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.

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    $\begingroup$ @GeraldEdgar It's not the same thing. I am saying mapping sets of positive measure to sets of measure zero instead of mapping sets of measure zero to sets of positive measure $\endgroup$
    – XT Chen
    Feb 24, 2020 at 14:40

1 Answer 1


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. $\Box$

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    $\begingroup$ In case one needs a paper reference, virtually the same construction is carried out in Real Analysis Exchange 22(1) (1996-97): 404–405 by Javier Fernández de Bobadilla de Olazabal. $\endgroup$ Feb 24, 2020 at 21:37
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    $\begingroup$ Furthermore, if you modify your construction just a little bit to make $g$ be a positive $C^\infty$ “bump” on each interval (connected component) of the complement of $C$, with all derivatives tending to $0$ at the edges of the interval, and with the height of the bumps tending fast enough to $0$, you can get $g$ hence $f$ to be both $C^\infty$, still with the same property. $\endgroup$
    – Gro-Tsen
    Feb 25, 2020 at 15:58

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