The canonical example of a function that is not Riemann integrable is the function f: [0,1] to R, such that f(x)=1 if x is rational and f(x)=0 if x is irrational ( i know some texts put this the other way around, but bear with me because i can reference at least one text that does not). Hence, for any partition P of [0,1], the upper sum is 1 and the lower sum is 0, so that f is not Riemann integrable. However, Lebesgue's theorem states that if f is a bounded function on a bounded set A, and we extend f to all of R by letting f(x)=0 for all x not in A, then f is Riemann integrable iff the set of points at which the extended f is discontinuous form a set of measure zero (Marsden, "Elementary Classical Analysis", pg 261). Thus, if f is the function defined above, the extension of f is zero everywhere except for the rational numbers r, such that 0 <= r <= 1. Now it is quite common knowledge that the set of rationals between 0 and 1 is a countable set and thus has measure zero. And since the extension of f is zero everywhere except at these rationals, we should be able to apply Lesbegue's theorem and conclude that f actually IS Riemann integrable on [0,1]. Or you could just use the simple Corollary given in Marsden, that a bounded function with a countable number of discontinuities is Riemann integrable. Aren't these results contradictory? I know many texts define f as 1 at irrationals and 0 at rationals, so that this issue does not arise. However the text "Measure Theory and Integration" by Taylor presents it the way I described. And besides, it's still a contradictory result, isn't it? What am I missing here?