Need help understanding proof that continuous functions are integrable

[chipotleaway]

Actually, the theorem is that functions that are uniformly continuous are Riemann integrable, but not enough room in the title!

I'm failing to see the motivation behind proof given in my lecturer's notes (page 35, Theorem 3.29) and also do not understand the steps.

1) First thing I'm not sure I understand is in the first line of the proof - the definition of L. I think this is the set of all upper Riemann sums corresponding to different partitions, because for a given partition P, isn't the upper Riemann sum of f over P unique?
Then taking the infimum of the set would be taking the smallest upper Riemann sum.

2) I have no idea how the chain of inequalities come about, not even the first one.

3) Isn't the proof done by line 3 of the inequalities (before 'Also')? What do the last 3 lines show?

Any help is much appreciated - these proofs are doing my head in.

 

[gopher_p]

Actually, the theorem is that functions that are uniformly continuous are Riemann integrable, but not enough room in the title!

I'm failing to see the motivation behind proof given in my lecturer's notes (page 35, Theorem 3.29) and also do not understand the steps.

I'm going to assume that all instances of ##\bar S## are typos and should have been ##\bar s##.

1) First thing I'm not sure I understand is in the first line of the proof - the definition of L. I think this is the set of all upper Riemann sums corresponding to different partitions, because for a given partition P, isn't the upper Riemann sum of f over P unique?
Then taking the infimum of the set would be taking the smallest upper Riemann sum.

##L## is not the set of all upper sums. Rather it is the infimum of that set.

Given a partition ##\mathcal{P}##, ##\bar s(\mathcal{P})## is a number. It's the supremum (you can prove it's actually the max) of all Riemann sums with sample points chosen according to the partition. I personally wouldn't use the word "unique" to describe a number. Maybe "well-defined" ...

2) I have no idea how the chain of inequalities come about, not even the first one.

The first inequality follows from the fact that ##\bar s(\mathcal{P})## is the max of all Riemann sums with sample points chosen according to ##\mathcal{P}##.

I have not worked out the details, but the second inequality almost certainly follows from the uniform continuity condition and the fact that ##\mathcal{P}_{\epsilon}'## is a refinement of ##\mathcal{P}##. While ##\bar s(\mathcal{P}_\epsilon ')\leq\bar s(\mathcal{P})## (a refinement results in a smaller upper sum), ##2\epsilon\Pi|b_j-a_j|## is (supposedly ... again I haven't checked the details) enough of a buffer to get the given inequality.

The third inequality follows from the fact that ##\bar s(\mathcal{P}_\epsilon ')\leq\bar s(\mathcal{P}_\epsilon )##, which is true because ... I'll let you figure that out. Hint: I already gave you the answer.

The last inequality is true because ##\bar s(\mathcal{P}_\epsilon)-L\leq\epsilon## from the definition of ##\mathcal{P}_\epsilon##.

3) Isn't the proof done by line 3 of the inequalities (before 'Also')? What do the last 3 lines show?

The first round of inequalities establishes an upper bound on ##\sum_{R\in\mathcal{P}}f(x(R))\int_R\ dy-L##. The second batch gives a lower bound. You need both to establish ##\sum_{R\in\mathcal{P}}f(x(R))\int_R\ dy-L\rightarrow 0##.

 

[chipotleaway]

Thanks a lot gopher, think I get it now. I'd thought the partition P had to be the one that gave the smallest Riemann sum and this was one of things that was making it confusing. By the way, the 2ε∏|b-a| term results from the previous lemma in the notes.

I still don't see how he came up with the proof though that applies for the majority of calculus/analysis proofs I work through - it's all stuff I could never come up with (proofs in linear algebra at this level, I find much more doable).