1. The problem statement, all variables and given/known data Let f(x,y) = 1 if x = 1/3 and y is rational, and let f(x,y) = 0 otherwise. Show that the double integral of f over the region Q = [0,1]x[0,1] in R2 exists (SSQ f dA exists) yet the integral from 0 to 1 of f(1/3, y) does not exist. (sorry for the weird way of writing, I'm not sure how to do integral signs) 2. Relevant equations 3. The attempt at a solution For the second part - a function is integrable if its set of discontinuities is finite or has zero content. However, irrational numbers are infinitely dense in any interval so the set of discontinuities are infinite and they cnanot be covered by a finite amoutn of intervals with arbitrarily small lengths, so the function f(1/3, y) is not integrable But the second part is kind of counter-intuitive to me. No matter how i partition that square, as long as the partitions have real lengths, then the lower riemann summ will always be 0 (since every square that contains a point where x= 1/3 and y is rational, will also contain points where x is not 1/3 and/or y is not rational) and the upper riemann summ will always be non-zero (since, any partitioning into squares of real lengths, will result in at least some squares containing points where x = 1/3 and y is rational). So, the lower integral and upper integral will not agree (one is zero, one non-zero), so how can the double integal over the region exist?