Suppose that $(\Omega,\mu)$ is a measure space. Let $\tau:\Omega\to\Omega$ is a measurable map such that $\mu\circ\tau^{-1}<<\mu$. Then $\tau$ s said to be null preserving. I want to prove the following. If $f:\Omega\to\mathbb R$ is measurable and $\mu(\{f\neq f\circ \tau\})=0,$ then there exists a measurable function $f^\prime$ such that $f^\prime=f^\prime \circ \tau$ and $\mu(f\neq f^\prime)=0.$ If we define $A:=\{x\in\Omega:f(x)=f(\tau(x))\}.$ I can prove that $A$ is $\tau$-invariant mod $\mu.$ A natural way to define $f^\prime$ would be $f^\prime=f1\_{B}$ where $B$ is $\tau$-invariant and $\mu(A\Delta B)=0.$ But I can not really see if it works. It will work definitely if we have $B\subseteq A.$ We can have $B$ to be the set $\cup\_{k=0}^\infty(A\setminus\ \cup\_{k=0}^\infty\tau^{-k}(A\setminus \tau^{-1}A)).$ Can we have that $B\subseteq A$? Also. I want to find some intuitive idea how the construction should be.



Null preserving transformation
I'll rename your $\Omega$ into $X$ to simplify typing.


Let $D=\{x\in X,~\forall n\in\mathbb{N}:~ f(\tau^n(x))=f(x)\}=\bigcap\limits\_{n\in\mathbb{N}\_0} \{x\in X: f(\tau^{n+1}(x))=f(\tau^n(x))\}$ $=\bigcap\limits\_{n\in\mathbb{N}\_0}\tau^{-n}(A)=X\backslash \bigcup\limits\_{n\in\mathbb{N}\_0}\tau^{-n}(X\backslash A)$. Since $\tau$ is null-preserving, $D$ is of full measure. Also, $\tau (D)\subset D$.


Define $g:X\to\mathbb{R}$ as follows. If for $x\in X$ there is $n\in \mathbb{N}\_0$ such that $\tau^n(x)\in D$, define $g(x)=f(\tau^n(x))$; otherwise, define $g(x)=0$. Note, that in the first case we get the same value, regardless of $n$ (as long as $\tau^n(x)\in D$).


Since $D$ is of full measure, it is clear that $g=f$ almost everywhere. Let us check $g=g\circ\tau$. If $x\in X$ is such that $\tau^n(x)\in D$, for some $n$, we get $g(x)=f(\tau^n(x))=f(\tau^{n+1}(x))=g(\tau(x))$. Otherwise, $\tau^n(x)\not\in D$, for every $n$, therefore $\tau^{n+1}(x)\not\in D$, for every $n$, and so $g(x)=0=g(\tau(x))$.


