# Is there a strictly increasing differentiable function maps positively measurable set to zero measure set?

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.

• @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 – XT Chen Feb 24 at 14:40

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. $$\Box$$

• 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. – Mateusz Kwaśnicki Feb 24 at 21:37
• 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. – Gro-Tsen Feb 25 at 15:58