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Can a Riemann-integrable function have uncountably many discountinuities in its domain of integration? (where obviously, 'countably infinitely many' would not be 'uncountably many'.)

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up vote 5 down vote accepted

Yes, the condition is as long as the number of discontinuities has Lebesgue measure zero, the function is Riemann integrable. Since there are plenty of sets that are uncountable with measure zero (like the Cantor set), the answer to your questions is yes a Riemann integrable function can have uncountably many discontinuities in its domain of integration.

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Given some function $f$, I can show it is Riemann integrable by picking a partition where the upper and lower sums get as close as I want. If $f$ is continuous, you do not need to be particularly clever about choosing the partition. The continuity will guarantee that if the partition is fine enough, the maximum and minimum on each interval get squeezed together.

When $f$ has a discontinuity, you need to be more careful about what your partition is doing. Clearly, on any subinterval containing the discontinuity, you can't make the maximum and minimum values arbitrarily close. Thus, the only way to make this difference between the upper and lower sums disappear is by repeatedly shrinking the subinterval where the discontinuity lies. If we say that the discontinuity lies in some subinterval $I$, and on that interval we have maximum values $M_{I}$ and minimum value $m_{I}$, then that quantity that this subinterval contributes to the difference between the upper and lower Riemann sums is $|I||M_{I} - m_{I}|$. Since we cannot shrink $|M_{I} - m_{I}|$, we must shrink $I$.

When $f$ has countably many discontinuities. The same trick, modified, will work. Successively build partitions that isolate more and more discontinuities into smaller and smaller subintervals. If you do this in such a way that you kill off the worst discontinuities first, in the limit you can have the Upper and Lower Riemann sums agree.

Now, you want to see what happens if $f$ has uncountably many discontinuities. Well, let's see what happens if we try the same trick as in the countable case. Let's try and isolate the discontinuities into smaller and smaller subintervals, killing off the worst discontinuities first and then working our way down to the smaller discontinuities. The immediate problem here is that if we isolate one additional new discontinuity in each new partition, we can only isolate countably many in the limit! So if we have any hope of squashing away the effect of the discontinuities, the set of those discontinuities must be rather clumped up, so that we can dispose of uncountably many at a time. Can you think of an uncountable set that can be covered by countably many intervals of arbitrarily small length?

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Nice description of the idea of the proof. – André Nicolas Feb 16 '13 at 0:40
Fantastic explanation-- thank you very much. – Ryan Feb 18 '13 at 6:10

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