# Null preserving transformation

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.

## 1 Answer

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 $$g(x)=0=g(\tau(x))$$.