1
$\begingroup$

Let $f:\mathbb R \rightarrow \mathbb R$ be a function such that $\lambda(I)=\lambda(f(I))$ for each interval $I \subseteq \mathbb R$. ($\lambda$ is Lebesgue measure here.) Let us call such functions ''nice'' functions.

Can we characterize nice functions?

For example;

$f(x)=x$ is a nice function.

$f(x)=2x$ is not a nice function. More generally if $m \in \mathbb R\setminus\{-1,1\}$ then $f(x)=mx$ is not a nice function.

Every translation is a nice function.

$$f(x)= \begin{cases} x+1; & x \in [-1,0) \newline x-1; & x \in [0,1) \newline x; & \text{elsewhere} \end{cases}$$ is a nice function.

Further, let $\mathcal A$ be the family of nice functions. Is the following conclusion true?

$f \in \mathcal A$ iff there exists a partition $I_{\alpha}$ of $\mathbb R$ into pairwise disjoint intervals such that $f=ax+b_{\alpha}$ on $I_{\alpha}$ for $a=-1$ or $1$ and for suitable $b_{\alpha}$.

$\endgroup$
1
$\begingroup$

Just a quick counterexample to your last question:

Let $C \subset \mathbb{R}$ be a fat Cantor set that is symmetric w.r.t. $0$ and define $f(x) = \begin{cases}x,& \text{if }x \in C\\\ -x,& \text{if }x \notin C \end{cases}.$

(More trivial counterexample would be $f(x) = \begin{cases}x,& \text{if }x \notin \mathbb{Q}\\\ 0,& \text{if }x \in \mathbb{Q} \end{cases}$, but changing things in a set of measure zero is not that satisfying.)

Edit: Perhaps even more satisfying example would be something like $f(\sum_{i=0}^\infty x_i2^{-i}) = \sum_{k=0}^\infty (1-x_{2k})2^{-2k}+\sum_{k=0}^\infty x_{2k+1}2^{-2k-1}$ with $x_i \in \{0,1\}$, i.e. switch every other digit in the binary representation (does not matter what you do at the endpoints of intervals).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.