For a Borel function when does there exist a set of full measure with measurable image Let $(X,\Sigma,\mu)$ be a standard Borel probability space. Consider a measurable and measure preserving self map $f:X\to X$. We know that $f(A)$ is analytic for any $A\in \Sigma$, but this does not necessarily mean that $f(A)$ is Borel measurable.

What are the weakest conditions we can impose on $X$ or $f$ to ensure the existence of some $A\in \Sigma$, such that $f(A)\in \Sigma$ and $\mu(A)=1$?

I say weakest, because we know that if $f$ is injective then $X$ would suffice (because if the complement of an analytic set is analytic then it is Borel).
EDIT: Michael's comment gives a very useful approach. I think surjectivity is still quite a strong condition, but if there was a weaker condition we could impose ensuring that the image contains a measurable set of full measure that would be grand.
 A: My comments above assumed a general measure space with $\mu(X)$ not necessarily 1. The case $\mu(X)=1$ is a bit easier. 
Assume: 


*

*$X$ is a measure space with sigma algebra $\Sigma$ and measure $\mu$. 

*$\mu(X)=1$

*$f:X\rightarrow X$ is a measurable and measure-preserving function.
Definition:
Let's say that Property P holds if there is a set $A \in \Sigma$ such that $\mu(A)=1$ and $f(A) \in \Sigma$. 
Claim 1:
Suppose $f(X) \in \Sigma$.  Then Property P holds.
Proof: Define $A=X$.  Then $A \in \Sigma$, $\mu(A)=1$, and $f(A) = f(X) \in \Sigma$. So Property P holds.  $\Box$
Claim 2:
Suppose the sigma algebra is complete, so that all subsets of measure-0 sets also have measure 0. Then Property P holds if and only if $f(X) \in \Sigma$.
Proof ($\Longleftarrow$): Suppose $f(X) \in \Sigma$.  Then Claim 1 ensures Property P holds. $\Box$. 
Proof ($\Longrightarrow$): Suppose Property P holds. Let $A$ be a set such that $A \in \Sigma$, $\mu(A)=1$, $f(A)\in \Sigma$. We know: 
$$ A \subseteq f^{-1}(f(A)) \quad (*)   $$
and since $f(A)$ is measurable we have $f^{-1}(f(A))$ is measurable and 
$$ 1= \mu(A) \overset{(a)}{\leq} \mu(f^{-1}(f(A))) \overset{(b)}= \mu(f(A))  \overset{(c)}{\leq} 1$$
where (a) holds by (*); (b) holds by the measure-preserving property of $f$; (c) holds because $f(A) \subseteq X$ and $\mu(X)=1$. It follows that $\mu(f(A))=1$ and so
$$ \mu(f(A)^c) = 0$$
But $f(X)^c \subseteq f(A)^c$ and so $\mu(f(X)^c)=0$ (by completeness).  Thus, $f(X)^c \in \Sigma$ and so $f(X)\in \Sigma$. $\Box$

