Double integral of piecewise function

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?
 
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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.
 
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