Proving Riemann Integrability of g∘f for Linear Functions

[O_o]

Homework Statement

Prove or give a counter example of the following statement:

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

Homework Equations

The Attempt at a Solution

I'm going to attempt to prove the statement is true.

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

Fix n \in \mathbb{N} such that \frac{1}{n} \leq a.

Fix \epsilon > 0

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

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

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

I was hoping someone could confirm my reasoning is okay or point out a place I made a mistake. Thanks

 

[RUber]

Looks like a solid argument to me.

 

[O_o]

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