Set of discontinuties with measure 0

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SUMMARY

The function f : [0,1]² -> R defined as f(x,y) = 1 if x ≠ 1/2, 1 if x = 1/2 and y is rational, and 0 if x = 1/2 and y is irrational is Riemann integrable. The set of discontinuities D = {x = 1/2} has measure 0, which confirms its integrability according to Fubini's Theorem. The confusion arose from the misunderstanding that the discontinuities were only at points (1/2,y) where y is irrational, but it was clarified that the discontinuities include all y in [0,1], forming a line with measure 0.

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Buri
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My professor gave this function in class as an example (for Fubini's Theorem)

f : [0,1]² -> R defined as:

f(x,y) = 1 if x not equal to 1/2, 1 if x = 1/2 and y rational, and 0 if x = 1/2 and y irrational.

My professor said that this function is Riemann integrable since the set of discontinuities D = {x = 1/2} has measure 0. However, I don't see why this is true. The function is discontinuous at all points of the form (1/2,y) where y is an irrational number in [0,1], but the irrationals in [0,1] are uncountable so I couldn't possibly cover it with countably many rectangles with total volume less than epsilon. Maybe it was meant to be y rational?

Thanks for the help!
 
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Wait never mind...the set of discontinuities is actually x = 1/2 and all y in [0,1] which is a line (and that has measure 0). Sometimes just asking a question helps me because I had just left my computer and realized where I was going wrong lol
 

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