- #1
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!
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!
