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Proving integration for a bounded increasing function

  1. Nov 8, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Suppose that f is a bounded, increasing function on [a,b]. If p is the partition of [a,b] into n equal sub intervals, compute Sp - sp and hence show f is integrable on [a,b]. What can you say about a decreasing function?


    2. Relevant equations

    We partition [a,b] into sub-intervals.

    For each i, we let : [itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

    Now, we define [itex]s_p = \sum_{i=1}^{n} m_i Δx_i[/itex] as the lower sum and [itex]S_p = \sum_{i=1}^{n} M_i Δx_i[/itex] as the upper sum.

    Some more info in my notes :

    Let [itex]M = sup \left\{{f(x)|x \in [a,b]}\right\}[/itex] and [itex]m = inf \left\{{f(x)|x \in [a,b]}\right\}[/itex]. Then we get sp ≤ M(b-a) so the set of all possible sp is bounded above.

    Let I = sup{sp} and J = inf{Sp}

    Definition : if I = J, then f(x) is integrable.

    Now another theorem I could use : For a bounded function f on [a,b]. f is integrable if and only if :

    [itex]\forall ε>0[/itex] there is a partition p of [a,b] such that Sp < sp + ε.

    3. The attempt at a solution

    So we are given that f is a bounded increasing function. This means that m ≤ f ≤ M for some lower bound m and upper bound M. For any partition p of [a,b] into n sub-intervals, we also have :

    [itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

    Now before I jump any further I want to confirm the direction I'm going in. I have two theorems I provided and I'm wondering which one is more appropriate to use here.
     
    Last edited: Nov 8, 2012
  2. jcsd
  3. Nov 8, 2012 #2

    Zondrina

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    I think I have an idea about how the information given relates, so continuing from my first post :

    So we are given that f is a bounded increasing function. This means that m ≤ f ≤ M for some lower bound m and upper bound M. For any partition p of [a,b] into n sub-intervals, we also have :

    [itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

    Now, writing out the upper sum we get : [itex]S_p = \sum_{i=1}^{n} M_iΔx_i[/itex] and the lower sum : [itex]s_p = \sum_{i=1}^{n} m_iΔx_i[/itex]

    So we get :

    [itex]S_p - s_p = \sum_{i=1}^{n} (M_i - m_i)Δx_i = \sum_{i=1}^{n} (f(x_i) - f(x_{i-1}))(\frac{b-a}{n})[/itex]

    I think this the path I'm trying to take.
     
  4. Nov 8, 2012 #3

    Zondrina

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    Sorry for all the posting, but since no one has responded I believe I have pieced this puzzle together. So I'll write it out cleanly here.

    So we are given that f is a bounded increasing function on [a,b]. This means that f(a) ≤ f(x) ≤ f(b) for any x in [a,b]. Hence f(a) is a lower bound for f and f(b) is an upper bound so that f is bounded. We need to show that for any ε>0, we can choose any partition p of [a,b] into n sub-intervals such that Sp - sp < ε. Suppose p is the partition [itex]\left\{{x_0, x_1, ..., x_n}\right\}[/itex].

    We let : [itex]L = \displaystyle\max_{1≤i≤n}(x_i - x_{i-1}) < \frac{ε}{f(b) - f(a)} = Q[/itex]

    Now, we define :

    [itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

    Now, writing out the upper sum we get : [itex]S_p = \sum_{i=1}^{n} M_iΔx_i[/itex] and the lower sum : [itex]s_p = \sum_{i=1}^{n} m_iΔx_i[/itex]

    So we get :

    [itex]S_p - s_p = \sum_{i=1}^{n} (M_i - m_i)Δx_i ≤ \sum_{i=1}^{n} (f(x_i) - f(x_{i-1}))L = (f(b) - f(a))L < (f(b) - f(a))Q = ε[/itex]

    Hence f is integrable on [a,b] since Sp - sp < ε.

    Is this good? It's my first time trying one of these.
     
    Last edited: Nov 8, 2012
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