For question 1, the “only if” part is easy. For the “if” part, it suffices to show that for an integrable f on and , there exists continuous h on such that and .

As f is integrable, there exists Partition such that the difference of the upper sum and the lower sum of is less than . Let and be the maximum among absoulte values of them. Let such that . Now it is time to define .

If , define on ; Otherwise define on and on . For the remaining intervals we make it linear. It is not difficult to check this is the desired one.

For question 2, the answer is negative. Let us contruct the sequence of continuous by the Cantor set. Instead of using the original Cantor set, we take the middle one forth on each interval each time, and we denote the remaining set at the n-th construction by . The resulting Cantor set has measure >0. Now we are going to construct convergigng to , discontinuous on .

Let be the length of each interval taken away on . Define on ; on the middle interval of each interval in , so that the lengths of the intervals on its both sides are (corrected by Milton). For the remaining undefined part of , we define linearly. This converges to which is 0 on and 1 otherwise. It is easy to check that this is discontinuous on , hence not R-integrable.

Actually, if you know a fact that there exists a function such that exists but is not R-integrable, it is easy to answer question 2 as we can let . The sequence of functions inside the limit is the desired one.

_________________Sometimes Truth is meanlingless;What means is how you believe in.

偶爾，真相並沒有意義；意義在於你怎樣相信。

Last edited by Milton on Thu Mar 03, 2005 12:33 am; edited 3 times in total