Let $g(t)$ be a strictly increasing differentiable function. Can it map positively measurable set to zero measurable set?


It's obviously that $\{g'>0\}$ is dense. If I can prove that the Lebesgue measure $m(\{g'=0\}) = 0$, then for every set with positive measure, there will be a positively measurable subset with $g'>\epsilon$ on it, and consequently maps the set to nonzero measure set(It's a theorem and I forget it's name).


The question derives from my textbook, which says if $g(t)$ is a strictly increasing differentiable function and Riemann integrable, and $f(x)$ is Riemann integrable, then
$$\int f(x) = \int f(g(t))g'(t)$$
All functions defined on suitable closed set.


It seems that $f(g(t))$ may even be not integrable and that is totally a typo. But with my intuition of measure theory, this might be true since $g$ is differentiable.



Is there a strictly increasing differentiable function maps positively measurable set to zero measure set?
There are strictly increasing $C^1$ functions that map sets of positive measure to sets of measure zero. Here is a construction:


Let $C\subset [0,1]$ be a Cantor set of positive measure. For a construction, see <https://en.wikipedia.org/wiki/Smith-Volterra-Cantor_set>. Let $g(x)=\operatorname{dist}(x,C)$. The function $g$ is clearly continuous and equal zero on $C$. In fact $g$ is a $1$-Lipschitz function. Let
$$
f(x)=\int\_0^x g(t)\, dt.
$$ 
The function $f$ is $C^1$ and it is strictly increasing. Indeed, if $y>x$, then
$$
f(y)-f(x)=\int\_x^y g(t)\, dx>0
$$
because the interval $[x,y]$ is not contained in the Cantor set $C$ and therefore it contains an interval where $g$ is positive.


On the other hand $f'=g=0$ on $C$ which has positive measure and $f(C)$ has measure zero since $m(f(C))=\int\_C f'(t)\, dt=\int\_C g(t)\, dt=0$.


As was pointed out by Mateusz Kwaśnicki in his comment, this construction gives the following result:



> 
> **Theorem.** Let $f$ be as above. Then there is a Riemann integrable function $h$ such that $h\circ f$ is not Riemann integrable.
> 
> 
> 


**Proof.** The set $f(C)$ is homeomorphic to the Cantor set ($f$ is strictly increasing so it is a homeomorphism) and has measure zero as explained above. Let 
$$
h(x)=\begin{cases}
1 & \text{if $x\in f(C)$}\\
0 & \text{if $x\not\in f(C)$.}
\end{cases}
$$
The function $h$ is Riemann integrable with the integral equal zero since it is bounded and continuous outside the set $f(C)$ of measure zero (because $\mathbb{R}\setminus f(C)$ is open and $h=0$ there). However,
$$
(h\circ f)(x)=\begin{cases}
1 & \text{if $x\in C$}\\
0 & \text{if $x\not\in C$.}
\end{cases}
$$
is not Riemann integrable since it is discontinuous on a set $C$ of positive measure. $\Box$


