# Is every measure preserving function ''almost surjective''?

Let $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ be two probability spaces and $\pi: X \rightarrow Y$ be a measure preserving function (not necessarily invertible). Can we conclude there exists a measurable set $Y'\subset Y$, $\nu(Y')=1$, such that $Y'\subset Img(\pi)$?

Can we conclude it using more assumptions? In my case, $Y = \{0,1\}^\mathbb{N}$, equipped with the product $\sigma$-algebra (of the discrete $\sigma$-algebra), and $\nu$ is the product Bernoulli measure. I think this probability space is not complete. The other probability space is anything.

In general, we don't know if $Img(\pi)$ is measurable or not (if it was, the question was over). But we know any set $S$ of positive measure won't be disjoint of $Img(\pi)$, because $\mu(\pi^{-1}(S)) = \nu(S) >0$, so $\pi^{-1}(S) \neq \emptyset$ and there exists $x \in X$ so that $\pi(x) \in S$.

Any help is appreciated. Thanks.

• For the second question, it sounds like you have an application - maybe you could explain why you think you need this result? – Dap Jan 19 '18 at 20:28
• Yes, I have. I'm asking this question because I am trying to check what are the properties of a (measure theoretical) semi-conjugacy between and "IID dynamical system" and a Bernoulli shift. Here, an "IID dynamical system" is to be understood as a measure preserving system $(f,\mu)$ having a non $\mu$-constant observable $\psi$ such that $(\psi \circ f^n)_n$ is IID (now, in purely probabilistic terms). – Lucas Amorim Jan 20 '18 at 16:51

The important thing is whether the image of $\pi$ is Lebesgue measurable, i.e. has inner measure $1.$ If it has inner measure less than $1$ then such a $Y'$ cannot exist.
Suppose, for instance, that $\nu$ is Lebesgue measure on $Y = [0, 1],$ that $X \subseteq [0, 1]$ is a non-Lebesgue-measurable set of outer measure $1$ (134D), that $\mu$ is the subspace measure $\nu_X$ on $X,$ and that $\phi(x) = x$ for $x \in X.$ Then $$\mu(\phi^{−1}(F)) = \nu^*(X \cap F) = \nu(F)$$ for every Lebesgue measurable set $F \subseteq Y.$
Since $\phi(X)=X$ is non-measurable, its inner measure is less than $1.$
• Dap, I was just looking around for what an inner measure is because I was taught with Folland's book and this concept is not in his expository text. I took a look and found out that a simple case of what you said can be shown: Let $(X,\mathcal{A},\mu) = ([0,1] \setminus \{0.5\}, \mathcal{B}_{[0,1] \setminus \{0.5\}}, Lesbegue)$ completed and $(Y, \mathcal{B}, \nu) = ([0,1], \{ \emptyset, [0,1] \}, \nu(\emptyset)=0$ $\nu([0,1])=1)$. Let $\pi(x) =x$. It is measure preserving. The image of $\pi$ is not measurable and does not contain another measurable of $\nu$-measure 1. Thanks! :) – Lucas Amorim Jan 20 '18 at 16:44