Riemann integral substitution with $e^x$ For a Riemann-integrable function $f$: $[1, e]\longrightarrow\mathbb{R}$ we have
$$\int_1^e f(x)dx=\int_0^1 f(e^x)e^x dx$$
and this easily follows from more general results of Lebesgue integration theory. It's also not hard to see that $(f\circ\exp)\exp$ is Riemann-integrable since it's a.e. continuous whenever $f$ is.
Can we give an elementary proof of this equality, i.e. one that doesn't go beyond the toolbox of elementary analysis and standard Riemann integration theory?
I guess the main obstacle is that $f$ is not necessarily the derivative of an $F$, but there just might be some clever argument relying heavily on the niceness of the $\exp$-function.

Well, the general approach for proving the integral substitution rule is an approximation argument. If $f$ is a piecewise-constant function that identity clearly holds, hence the problem boils down to proving that any Riemann-integrable function can be approximated with a piecewise-constant function in a effective way. There are some natural candidates:
$$ f_n(x) = f\left(n\left\lfloor\frac{x}{n}\right\rfloor\right), $$
for instance. Can you fill the missing details?
