Let $F:[0,1]\to[0,1]$ be a Lebesgue measure preserving function. Is $F$ almost surjective, i.e., the image of $F$ has interior measure one?


This question is motivated by the following observation. If $F$ satisfies the above and $X$ is uniformly distributed over $[0,1]$, then $F(X)\sim Unif[0,1]$, i.e., $F(X)$ seems to appear almost everywhere in [0,1]. I believe the answer is no, but a counterexample is nontrivial.


A relating post is <https://math.stackexchange.com/questions/2612075/is-every-measure-preserving-function-almost-surjective> but the counter example therein does not apply to my question. (The function constructed there has different domain and range.)



Is measure preserving function almost surjective?
Yes, by Luzin's theorem. Fix $\varepsilon>0$ and take a compact subset $K$ of measure at least $1-\varepsilon$ such that $F$ is continuous on $K$. Then $F(K)$ is a compact set of at least the same measure as $K$, since $F^{-1}(F(K))\supset K$. So, for any $\varepsilon>0$, $F([0,1])$ contains a measurable subset of measure $\geqslant 1-\varepsilon$. This implies that inner measure of $F([0,1])$ is 1.


