1
$\begingroup$

I don't want to comment on an old question, so I'm asking a new one. The question I'm referring to is Absolutely Continuous and Strictly Increasing on a Subinterval. Specifically, I'm concerned about the second part of the question. I can easily see that $G$ is absolutely continuous, but I'm having a bit of trouble proving that it's not monotone on any subinterval of $[0,1]$. Would a proof by contradiction work? Any hints would be much appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

$G(x)=\int_0^x \chi_A(t) - \chi_{[0,1]\setminus A}(t)\,dt $, so $G'(x) = \chi_A(x) - \chi_{[0,1]\setminus A}(x)$ almost everywhere. (Are you familiar with this?) By the hypothesis on $A$, this implies that in every subinterval of $[0,1]$, $G'$ takes the values $1$ and $-1$. This implies that $G$ is nonmonotone in every subinterval of $[0,1]$.

$\endgroup$
3
  • $\begingroup$ Beautiful. How did you think of this solution? I probably would've never thought of something like that no matter how hard I tried. $\endgroup$
    – Aden Dong
    Dec 3, 2014 at 16:06
  • $\begingroup$ Absolutely continuous functions are precisely those that are the integrals of their derivatives, and that's one of the most useful things about them, so it's always on my radar when they come up. I was initially thinking of working explicitly with Lebesgue measures of sets, but hadn't gotten far in terms of turning intuition into an actual argument before using the derivative occurred to me. $\endgroup$ Dec 3, 2014 at 17:57
  • $\begingroup$ Ahh I see. Thanks for the insight. $\endgroup$
    – Aden Dong
    Dec 3, 2014 at 18:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .