# A problem with a Borel set, a continuous function with a fixed point and a non-null fixed set, II

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 $C^0$.

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 $f(A)\cap A=\emptyset$.
• @nicomezi the set $\{x_0 - b | b\in B \}$ – supinf Aug 30 '18 at 13:13
• 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... – Sebastien Palcoux Aug 31 '18 at 18:08