute both double integrals of f(x,y) over [0,1]x[0,1]


Homework Equations





The Attempt at a Solution



I'm tempted to say that we can do the dydx integral since when x is rational, integrating y gives squares of area=1 and when x is irrational, we get triangles of area 1, so when we integrate over x, we just get 1. Then, since f(x,y) is bounded, the other integral has the same value.

Is this reasoning any good at all, or am I just crazy?
 
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(disclaimer... I don't know too much about it myself, so apologies if i confuse the issue)

but could this have something to do with Lebesgue measure, and one of those sets having Lebesgue measure of zero?
 
Last edited:
Dr. Lady said:

Homework Statement


Let f(x,y)= 1 if x is rational, 2*y if x is irrational
Compute both double integrals of f(x,y) over [0,1]x[0,1]


Homework Equations





The Attempt at a Solution



I'm tempted to say that we can do the dydx integral since when x is rational, integrating y gives squares of area=1 and when x is irrational, we get triangles of area 1, so when we integrate over x, we just get 1. Then, since f(x,y) is bounded, the other integral has the same value.

Is this reasoning any good at all, or am I just crazy?
I don't know what you mean by "integrating y gives squares of area 1" or "we get triangles of area 1" integrating with respect to y gives a number as does integrating first with respect to x. If you were doing this with as a Riemann integral the crucial point would be that any region in the plane, no matter how small, contains points (x,y) in which x is rational as well as points in which x is irrational. If you are doing this as a Lebesque integral, You can, as lanedance suggests, divide the region into the set {(x, y)| x rational} and {(x,y)| x irrational}. and the second set has two dimensional Lebesque measure 0.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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