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Help with a proof

  1. Mar 6, 2005 #1
    Let f = 0 if x=1/n for all n belonging to positive integers
    and 1 otherwise

    i need to prove that the integral of f over 0 to 1 exists and that the integral is 1.

    first of all i cant understand the function. (great start) If x = 1, then x is in the form 1/n then f(1) = 0.

    If i picked a partition [a,1] where a<1 then the upper sum would be U(f,P) = 0 (1-a) = 0 (right?)

    and lower sum would vary because A could be either of the form (1/n) of not of that form

    if a was of the form 1/n then L(f,P) = 0(1-a) = 0
    if a was not of the form 1,n then L(f,P) = 1(1-a) = 1-a

    stuck right here what should ido
    i'd realy \like to know if i am in right in terms of the fucntions value

    any help would be greatly appreciated
     
  2. jcsd
  3. Mar 6, 2005 #2

    Hurkyl

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    Just to be clear, you're saying the function is defined on the interval [0, 1] by:

    [tex]
    f(x) := \left\{
    \begin{array}{ll}
    0 \quad & x = 1/n \, \mathrf{for some} \, n \in \mathbb{N} \\
    1 \quad & \mathrf{otherwise}
    \end{array}
    [/tex]

    Since you're having trouble understanding the function, let's start with that first.

    What is f(1/2)? f(1/6)? f(3/4)? [itex]f(\pi/4)[/itex]? f(√2 / √8)?
     
    Last edited: Mar 6, 2005
  4. Mar 6, 2005 #3

    all values of x that are fractions of the form 1/2 are zero so f(1/2) = f(1/6) = 0 because no matter what the fraction you can lawys write it with a natural in the denom right? however the function at the square roots would be 1 becasue they are irrational

    as for 3/4 and n/4 they are also 1 because 1/0.75 is not a natural and 1/(n/4) = 4/n only if n is a multiple of 4
     
    Last edited: Mar 6, 2005
  5. Mar 6, 2005 #4

    Hurkyl

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    This is almost correct. Note that f(n/4) = 0 if n = 1, or 1/2, or ...


    Now, I actually wrote [itex]f(\pi/4)[/itex], but the default font renders pi as π, so it was confusing... but I'm happy about the mistake because seeing you write the above made it more clear that you're understanding how it works.

    Now, it's actually the case that f(√2 / √8) = 1! Can you figure out why?
     
    Last edited: Mar 6, 2005
  6. Mar 6, 2005 #5
    becasue the division of two irrationals gives an irrational number?
    i would think you suppose that two rationals divided would give a rational so then two irarationals divided should give an irrational.

    but isnt root2/ root8 = root 2 / 2 root 2 = 1/2??
     
  7. Mar 6, 2005 #6

    AKG

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    You can easily show the set of discontinuities to be countable, as they can very obviously be indexed by the naturals. Since it is a countable set of points, it has measure 0, so f is integrable. Alternatively, the upper sum is:

    [tex]U(f, P) = \sum _{S \in P} M_S(f)v(S)[/tex]

    Where v is the volume function, S is a sub-rectangle (in this case, sub-interval) of P, and [itex]M_S(f) = \sup \{f(x) : x \in S \}[/itex], i.e. the maximum value that f take on in the interval S. clearly, [itex]M_S(f) = 1[/itex] for all S, so:

    [tex]U(f, P) = \sum _{S \in P}v(S) = v([0, 1]) = 1[/tex]

    Therefore, the upper integral, UI(f) = 1, since:

    [tex]UI(f) = \inf _P \{U(f, P)\} = \inf _P \{1\} = 1[/tex]

    Now, if you are allowed to use the argument I gave initially to prove that f is integrable, then the above is sufficient to show that the integral is 1. On the other hand, if you have to show integrability by showing that the lower integral equals the upper integral, then it remains to show that the lower integral is 1:

    [tex]L(f, P) = \sum _{S \in P} m_S(f)v(S)[/tex]

    Suppose the lower integral were not 1, suppose it is 1 - e. Choose a partition P so that it starts with the sub-interval [itex]S_1 = [0, e/2][/itex]. Now, on the interval [e/2, 1], there are finitely many points of the form 1/n. So, partition [e/2, 1] by placing small intervals around the points of the form 1/n. For such a partition:

    [tex]L(f, P) = \sum _{S \in P} m_S(f)v(S) = 0(e/2) + \sum _{S \in (P - \{S_1\})} m_S(f)v(S)[/tex]

    Since we can choose the rectangles surrounding the finitely many points of the form 1/n in [e/2, 1] to be arbitrarily small, i.e. we can take "better" and "better" partitions with smaller and smaller intervals surrounding these points, the rectangles on which [itex]m_S(f) = 0[/itex] become smaller and smaller, and since there are finitely many of them, we know that if we take the supremum of L(f, P) over all partitions P of this form, we must get a value of 1 - e/2. This contradicts the assumption that the supremum of L(f, P) over all partitions P (of any form) is 1 - e (note the the supremum over these L(f, P) is exactly, by definition, the lower integral), so the supremum must be 1, as required.
     
  8. Mar 6, 2005 #7

    Hurkyl

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    Er, bleh, typo. I was trying to contradict when you said " however the function at the square roots would be 1 becasue they are irrational" and I said 1 myself! Yes, f(√2 / √8) = 0.


    Anyways, I think you understand the function now. Do you know where it is discontinuous?
     
  9. Mar 6, 2005 #8

    Hurkyl

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    PS. don't do the entire problem for him. :tongue2: One learns more by getting hints and filling in the missing steps than reading how someone else solved the problem.
     
  10. Mar 6, 2005 #9
    what is a volume function?? (note i ma in a first year intro to real analysis course)

    p.s. i probably know this as the width of the interval

    so the function is 1 whenever x is not of the form a/n and n is not an integer multiple of a. i understand the function now :smile:
    as for hte proof now

    what i posted in post #1 is erroneous in some way??

    p.s. coirrected

    yes the upper sum is 1. So i cannot compute the lower sum because that would give a zer owouldnt it ?/

    and bascially your proof states that the function is 1 for the most part but the discontuities are finite because the set of 1/n is finite so they are removable .
     
    Last edited: Mar 6, 2005
  11. Mar 6, 2005 #10

    AKG

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    The volume function in 1 dimension is just the length of the interval. Good point Hurkyl, my bad.
     
  12. Mar 6, 2005 #11

    Hurkyl

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    Right, so you see that already...

    Also, [a, 1] is not a partition. A partition of the interval [0, 1] requires a collection of subintervals that cover the whole thing. For instance, {[0, 1/3], [1/3, π/4], [π/4, 1]} is a partition, though you would usually write it as the sequence (0, 1/3, π/4, 1).
     
  13. Mar 6, 2005 #12
    i dont undersatnd the step
    [tex]L(f, P) = \sum _{S \in P} m_S(f)v(S) = 0(e/2) + \sum _{S \in (P - \{S_1\})} m_S(f)v(S)[/tex]

    where does 0(e/2 ) comes from??
     
  14. Mar 6, 2005 #13

    AKG

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    Well, the sum is over all subintervals S in P. Any partition has finitely many sub-intervals, so we can write P = {S1, S2, ..., Sm} (so it has m sub-rectangles). The sum over all S in P is the same as the "sum" over just S1 plus the sum over all the remaining sub-rectangles S2, S3, ..., Sm. The sum over just S1 is 0 x e/2. 0 is the minimum value of f on that interval, and e/2 is the "volume" (or length) of that interval.
     
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