In this note, we prove that the set of continuous point of Riemann integrable functions on some interval is dense in . Our proof start with the following simple observation.
Lemma: Assume that is a partition of such that
for some ; then there exists some index such that where and are the supremum and infimum of over the subinterval .
We now prove this result.
Proof of Lemma: By contradiction, we would have for all ; hence
which gives us a contradiction.
We now state our main result:
Theorem. Let be Riemann integrable over . Define
Then is dense in .
We now prove the theorem.
Proof of Theorem: Let be a partition of such that , where and are the upper and lower Darboux sums of . By the lemma, there is some index for which .
Step 1. Narrow to obtain a subinterval whereas is relatively small.
To see this, we first select
If happens to be , then we immediately get . Otherwise, it will be the case that
where (resp. ) and (resp. ) and also that
where (resp. ) and (resp. ). Thus, in any case, there would exist and
where . Since is integrable over , it is integrable over .
Step 2. Repeat the above argument to get a sequence of strictly nested subintervals.
The very same reasoning but with gives us , with and with
Continuing the process gives us a sequence of nested closed intervals such that
By the Cantor theorem,
Step 3. Existence of points of continuity in every subinterval.
Let then . Since the sequences of and are strictly monotonic, it follows that is none of the extrema. Let be given, and choose such that . Keep in mind that for any .
Let such that if , then . Since , for each , it follows that if then
and thus is continuous at . Since is integrable over it also is integrable over and . Set .
By the previous reasoning, we obtain two new different points of continuity, and . Having repeated this times we would obtain continuity points. Thus cannot be continuous over a finite number of points , for it would mean there exists such that for each . It follows is infinite.
Step 4. is dense in .
Let and consider the sequence
The previous argument means there exists for each a with . But then given we can take obtain an large enough such that whenever and thus
for , that is, . Since was arbitrarily chosen, is dense over . This completes the proof.