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Analysis Riemann Integral problem

  1. Jun 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose α(x) increases on [a,b] a≤ x_0 ≤b, α is continuous at x_0,
    f(x_0) =1 , at all other x in [a,b] f(x)=0.
    denote ('x knot' as x_0)


    Prove that f is Riemann Integrable and that ∫fdα=0.



    2. Relevant equations
    Can anyone check my proof or suggest a good method to show that inf U(P,f,α)= 0?
    I have proved that the lower limit is equal to zero. Now I just need to prove that the upper limit is equal to zero or that f is Riemann Integrable.



    3. The attempt at a solution

    Here's my attempt at proving that : U(P,f,α)= 0:

    α continuous at x_0 ⇒ for each ε>0 there exists a δ>0 s.t for q in [a,b] if
    ⎮α(x_0)-α(q)⎮<ε then ⎮x_0 - q⎮<δ
    Pick elements p<x_0<q in the neighborhood of radius δ about x_0 we can then choose a partition such that
    Δα_i=(α(q)-α(p))/n
    this is true for any segment (x_i-1,x_i) s.t ⎮x_i -x_0⎮<δ
    Now choose a partition P of [a,b] with the above partition in the neighborhood of x_0 and arbitrarily let a=x_1 and b=b_2.
    The definition of f(x) implies that the only segment of the partition P where Σsupf(x) is not equal to zero is a segment in the neighborhood of radius δ about x_0.
    There supf(x)=1
    so Σsupf(x)Δα_i = (α(q) -α(p))/n
    This being true for all n in N we can take n very large to get zero.

    Does this work?
    If not can anyone give a hint?
    Thank you
     
  2. jcsd
  3. Jun 28, 2009 #2
    Actually I think I may have found a different proof:

    From the definition of f it can be shown that f is continuous at every point in [a,b] except for x_0. Therefore f is discontinuous at only finitely many points in [a,b].
    Again from the definition f is bounded.

    alpha is given to be continuous at x_0.

    Therefore by (theorem 6.10 in Rudin) f is Riemann integrable at alpha. (1)
    The lower Riemann integral of f is equal to zero (easily proved). (2)
    Therefore by (1) and (2) the total Riemann integral of f in [a,b] is equal to zero.
     
  4. Jun 29, 2009 #3

    HallsofIvy

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    There is no "f" in the statement of your problem! Do you mean "a"?

    There is no "alpha" in the statement of your problem!

    What does "at alpha" mean? Is alpha a function or a point?

     
  5. Jun 29, 2009 #4

    quasar987

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  6. Jun 29, 2009 #5

    The f is the f(x) defined in the statement. Sorry if that was not clear.

    I used the greek letter alpha "α" in the statement of the problem. I see that the font is kind of ambiguous.

    Both f and alpha "α" are functions defined on [a,b].
     
  7. Jun 29, 2009 #6
  8. Jun 30, 2009 #7
    I would probably just use a five point partition P_n consisting only of a, r_n, x_0, s_n, and b, where r_n=x_0 - 1/n, and s_n=x_0 + 1/n. Then write out the upper sum U(P_n,f,alpha). Then let n approach infinity.
     
  9. Jun 30, 2009 #8
    But in the intervals [a,r_n] and [s_n,b] as n approaches infinity r_n and s_n approach x_0. Then we could take f(x_0) as supf(x) in those intervals, that is if n goes to infinity.
    So wouldn't the sum be equal to 2?

    If that is not the case could you please explain?
    thanks
     
  10. Jun 30, 2009 #9
    No, because on those two intervals f is identically zero.
     
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