Suppose $\theta$ is measurable and $A\in\F$ satisfies $\mu(\theta^{-1}(A)\D A)=0$. Then there is a subset $B$ in the $\mu$-completion $\F^\mu$ of $\F$ such that $\mu(A\D B)=0$ and $\theta^{-1}(B)=B$.
We have: $\mu(\theta^{-n-1}(A)\D\theta^{-n}(A))=0$ and $\mu(\theta^{-n}(A)\D\theta^{-m}(A))=0$. Put
$$
A_{max}\colon=\bigcup_{n=0}^\infty\theta^{-n}(A)
\quad\mbox{and}\quad
A_{min}\colon=\bigcap_{n=0}^\infty\theta^{-n}(A)~.
$$
Then $A_{min}\sbe A\sbe A_{max}$ and $\theta^{-1}(A_{max})\sbe A_{max}$ and $\theta^{-1}(A_{min})\spe A_{min}$. Moreover
$$
\mu(A_{max}\sm A_{min})
=\mu\Big(\bigcup_n\theta^{-n}(A)\cap\bigcup_m\theta^{-m}(A^c)\Big)
\leq\sum_{n,m}\mu(\theta^{-n}(A)\D\theta^{-m}(A))
=0~.
$$
Now define
$$
{\cal C}\colon=\{C\sbe\O:A_{min}\sbe C\sbe
A_{max},\theta^{-1}(C)\spe C\}
$$
and put $B\colon=\bigcup\{C:C\in{\cal C}\}$. We obviously have
$$
\theta^{-1}(B)
=\bigcup\{\theta^{-1}C:C\in{\cal C}\}
\spe\bigcup\{C:C\in{\cal C}\}
=B~.
$$
On the other hand we get for all $C\in{\cal C}$:
$$
A_{min}
\sbe\theta^{-1}(A_{min})
\sbe\theta^{-1}(C),\quad
A_{max}
\spe\theta^{-1}(A_{max})
\spe\theta^{-1}(C)
$$
and $\theta^{-1}\theta^{-1}(C)\spe\theta^{-1}(C)$, i.e. $\theta^{-1}(C)\in{\cal C}$.