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Homework Help: Uniform partition is the only one that matters

  1. Apr 23, 2013 #1
    1. The problem statement, all variables and given/known data
    We had to prove that an integral didn't exist by computing upper and lower sums. It was for

    f(x)=x for rational
    0 for irrational

    2. Relevant equations

    3. The attempt at a solution

    Lower was easy, it's just 0 (inf of any interval is 0). But the upper I had a hard time with. I was trying to compute it for any arbitrary partition. Then I basically said that it's the same for f(x)=x, which works.

    But the TA today said that the only partition that matters is the uniform one. That if we compute the upper sum for the uniform partition, it will be the inf of the upper sums.

    Kind of my understanding of why is that as n->inf, the partition turns into a refinement of every single possible partition. Thus is the inf. Correct? Why even bother talking about other partitions then? If the uniform is the only one that matters...
  2. jcsd
  3. Apr 23, 2013 #2


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    Whether the partition is uniform or not doesn't make any particular difference. The lower bound is zero because every interval of finite size contains an irrational. The upper is nonzero because every interval of finite size contains a nonzero rational. So yes, the upper sums of f(x) are the same as the upper sums of f(x)=x. What's important isn't that the partition is uniform. It's that both the rationals and irrationals are dense. You should mention that someplace.
  4. Apr 23, 2013 #3
    Of course I used density. I thought that went without saying.

    But my question is how do we compute the upper sum? The problem specifically asks to compute the upper sum, as in get a number. The TA said just use the uniform partition. My understanding is that this works because it's a refinement of ALL possible partitions. Am I correct in that assertion?

    If the question was "show that this integral doesn't exist", you're right the partition wouldn't matter because the upper sum would be non-zero. But that's not the problem. The problem asks "find the upper and lower sums".
  5. Apr 23, 2013 #4
    Maybe I should have made that clearer. Because you're right, IF the problem was just asking to show that it didn't exist, I would just have to show that it's non-zero. But it's asking to explicitly compute the sums.
  6. Apr 23, 2013 #5


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    The contribution to the upper sum over any interval [a,b] is b*(b-a) if a>=0, isn't it? Even if b is irrational. That's just density again. As you said, the upper sums are the same as f(x)=x. The lower sums are the same as f(x)=0.
  7. Apr 23, 2013 #6
    That's correct. That still doesn't answer my question.

    Yes, over any interval, the contribution is b*(b-a) because b is the sup. I get that. But I'm talking about the upper integral.

    There's a theorem we have that if we can find ONE sequence of partitions such that U(f,Pn)-L(f,Pn) goes to zero, then we know that f is integrable. However is it true that if we can find ONE sequence of partitions such that it DOESN'T go to zero, f is NOT integrable.

    I thought the negation of there exists is for all. So we have to show that for ALL sequences of partitions U(f,Pn)-L(f,Pn) doesn't go to zero. But if we show that this isn't true for the uniform partition, we show that the integral doesn't exist?
  8. Apr 23, 2013 #7
    I guess the clearest way to state this is:


    [itex] \sum sup(I_{k})*(b-a)/n \neq \sum inf(I_{k})*(b-a)/n[/itex] imply that the upper and lower sums are not equal?

    Where the Ik's are the uniform intervals
    Last edited: Apr 23, 2013
  9. Apr 23, 2013 #8
    For ANY arbitrary function, not just the one given. If it helps, I'm not confused about the function. I get why the upper and lower sums are different. The upper sum is the same for f(x)=x, the lower is just 0. That's NOT my question.
  10. Apr 23, 2013 #9


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    That's not clear at all. If you pick [a,b] to split into partitions where a>0 b>a, then all of the lower sums are zero, no matter what the partition. All of the upper sums are nonzero and greater than a*(b-a). Doesn't that prove it's not integrable?
  11. Apr 23, 2013 #10
    That does prove that it's not integrable, but again that's not my question (or even the problem I had to do for homework). And I cleaned up some of the notation.

    You're right, the lower sum is zero regardless of partition. I agree, and that's easy to show. But is the upper sum of ANY BOUNDED FUNCTION always reached by computing the uniform partition?
  12. Apr 23, 2013 #11
    The problem was in two steps:
    Compute the upper and lower sums.
    Then show that the integral doesn't exist.

    The first part is where I'm stuck. The lower is 0 clearly. But to compute the upper, can I just use the uniform partition? How do I know that the uniform partition is the infimum?
  13. Apr 23, 2013 #12


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    No, you can find it with any sequence of partitions whose mesh size goes to zero. It doesn't HAVE to be uniform. If the upper limit exists for uniform partition sequences, then it will also exist for nonuniform partition sequences. And vice versa.
  14. Apr 23, 2013 #13
    Awesome, that answered my question perfectly. So as long as the length of the intervals goes to 0, the sum will be the same regardless?

    On another note, is there a proof of this? I'd be curious to see the proof. I'm not asking you to prove it, but if you know of somewhere that a proof exists or you have a brief sketch, I'd really appreciate it. :)
  15. Apr 23, 2013 #14


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    Uniform partitions are a subset of all partitions. That immediately tells you the the inf over uniform sums is greater than or equal to that the inf over nonuniform partitions. If the endpoints are rational, then you can always refine to a uniform partition with rational endpoints. That would show you the inf of the upper sums over a rational nonuniform partition is greater than or equal the rational uniform, since if you refine the value of the upper sum will drop. The final step would be to show you can approximate a nonrational partition pretty well by a rational partition. Since the rationals are dense. I've never seen a proof of it, but it should go something like that.
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