Integration by substitution's theorem only with $f$ Riemann integrable and $g$ a monotone function. I'm trying to solve this problem: Let $f\colon\left[a,b\right]\to\mathbb{R}$
be a Riemann integrable function and $g\colon\left[c,d\right]\to\mathbb{R}$
be a monotone function such that $g'$ is Riemann integrable. Prove
that if $g\left(\left[c,d\right]\right)\subseteq\left[a,b\right]$,
then $\int_{g\left(c\right)}^{g\left(d\right)}f\left(x\right)dx=\int_{c}^{d}f\left(g\left(t\right)\right)\cdot g'\left(t\right)dt$.
This is similar to the integration by substitution's theorem, but in that theorem
is given the hypothesis that $f$ is continuous and therefore it has
antiderivative $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$ and
the prove is easy applying the change rule to the function $F\left(g\left(t\right)\right)$.
But I don't know how to use the fact that $g$ is monotone in the
problem, because in this case we only know that $f$ is Riemann integrable
and may be it has no antiderivative. Could you help me or give me
some suggestions?
Thanks.
 A: Assuming only that $f$ is Riemann integrable and $g$ is differentiable and monotone, this is straightforward to prove with the additional condition that $g'$ is continuous on $[c,d]$.
We assume without loss of generality that $g$ is non-decreasing. 
Consider a partition $P: c = x_0 < x_1 < \ldots < x_n = d$ and the Riemann sum 
$$\tag{1}S(P,fg')= \sum_{j=1}^n f(g(\xi_j))g'(\xi_j)(x_j - x_{j-1})$$
using intermediate points $\xi_j \in [x_{j-1},x_j]$. 
If $g$ is increasing, then we have a partition $P'$ of $[g(c),g(d)]$ given by 
$$g(c) = g(x_0) < g(x_1) < \ldots < g(x_n) = g(d),$$
Using the intermediate points $g(\xi_j)$, we have a Riemann sum for the integral of $f$ over $[g(c),g(d)]$ taking the form
$$S(P',f) = \sum_{j=1}^n f(g(\xi_j))(\,g(x_j) - g(x_{j-1})\,),$$
where the monotonicity of $g$ is needed to ensure that $g(\xi_j) \in [g(x_{j-1}), g(x_j)]$.
Applying the mean value theorem, there exist points $\eta_j \in (x_{j-1},x_j)$ such that 
$$\tag{2}S(P',f) = \sum_{j=1}^n f(g(\xi_j))g'(\eta_j)(x_j - x_{j-1})$$
Thus,
$$\tag{3}\left|\int_{g(c)}^{g(d)} f(x) \, dx - S(P,fg') \right| \leqslant \left|\int_{g(c)}^{g(d)} f(x) \, dx - S(P',f) \right|+ \left|S(P',f) - S(P,fg') \right|$$
Since $f$ is Riemann integrable, for any $\epsilon > 0$ there exists $\delta_1$ such that if $\|P'\| < \delta_1$, then the first term on the RHS of (3) is less than $\epsilon/2$.
Using (1) and (2), we also have for the second term on the RHS of (3),
$$\left|S(P',f) - S(P,fg') \right| = \left|\sum_{j=1}^n f(g(\xi_j))\left(\, g'(\xi_j)-g'(\eta_j)\,\right)(x_j - x_{j-1}) \right|  \\ \leqslant \sum_{j=1}^n| f(g(\xi_j))|\left|\, g'(\xi_j)-g'(\eta_j)\,\right|(x_j - x_{j-1})$$
Since $f\circ g$ is bounded and $g'$ is continuous and, hence, uniformly continuous on $[c,d]$, we can find $\delta_2 > 0$ such that if $\|P\| < \delta_2$ then $\left|S(P',f) - S(P,fg') \right| < \epsilon/2$. Also by continuity of $g$ there exists $\delta_3$ such that if $\|P\| < \delta_3$, then $\|P'\| < \delta_1$.
Therefore, if $\|P\| < \min(\delta_2, \delta_3)$ then it follows that
$$\left|\int_{g(c)}^{g(d)} f(x) \, dx - S(P,fg') \right|< \epsilon,$$
which proves that $(f\circ g) g'$ is integrable and
$$\int_{g(c)}^{g(d)} f(x) \, dx= \int_{c}^{d} f(g(t))g'(t) \, dt$$
This can also be proved more generally if $g'$ is not assumed to be continuous, but it is more difficult.  A reference is H. Kestelman, Mathematical Gazette, 45 [1961], 17-33.
