Since Q is dense in R, we have that supx∈[xi,xi+1]f(x)=1 and infx∈[xi,xi+1]f(x)=0 on any interval [xi,xi+1]⊂R, therefore the upper and lower sums don't converge to the same limit. Hence f is not Riemann-integrable.
It does not imply that every discontinues function is non-integrable because g(x)=x for x∈[0,1) and g(x)=x+1 for x∈[1,2] is integrable on [0,2] and have discontinuity at x=1.
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