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Equal partition Integral problem

  1. Jul 20, 2013 #1
    This is spivak's calculus 2nd edition #12b.

    The question in part a defines Pn as a partition of [a,b] into n equal intervals.

    The question in part b states: Find an integrable function f on [0,1] and an ε > 0 such that U(f,Pn)-∫0 →1 (f) > ε for all n.

    I'm made no progress on this at all.
    Part (a) asks to prove that there is no such ε if f is continuous, and I proved that using uniform continuity. I've been trying to create a function that has little jumps at all the rational numbers and none anywhere else to make a counterexample, but that gets me nowhere.

    Also, I learned that apparently there are different types of integrals, so I think that this book uses the Riemann Stieltjes Integral.
     
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  3. Jul 21, 2013 #2

    HallsofIvy

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    What about taking f to be 0 if x is irrational, 1 if x is rational? That should work.
     
  4. Jul 21, 2013 #3
    That function f isn't integrable. Since in every interval there is some x so f(x)=1 and some x so f(x)=0, the upper sum is 1 and the lower sum is 0.
     
  5. Jul 21, 2013 #4

    WannabeNewton

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    Can you modify what Halls wrote down in any way?
     
  6. Jul 21, 2013 #5

    micromass

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    A small hint:

    [Broken]
     
    Last edited by a moderator: May 6, 2017
  7. Jul 21, 2013 #6
    There are two things I can think of. One is an example they use a lot in the book which is 0 for irrationals and 1/q for a rational with p/q in lowest form. But the upper bound can be arbitrarily close to 0 for equal partition.

    The second is let the value be 1 for all rationals of the form 1/m and 0 otherwise. That is definitely integrable but it's hard to tell if equal partitions could show it's integrable.

    Also, I must be missing something, but how is popcorn a hint?
     
  8. Jul 21, 2013 #7

    micromass

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    Clearly ##\int_0^1 f = 0## in both cases. Can you calculate ##U(f,P_n)## for both functions?

    http://en.wikipedia.org/wiki/Popcorn_function
     
  9. Jul 21, 2013 #8
    It's really hard to tell. I thought it was obvious earlier but I was only thinking of the values of the function at the end points.
     
  10. Jul 22, 2013 #9

    HallsofIvy

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    Oh, right. You did say "Riemann Stieltjes Integral".
     
  11. Jul 26, 2013 #10
    @Micromass

    I'm fairly sure for the first function ##\inf(U(f,P_n))=0##.

    A non-rigorous proof is as follows:
    Let's say we want to find ##n## such that ##U(f,P_n)<\frac{1}{m}## Note that there's only a finite number of rationals ##x## such that ##f(x) \ge \frac{1}{m}##. Each of these can only be in a maximum of two intervals in the partition ##P_n##. Every other interval will have a max value of ##\frac{1}{m+1}## or less. Thus if you make the intervals small enough, eventually all the intervals with max values ##\frac{1}{m+1}## or less will "outweigh" the small number of intervals with max values greater than ##\frac{1}{m+1}##.

    You can do a similar kind of proof for the second function to show ##\inf(U(f,P_n))=0##.

    While of course it's not rigorous (I don't want to go through the trouble of working with these fractions), is there any problem with my argument?

    Also, very strangely, in the appendix to the chapter, the book says this:

    "The moral of this tale is that anything which looks like a good approximation to an integral really is, provided that all the lengths ##t_i - t_{i-1}## of the intervals in the partition are small enough. We've proved that this only for continuous ##f##, but it actually holds for any integrable ##f##."

    This is said after proving that if ##f## is continuous on ##[a,b]##, then for every ##\epsilon > 0## there is a ##\delta > 0## such that for a partition ##P = \{t_0,\cdots,t_n\}## of ##[a,b]##, with each piece of the partition with a width smaller than ##\delta## then ##\mid \sum_{i=1}^n f(x_i)(t_i-t_{i-1}) - \int_a^b f(x)\,dx. \mid < \epsilon ##.


    How is possible for the problem in my original post and this statement to be true?
     
    Last edited: Jul 26, 2013
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