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Riemann Integrability

  1. Feb 5, 2015 #1

    O_o

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    1. The problem statement, all variables and given/known data
    Prove or give a counter example of the following statement:

    If [itex]f: [a,b] \to [c,d][/itex] is linear and [itex] g:[c,d] \to \mathbb{R} [/itex] is Riemann integrable then [itex] g \circ f [/itex] is Riemann integrable

    2. Relevant equations


    3. The attempt at a solution
    I'm going to attempt to prove the statement is true.

    Let [itex] f(x) = ax + b [/itex]. I'm going to assume [itex] a > 0 [/itex] and do a similar proof for [itex] a < 0 [/itex] if this one is alright.

    Fix [itex] n \in \mathbb{N} [/itex] such that [itex] \frac{1}{n} \leq a [/itex].

    Fix [itex] \epsilon > 0[/itex]

    Since g is Riemann integrable [itex] \exists P = \{y_0, y_1, ...., y_n\} [/itex] such that [itex] U(g,P) - L(g,P) \lt \frac{\epsilon}{n}[/itex]. Where [itex] P [/itex] is a partition of [itex] [c,d] [/itex].

    Let [itex] Q = \{x_0, x_1, ...., x_n\} [/itex] where [itex] x_i = \frac{y_i - b}{a} [/itex]. Since [itex] f(x) [/itex] is strictly increasing, [itex] x \in [x_{i-1}, x_i] \implies f(x) \in [y_{i-1}, y_i] [/itex].

    This means [itex] M_i = \sup\{ g(y): y \in [y_{i-1}, y_i]\} = \sup\{g(f(x)): x \in [x_{i-1}, x_i]\} [/itex] and likewise for the infimum over the interval, which I will label [itex] m_i [/itex].
    [tex] \implies \\ U(g \circ f ,Q) - L(g \circ f ,Q) = \sum_{i=1}^n \left(M_i - m_i\right) (x_i - x_{i-1}) \\
    = \sum_{i=1}^n \left(M_i - m_i\right) \frac{(y_i - y_{i-1})}{a} \\
    \leq \sum_{i=1}^n \left(M_i - m_i\right) (y_i - y_{i-1}) n \\
    = n \left(U(g,P) - L(g,P)\right) \\
    \leq n \frac{\epsilon}{n} = \epsilon
    [/tex]

    I was hoping someone could confirm my reasoning is okay or point out a place I made a mistake. Thanks
     
  2. jcsd
  3. Feb 6, 2015 #2

    RUber

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    Homework Helper

    Looks like a solid argument to me.
     
  4. Feb 7, 2015 #3

    O_o

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    Cheers. Thanks for taking the time to read it over.
     
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