Is measure preserving function almost surjective? 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.)
 A: A few remarks (where $m$ is the Lebesgue measure):


*

*If $F[0,1]$ is a Lebesgue measurable set, then $F[0,1]$ has Lebesgue measure equal to one (you can read Fedor Petrov's comment for this post).

*Because of Birkhoff's Ergodic Theorem: For every $[a,b]\subset [0,1]$ $$
\int_0^1\left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{[a,b]}(F^i (x))\right)dm=m([a,b]).
$$

*If $F[0,1]$ is a Lebesgue measurable set and $F$ is also Ergodic, then for $m$-a.e. $x$ $$
1=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{F[0,1]}(F^i (x))=m(F[0,1]), 
$$
which is another proof that $m(F[0,1])=1.$

*If $F[0,1]$ is a Lebesgue measurable set, then $$
1=\int_0^1\left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{F[0,1]}(F^i (x))\right)dm=m(F[0,1]), 
$$
which is another proof that $m(F[0,1])=1.$

*If $T$ is ergodic and $K\subset [0,1]\setminus F[0,1]$ is a Lebesgue measurable set, then  $m(K)=0.$ By contradiction, we would obtain that for $m$-a.e. $x$ $$
0=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{K}(F^i (x))=\mathcal{Leb}(K)\neq 0. 
$$

*If $K\subset [0,1]\setminus F[0,1]$ is a Lebesgue measurable set, then  $m(K)=0.$ By contradiction, we would obtain that 
$$
0=\int_0^1\left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{K}(F^i (x))\right)d m=m(K)\neq 0. 
$$

*Let $\mathcal{A}=\{K:K\subset F[0,1], K \mbox{ measurable }\}$ and $\mathcal{B}=\{K:K\subset [0,1]\setminus F[0,1], K \mbox{ measurable }\}.$ Then $\sup_{K\in \mathcal{A}} m(K)=1.$ By contradiction, if not, by sigma aditivity
$$
1=m[0,1]=\sup_{K\in \mathcal{A}} m(K)+\sup_{K\in \mathcal{B}} m(K)=\sup_{K\in \mathcal{A}} m(K)<1,
$$
which is another proof that the image of $F$ has inner measure one.

A: 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.
