This post is the improvement of A problem with a Borel set, a continuous function with a fixed point and a non-null fixed set by natural additional assumptions avoiding some easy counter-examples.

Let $\mu$ be the Lebesgue measure on $[0,1]$ and $A$ be a Borel subset of $[0,1]$.
Consider a point $x_0 \in A$ such that for any neighborhood $V$ of $x_0$ in $[0,1]$, we have:

  • $\mu([0,x_0] \cap V \cap A)>0$,
  • $\mu([x_0,1] \cap V \cap A)>0$.

Let $f: [0,1] \rightarrow [0,1]$ be a polynomial function such that:

  • $f(x_0) = x_0$,
  • $\mu(f(A))>0$.

Question: Is it true that $\mu(f(A) \cap A)>0$ ?

Remark: If possible, we can consider the more general cases $f \in C^{\infty}$, $C^1$ or even $C^0$.


I think I found a counterexample.

Consider the function $f(x) = x/2+1/4$ and $x_0=1/2$. For our construction, we need the set $$ B= \bigcup_{n\geq 3} (2^{-2n-1},2^{-2n}). $$ Notice that $\frac12 B \cap B = \emptyset$ and that $B$ contains open intervals arbitrarily close to $0$. We choose $$ A= (x_0+B) \cup (x_0-B). $$ This symmetry guarantees the conditions involving the intervals $[0,x_0],[x_0,1]$. Then, it can be shown that $$ f(A) = (x_0 +B/2) \cup (x_0 - B/2). $$ Then it follows from the above that $f(A)\cap A=\emptyset$.

  • $\begingroup$ @nicomezi the set $\{x_0 - b | b\in B \}$ $\endgroup$ – supinf Aug 30 '18 at 13:13
  • $\begingroup$ Thank you.... I am running at low speed today. Your construction seems fine but I am not sure. $\endgroup$ – nicomezi Aug 30 '18 at 13:18
  • 1
    $\begingroup$ Correct! Your construction extends to any contraction mapping. $\endgroup$ – Sebastien Palcoux Aug 30 '18 at 14:23
  • $\begingroup$ Perhaps what I had in mind when I wrote this question is the following: if $f$ is ergodic and if $A,B$ are Borel sets with $\mu(A)>0$ and $\mu(B)>0$, then by definition of ergodicity there is $n>0$ such that $\mu(f^n(A) \cap B)>0$. I don't know if this can help for an other natural improvement of the question... $\endgroup$ – Sebastien Palcoux Aug 31 '18 at 18:08

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