TITLE: Strictly increasing, absolutely continuous function with vanishing derivative
QUESTION [9 upvotes]: The following problem comes from Folland's Real Analysis:
$A\subset [0,1]$ is a Borel set which has the property that $0<m(A\cap I)<m(I)$ for every interval $I\subset [0,1]$. We define the functions
$F(x)=m([0,x]\cap A)$
$G(x)=m([0,x]\cap A)-m([0,x] -A)$
Its fairly straightforward to show

$F$ is strictly increasing and absolutely continuous
$G$ is absolutely continuous

We wish to show that

$F'=0$ on some set of positive measure
$G$ is not monotone on any interval

Any thoughts?

REPLY [2 votes]: We have $F(x)=\int_{-\infty}^x\chi_A(t)dt$, where $\chi_A\in L^1(m)$ since $A$ is a Borel subset of $[0,1]$. Consequently (see, for example, Folland, Real Analysis, Corollary 3.33) $F$ is absolutely continuous and $F'=\chi_A$ a.e. Thus $F'=0$ a.e. on $[0,1]\backslash A$. Since $m(A)<m([0,1])$ by assumption, $F'=0$ on a set of positive measure. (If $0\le x<y\le1$, then $F(x)-F(y)=m((x,y]\cap A)>0$ by assumption, so $F$ is strictly increasing on $[0,1].$)
Likewise, $G(x)=\int_{-\infty}^x(\chi_A-\chi_{[0,1]\setminus A})(t)dt$, so $G$ is absolutely continuous and $G'=\chi_A-\chi_{[0,1]\setminus A}$ a.e. If $I$ is any subinterval of $[0,1]$, then both $I\cap A$ and $I\setminus A$ have positive measure, since $0<m(A\cap I)<m(A)$ by assumption. Thus $G'$ assumes both values $1$ and $-1$ on $I$, whence $G$ is not monotone on $I.$