Natural extensions in ergodic theory / Measurability question A useful "abstract nonsense" construction in ergodic theory takes a measure-preserving transformation
$T$ of a probability space $(X,\mathcal B,\mu)$ and extends it to an invertible measure-preserving transformation $\bar T$ of a probability space $(\bar X,\bar{\mathcal B},\bar\mu)$. 
One description of this is in Omri Sarig's notes (section 1.6.4). In his construction he needs to make the assumption that $T(X)=X$, or the weaker assumption, $T(X)$ is measurable. My question is whether this is automatic for Lebesgue spaces. 
Hence my precise question:
If $T$ is a measure-preserving transformation of $[0,1]$ (equipped with Lebesgue measure and the $\sigma$-algebra of Lebesgue measurable sets), is $T([0,1])$ necessarily measurable?

Note: $[0,1]\setminus T([0,1])$ does not contain any measurable sets of positive measure by the Poincaré recurrence theorem, so $T([0,1])$ is certainly of outer measure 1.
 A: I think the answer is yes.
Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $[0,1]$ and $\mathcal{L}$ the Lebesgue $\sigma$-algebra.  Suppose $T$ is $(\mathcal{L}, \mathcal{L})$-measurable and measure preserving.  In particular $T$ is $(\mathcal{L}, \mathcal{B})$-measurable (the usual sense of "Lebesgue measurable") and so it is almost everywhere equal to a Borel ($(\mathcal{B}, \mathcal{B})$-measurable) map $S$.  That is, there is a measurable set $E$ of measure 1 such that $S = T$ on $E$.  Moreover, without loss of generality we may assume $E$ is Borel.
Now since $E$ is a Borel set and $S$ is a Borel map, $S(E) = T(E)$ is analytic and in particular Lebesgue measurable.  Moreover, $E \subset T^{-1}(T(E))$, so $T^{-1}(T(E))$ has measure 1.  But $T$ was measure preserving, so $T(E)$ has measure 1.  Since $T(E) \subset T([0,1])$, we conclude that $T([0,1])$ has measure 1 and in particular is Lebesgue measurable.
The same argument should work if $[0,1]$ is replaced by any Polish space equipped with the completion of a Borel measure.
