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.

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Reblogged this on daodat22121996 and commented:

Something …. really “dense” 🙂

Comment by narutomath96 — April 5, 2015 @ 22:54

? 🙂

Comment by Van Hoang Nguyen — December 13, 2016 @ 16:49

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