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Riemann Integrability

  • Thread starter O_o
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  • #1
O_o
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Homework Statement


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

Homework Equations




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
 

Answers and Replies

  • #2
RUber
Homework Helper
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Looks like a solid argument to me.
 
  • #3
O_o
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Cheers. Thanks for taking the time to read it over.
 

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