Suppose that f is continuous a.e. Let e > 0. We know that for almost every x there exists d > 0 such that |x ' -x| < dÞ |f(x ' )-f(x)| < e. Let A_n be the set of x such that d = 1/n works. We see that A_n is increasing (that is A_n is contained in A_n+1 for all n) and the union of A_n has measure 1.

Fix 0 < c < 1. Since measure is a continuous set function we have lim_n®¥m(A_n)=m(A)=1 (here m(S) stands for the measure of the set S) so there exists n such that m(A_n) ³ c.

In other words for any e > 0 and any 0 < c < 1 we can find an n such that the set of x such that:

|x ' -x| < 1/n Þ |f(x ' )-f(x)| < e (*)

has measure at least c.

Now take a dissection of [0,1] into intervals of length < 1/n. The sum of the lengths of intervals which have an x with the property (*) certainly is at least c - since their union contains the set of x with property (*) which has measure c. In each of these intervals there exists x with the property (*) so f varies by less than 2e in each of these intervals.

Now let's estimate the difference between the upper and lower sum. What about the contribution of the intervals which have an x with (*)? Well f varies by less than 2e in each of these intervals and the sum of the lengths of these intervals cannot exceed 1 so the maximum contribution to the difference between the lower and upper sum is 2e. What about the contribution of the intervals which don't have an x with property (*)? Well the sum of the lengths of these intervals is at most (1-c) and f varies by no more than S-s in these intervals. (Where S=sup f, s=inf f.) Hence the maximum contribution is (1-c)(S-s).

Hence the difference between the lower and upper sum is at most 2e+ (1-c)(S-s).

To recap: given e > 0 and 0 < c < 1 we can find n such that any dissection whose intervals have size < 1/n has the property that difference between the lower and upper sums is < 2e+ (1-c)(S-s). This can be made arbitrarily small by taking e small and c close to 1. So we're done one way.

Suppose f is not continuous a.e. Let A_n be the set of x such that: there doesn't exist d > 0 with |x ' -x| < dÞ |f(x)-f(x ' )| < 1/n. The union of the A_n is the set on which f is discontinuous. Since their union does not have measure zero there exists an m such that A_m does not have measure zero. In particular there exists a c > 0 such that A_m cannot be covered by disjoint intervals whose lengths sum to < c.

Dissect [0,1] into n intervals of length 1/n. A_m must intersect at least n c of these intervals (not counting intersections at endpoints). f varies by at least 1/m on these intervals. So the difference between the upper and lower sums is ³ 1/m ×n c×1/n=c/m, for any n. If f were Riemann integrable then the difference would tend to zero as n®¥ which it doesn't. Hence f is not Riemann integrable.