Examples
1) If we use $X=\{1, 2, 3, 4\}$, $\Sigma = \{\phi, X, \{1, 2\}, \{3, 4\} \}$, $f(1)=f(2)=2, f(3)=f(4)=4$, $\mu(\{1,2\})=\mu(\{3,4\})=1/2$, then $\mu(X)=1$, $f$ is measurable and measure-preserving, this is a complete measure space, but Property P does not hold because $f(X) = \{2, 4\} \notin \Sigma$.
2) Same $X$ and $\Sigma$ as before, but define $f:X\rightarrow X$ by $f(1)=1, f(2)=2, f(3)=4, f(4)=4$.  Define $\mu(\{1,2\})=1, \mu(\{3,4\})=0$.  This is a non-complete measure space with $\mu(X)=1$, $f$ is measurable and measure-preserving, and Property P holds with respect to the set $A = \{1, 2\}$. However, $f(X)$ is not measurable.
Now, if we complete the measure by adding the null sets $\{3\}$ and $\{4\}$ to $\Sigma$, then indeed $f(X)$ is measurable with measure 1. 
A: Here is the general case (for possibly incomplete measure spaces). Recall that Property P is defined in my other answer. 
Let $(X, \Sigma, \mu)$ be the original measure space.  Let $(X, \Sigma_c, \mu_c)$ be the completed measure space. Note that $\mu(B)=\mu_c(B)$ for all $B \in \Sigma$. 
Claim:
Property P holds if and only if $f(X) \in \Sigma_c$. 
To prove the claim we first have some lemmas. 
Lemma 1:
Every element of $\Sigma_c$ can be written as $R \cup H$ where $R \in \Sigma$ and $H$ is a subset of a measure-0 set in $\Sigma$. 
Proof:  (see wikipedia for this lemma here): 
https://en.wikipedia.org/wiki/Complete_measure
Lemma 2:
If $f$ is measurable and measure-preserving under $(X, \Sigma, \mu)$, then it is also measurable and measure-preserving under $(X, \Sigma_c, \mu_c)$. 
Proof:  Let $S \in \Sigma_c$.  We want to show $f^{-1}(S) \in \Sigma_c$ and $\mu_c(f^{-1}(S)) = \mu_c(S)$.  
We know $S= R \cup H$ for $R \in \Sigma$ and $H \subseteq Z$ for some $Z \in \Sigma$ with $\mu(Z)=0$.  So 
$$f^{-1}(S) = f^{-1}(R \cup H) = f^{-1}(R) \cup f^{-1}(H) \quad (*) $$
But $f^{-1}(H) \subseteq f^{-1}(Z)$ and $\mu(f^{-1}(Z))=\mu(Z)=0$ by the measure-preserving property.  So $f^{-1}(H)$ is a subset of a measure zero set and so $f^{-1}(H) \in \Sigma_c$.  Also, $f^{-1}(R) \in \Sigma \subseteq \Sigma_c$. Thus, (*) shows that $f^{-1}(S)$ is the union of two sets in $\Sigma_c$, so $f^{-1}(S) \in \Sigma_c$. 
Next, note that since $S = R \cup H$ with $\mu_c(H)=0$, we know 
$$\mu_c(S)=\mu_c(R) \quad (**) $$
On the other hand, since (*  ) writes $f^{-1}(S)$ as the union of two sets in $\Sigma_c$ we get:
\begin{align}
\mu_c(f^{-1}(S))  &\overset{(a)}{\leq}  \mu_c(f^{-1}(R)) + \mu_c(f^{-1}(H)) \\
&\overset{(b)}{\leq} \mu_c(f^{-1}(R)) + \mu_c(f^{-1}(Z)) \\
&\overset{(c)}{=} \mu(f^{-1}(R)) + \mu(f^{-1}(Z))\\
&\overset{(d)}{=} 
\mu(R) + \mu(Z) \\
&\overset{(e)}{=} \mu(R) + 0 \\
&\overset{(f)}{=} \mu_c(R)  \\
&\overset{(g)}{=} \mu_c(S)  
\end{align}
where (a) holds by (*); (b) holds because $H\subseteq Z$; (c) holds because $R$ and $Z$ are both in $\Sigma$; (d) holds because $f$ is measure-preserving in the original measure space; (e) holds because $\mu(Z)=0$; (f) holds because $R \in \Sigma$; (g) holds by (**).
$\Box$
Proof of claim that  Property P holds if and only if $f(X) \in \Sigma_c$: 
Proof ($\Longrightarrow$): Suppose Property P holds.  Then there is a set $A$ such that $A \in \Sigma$, $\mu(A)=1$, and $f(A) \in \Sigma$.  Since every element of $\Sigma$ is also in $\Sigma_c$, and $1=\mu(A)=\mu_c(A)$, it follows that property P holds in the completed measure space.  Since $f$ is measurable and measure-preserving in the completed measure space, and since property P holds in the completed measure space, by Claim 2 in my previous answer it follows that $f(X) \in \Sigma_c$. $\Box$
Proof ($\Longleftarrow$): Suppose $f(X) \in \Sigma_c$.  Then $f(X) = R \cup H$ where $R \in \Sigma$ and $H\subseteq Z$ where $Z \in \Sigma$ and $\mu(Z)=0$.  Then $\mu_c(H)=0$ and so $\mu_c(f(X)) = \mu_c(R) = \mu(R)$. 
However, we have $f^{-1}(f(X)) = X$, and since $f$ is measure-preserving in the completed measure space we have
$$1=\mu(X)=\mu_c(X) = \mu_c(f^{-1}(f(X)) = \mu_c(f(X)) =  \mu(R)$$
So define $A = f^{-1}(R)$.  Since $R \in \Sigma$ and $f$ is measurable, we know $A \in \Sigma$.  Since $f$ is measure-preserving we know $\mu(A)=\mu(f^{-1}(R))=\mu(R)=1$. Finally, we know $f(A)=R \in \Sigma$. Thus, Property P holds in the original measure space. 
(To show that $f(A)=R$ note that $f(A) = f(f^{-1}(R)) \subseteq R$.  However, since $R\subseteq f(X)$, every element of $R$ can be mapped to by $f$, so $f(f^{-1}(R))=R$.) $\Box$
