0

Is it possible to construct a homeomorphism $f : [0,1]\to [0,1]$ that does not satisfy Lusin N property? i.e. $f$ does not always map zero-measure sets to zero-measure sets.

I somehow believe we can construct such examples. (play with the Cantor function for example)

I'm aware that there is a similar question here, aiming in the other direction. (prove the statement right, possibly with stronger condition)

Jordan
  • 678
Expialidocius
  • 63
  • 3
    It is well known that there is a strictly increasing continuous function $f:[0,1]\to [0,1]$ with $f'=0$ a.e. This is a homeomorphism and it does not have Lusin N property because it is of bounded variation and not absolutely continuous. Ref. https://en.wikipedia.org/wiki/Luzin_N_property – Kavi Rama Murthy Jun 09 '25 at 09:56
  • 1
    The Cantor function $f$ already sends already sends the zero-measure Cantor set to $[0,1]$ surjectively. $\frac12(x+f)$ is a homeomprhism which sends the Cantor set to a set of measure $\frac12$. – Sassatelli Giulio Jun 09 '25 at 10:52
  • @SassatelliGiulio Why $\mu(g(C)) \geqslant \frac{1}{2}$ for $g = \frac{1}{2}(x+f)$ and $C=$ Cantor set. – Expialidocius Jun 09 '25 at 11:34
  • I think Kavi Rama Murthy's comment solves the problem. I add a link for the construction of one such function. – Expialidocius Jun 09 '25 at 11:37
  • 2
    @Expialidocius Because its complement is sent to a set of measure $\frac12$. – Sassatelli Giulio Jun 09 '25 at 11:53
  • 1
    For $x \in C$, $$f(x) = \frac{1}{2} c(x) + \frac{1}{2} x$$

    For $x \in I_k = (a_k, b_k)$: $$f(x) = f(a_k) + \frac{f(b_k) - f(a_k)}{b_k - a_k} (x - a_k)$$

     – QwQ Jun 09 '25 at 13:34

0 Answers0