Can a Dense Set Help Prove Integrability?

[kingwinner]

http://www.geocities.com/asdfasdf23135/advcal16.JPG

I really need some help on this question. I have attached a theorem that I think should be helpful, but I still have no idea how to prove (or disprove) the integrability.

Also, for part (ii), first of all I would like to undertsand the question. Is k a fixed integer or is k an arbitrary integer? Will this affect the answer?

I am sure someone here knows how to solve this problem. Please give me some hints so that I can finish it, too!

Thanks a lot!:)

 

[Dick]

On the second one k is intended to range through all of the integers. So it's an unbounded set and can't be covered by any finite set of rectangles. So it's not 'zero content' in the sense of the other exercise. My suggestion would be to use the definition of Riemann integrable directly, define upper and lower sums and try to make them converge to each other.

 

[kingwinner]

On the second one k is intended to range through all of the integers. So it's an unbounded set and can't be covered by any finite set of rectangles. So it's not 'zero content' in the sense of the other exercise.

(ii) But the question says FOR SOME k, I am not sure what it means...
It says "for some", so shouldn't k is fixed? (i.e. we are GIVEN an integer k?) I am very confused by this kind of wording...

My suggestion would be to use the definition of Riemann integrable directly, define upper and lower sums and try to make them converge to each other.

Define upper and lower Riemann sums using what partition?

More hints would be nice...

Thanks!

 

[Dick]

Yes, it does say for 'some k'. That's a little ambiguous. If it means for a fixed k, then it's basically the same as the first problem. Might be worth asking about. For the first one think about a partition that includes your f=1 points but is very narrow around each point. You are going to want to show you can find a partition such that the lower sum is zero and the upper sum is as small as you like. In the same spirit as the zero content proofs.

 

[kingwinner]

(i) Is this going to work?

(ii) I asked my instructor and he says that k is NOT fixed, but I think it is horrible wording...
Now, taking k as NOT fixed, is it integrable on [0,1] or not? And how can I prove it? (I am particularly scared about the proof if the answer is no since none of the theorems in my textbook seem to apply...)

Thanks for helping!

 

[EnumaElish]

Think of it this way: (i) is a special case of (ii), with k = 1.

You can make up as many special cases as there are positive integers. But (ii) asks you to make the general case by treating k as "any positive integer." k is not a special integer, it can be any integer. But while you are solving the problem, you should treat it as a constant integer number. Like Dick posted, (ii) is pretty much like (i), except "k" replaces "1," so the function is discontinuous at different points (except when k=1) relative to (i). That's the only difference (and when k=1, there is no difference).

Does that make it clear?

 

[kingwinner]

Um...I am very pulzzed...(sorry)

Are (i) and (ii) actually integrable on [0,1]? I would like to know this so that I can think about the proof in the right direction.

 

[Dick]

If k is not fixed then I would assume that to mean the values of x include ALL k. This makes i) and ii) quite different. It is useful to know in what direction to push a proof. i) is Riemann integrable. ii) is not. The values of x in the first case are discrete points, in ii) the x are dense set of the reals (like the rationals).

 

[kingwinner]

If k is not fixed then I would assume that to mean the values of x include ALL k. This makes i) and ii) quite different. It is useful to know in what direction to push a proof. i) is Riemann integrable. ii) is not. The values of x in the first case are discrete points, in ii) the x are dense set of the reals (like the rationals).

For ii), WHY is the set S={k/2^n|k E Z, n=0,1,2,...} a dense set of [0,1]? Why would this fact make f not integrable on [0,1]?

To prove that someone is not integrable, the only thing I can think of is to use the lemma on the last line:
http://www.geocities.com/asdfasdf23135/advcal17.JPG

But how?

 

[Dick]

Because if k assumes all integer values and n can be ever so large then every interval [a,b] will contain a point where f(x)=1 and another point where f(x)=0. Since for sufficiently large n, 2^(-n) is less than b-a.

 

[kingwinner]

Because if k assumes all integer values and n can be ever so large then every interval [a,b] will contain a point where f(x)=1 and another point where f(x)=0. Since for sufficiently large n, 2^(-n) is less than b-a.

OK! Now ,how can I prove that f is not integrable on [0,1]?

 

[Dick]

Reread that answer about what values f takes in any rectangle. What does that mean for upper and lower sums?

 

[EnumaElish]

If k is not fixed then I would assume that to mean the values of x include ALL k. This makes i) and ii) quite different. It is useful to know in what direction to push a proof. i) is Riemann integrable. ii) is not. The values of x in the first case are discrete points, in ii) the x are dense set of the reals (like the rationals).

This would make the problem rather unconventional, at least relative to the "standard" interpretation that k is a constant integer number.

 

[Dick]

This would make the problem rather unconventional, at least relative to the "standard" interpretation that k is a constant integer number.

In post 5 the instructor told the OP that k is not constant. That doesn't make the problem all that unconventional. It's now basically the same proof as showing that the characteristic function of the rationals is not Riemann integrable.

 

[kingwinner]

Because if k assumes all integer values and n can be ever so large then every interval [a,b] will contain a point where f(x)=1 and another point where f(x)=0. Since for sufficiently large n, 2^(-n) is less than b-a.

I still don't understand this completely. Just what does it mean by "dense" intuitvely?
I know that the rationals and irrationals are dense in the reals, but nothing else...nor does my textbook ever explain the word "dense"...

 

[Dick]

For the purposes of this problem, it means EVERY interval contain members of the set.

 

[EnumaElish]

http://en.wikipedia.org/wiki/Dense_set