Prove an alternative formulation of integration by substitution theorem I'm trying to prove the following formulation of the integration by substitution theorem:

Let $f:[a,b] \rightarrow \mathbb{R}$ be Riemann-integrable function, and $\phi:[\alpha,\beta]\rightarrow [a,b]$ an affine function, that is $\phi(t)=mt+n$, such that $\phi(\alpha)=a$ and $\phi(\beta)=b$.
Prove: $$\int _a^b\:f(x)dx=\int _\alpha^\beta\:f(\phi(t))\phi'(t)dt$$

I tried to work it out with Riemann and Darboux sums, but got stuck.
Any help appreciated.
 A: We have only that $f$ is Riemann integrable and not necessarily continuous.
Proving with Riemann sums is not very elegant but straightforward since $\phi$ is monotonic and continuously differentiable. 
For any partition  $P = (t_0,t_1, \ldots, t_n)$ of $[\alpha, \beta]$, there is a corresponding partition $P' = (x_0, x_1, \ldots, x_n)$ of $[a,b]$ where $x_k = \phi(t_k)$.  Furthermore, if $P$ is tagged with points $\xi_k \in [t_{k-1},t_k]$ then the partition $P'$ has the corresponding tags $\eta_k = \phi(\xi_k) \in [x_{k-1},x_k].$
Consider the Riemann sums for the function $(f \circ \phi) \phi':t \mapsto f(\phi(t))\,\phi'(t)$ corresponding to the tagged partition $P$ and for $f$ corresponding to $P'$:
$$S(P, (f \circ\phi) \phi') = \sum_{k=1}^n f(\phi(\xi_k))\,\phi'(\xi_k) (t_k - t_{k-1}), \\  S(P',f) = \sum_{k=1}^nf(\eta_k)\, (x_k - x_{k-1}) =  \sum_{k=1}^nf(\phi(\xi_k))\, (\phi(t_k) - \phi(t_{k-1})) .$$ 
Applying the mean value theorem there exists points $\theta_k \in [t_{k-1},t_k]$ such that
$$S(P',f) =  \sum_{k=1}^nf(\phi(\xi_k))\, \phi'(\theta_k) ( t_k - t_{k-1}) $$
We have
$$\left|S(P, (f \circ\phi) \phi') - \int_a^b f(x) \, dx\right| \\ \leqslant \left|S(P, (f \circ\phi) \phi') - S(P',f)\right| + \left|S(P', f ) - \int_a^b f(x) \, dx\right| \\ \leqslant  \left|\sum_{k=1}^nf(\phi(\xi_k))\, (\, \phi'(\xi_k) - \phi'(\theta_k)\, ) ( t_k - t_{k-1})  \right| + \left|S(P', f ) - \int_a^b f(x) \, dx\right| \\ \leqslant \sum_{k=1}^n |f(\phi(\xi_k))|\, |\, \phi'(\xi_k) - \phi'(\theta_k)\, | ( t_k - t_{k-1}) + \left|S(P', f ) - \int_a^b f(x) \, dx\right| $$
Since $f$ is bounded, there exists $M$ such that $|f(x) \leqslant M$.  Since $\phi'$ is continuous, for any $\epsilon > 0$ there exists $\delta_1 > 0$ such that $|y-z| < \delta_1$ implies $|\phi'(y) - \phi'(z)| < \epsilon/(2M (\beta- \alpha))$.
Hence, if $\|P\| < \delta_1$, we have
$$\sum_{k=1}^n |f(\phi(\xi_k))|\, |\, \phi'(\xi_k) - \phi'(\theta_k)\, | ( t_k - t_{k-1}) < \frac{\epsilon}{2}$$
Since $f$ is integrable there exists $\delta > 0$ such that if $\|P'\| < \delta$, then 
$$\left|S(P', f ) - \int_a^b f(x) \, dx\right| < \frac{\epsilon}{2}.$$
There exists $\delta_2 >0$ such that if $\|P\| < \delta_2$ then $\|P'\| < \delta$. This is true because $\phi$ is uniformly continuous.
Therefore, if $\|P\| < \min(\delta_1, \delta_2)$ then
$$\left|S(P, (f \circ \phi) \, \phi') - \int_a^b f(x) \, dx \right| < \epsilon,$$
and
$$\int_\alpha^ \beta f(\phi(t)) \, \phi'(t) \, dt = \int_a^b f(x) \, dx.$$
