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Proving that a discontinuous function is integrable

  1. Jul 15, 2011 #1
    I was self-studying from a Calculus book and found the following question:

    1. The problem statement, all variables and given/known data
    U is the function defined by
    [tex]U(x)=\begin{cases}
    0 & \text{ if } x<0 \\
    1 & \text{ if } x\geq 0
    \end{cases}[/tex]
    Let [a, b] be an interval such that a < 0 < b. Show that, even though the function U is discontinuous in [a, b], it is integrable in [a, b] and
    [tex]\int_a^bU(x)dx=b[/tex]

    2. Relevant equations
    Definition of definite integral:
    [tex]\int_a^bf(x)dx=\lim_{||\Delta||\to 0}\sum_{i=1}^n f(\xi_i)\Delta_ix[/tex]
    where Δ is a partition, Δix is the i-th subinterval of the partition Δ, that is, the subinterval [xi-1, xi], ξi is any number such that xi-1 ≤ ξi ≤ xi and ||Δ|| is the norm of the partition (the largest subinterval of the partition).

    3. The attempt at a solution
    I'm not very sure on how to start, and the attempt below seems inconsistent. Could someone point to me what may be wrong with it and suggest corrections?
    To show this, I have to show that (according to the definition given by the book I'm using) there is a number L which satisfies the condition that, for any ε > 0, there is a δ > 0 such that
    [tex]\left | \sum_{i=1}^n U(\xi_i)\Delta_ix \ \right | < \epsilon[/tex]
    for every partition Δ for which ||Δ|| < δ and for any ξi in the closed interval [xi-1, xi], i = 1, 2, ..., n.
    Let Δ be a partition of the interval [a,b]. There is a discontinuity at x = 0. So, I will divide the whole sum into two sums: S1 over the interval [a, 0] and S2 over the interval [0, b]. For this, I will specify that x = 0 is a partition point, that is, xk = 0 is at an extreme of a subinterval Δkx.
    Calculation of S1: Suppose that in the interval [a, 0] there is no subinterval for which ξi = 0 (that is, for every ξi, U(ξi) = 0). So, the sum of all the U(ξiix in [b, 0) is 0. Thus, S1 = 0.
    Now, for S2: Suppose Δjx is the first subinterval for which ξj belongs to [0, b]. Thus, for every ξi in the interval [0, b], U(ξi) = 1; so, the whole Riemann sum will become:
    [tex]\sum_{i=1}^n f(\xi_i)\Delta_ix = S_1 + S_2 = 0+\sum_{i=j}^n 1\times\Delta_ix=\sum_{i=j}^n \Delta_ix=b[/tex]
    So:
    [tex]\sum_{i=1}^n f(\xi_i)\Delta_ix - b = 0[/tex]
    and, for every ε > 0, there is a δ > 0 such that
    [tex]\left | \sum_{i=1}^n f(\xi_i)\Delta_ix - b \right | < \epsilon[/tex] whenever ||Δ|| < δ
    and, thus:
    [tex]\lim_{||\Delta||\to 0} \sum_{i=1}^n U(\xi_i)\Delta_ix = b[/tex]

    I know this is not complete, because, above, I chose x = 0 to be at an extreme of a subinterval. If x = 0 were in the middle of a subinterval Δkx (that is, if xk-1 ≤ 0 ≤ xk), there would be situations where the Riemann sum would be less than b, and situations in which it would be greater than b, depending on the choice of ξk. But I'm not sure how to handle these situations.

    Thank you in advance.
     
    Last edited: Jul 15, 2011
  2. jcsd
  3. Jul 15, 2011 #2

    micromass

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    Hi pc2-brazil! :smile:

    Your proof is quite good, but is has some holes.

    Typo: you forgot to do something with the L in your definition.

    OK, fine. But can you do that?? The above definition says it must hold for every partition. But you only take the partition here such that 0 is a partition point. Why can you choose 0 to be a partition point?

    Close, but not quite: U(0)=1, so if we pick [itex]\xi_i=0[/itex], then we would get 1...

    Rest seems ok!

    Your proof is very good, just try to fill in the holes :smile:
     
  4. Jul 21, 2011 #3
    I'm still not sure how I could include every situation. Maybe I should try to prove them all separately?
    The problem is that, above, I chose x = 0 to be at an extreme of a subinterval (as I mentioned in the first post). If x = 0 were in the middle of a subinterval Δkx (that is, if xk-1 ≤ 0 ≤ xk), there would be situations where the Riemann sum would be less than b, and situations in which it would be greater than b, depending on the choice of ξk. But I'm not sure how to handle all these situations.
     
  5. Jul 21, 2011 #4

    micromass

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    The thing is that the subinterval around zero will only contribute a factor [itex]\Delta x_k[/itex] to the entire sum. And in the limit, this factor will vanish!
     
  6. Jul 22, 2011 #5
    Does your textbook have
    [itex] L-\epsilon < \left | \sum_{i=1}^n f(\xi_i)\Delta_ix - b \right | < L + \epsilon [/itex]
    or something similar as part of the definition of an integral?


    micromass is definitely going the right direction.
    Some bigger hints/ ideas.
    Let k be the [xk, xk+1] containing x=0.
    What is the sum for this interval?
    [itex] ||\Delta|| [/itex] is the width of the largest or widest subinterval, often called the norm. As ||\Delta|| ->0, what happens to your formula for the sum or area above that interval?

    Minor point: technically, the sum or area above the kth interval is less than
    We can probably simply say area of kth subinterval is less than ....

    In summary, it would be helpful to break it down into three sums. The two you've already got, plus a 3rd "sum" over the single interval containing 0.


    All that being said, in graduate school, we used a different definition. The partition could be specified, and we showed the limit inequality holds for any partition finer than the specified partition, eg. intervals could be subdivided, but every subpartition has at least the same partition points. I may be going past Riemann integrals from introductory Calculus texts into Lebesque integrals from graduate school though.
     
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