Suppose we have a probability space $(X,\mathcal{B},\mu)$. Suppose $T:X\to X$ is non-singular. That is, $T$ is measurable and $\mu(N)=0$ if and only if $\mu(T^{-1}N)=0$.
Can we deduce that for any $p>1$ and $v\in L^p(X)$, $\|v\circ T\|_p\leq \|v\|_p$?
Just a side remark: Note that non-singularity is a weaker notion of invariance - if $T$ was $\mu$-invariant, that is $\mu(A)=\mu(T^{-1}A)$ for all $A$ measurable, then trivially $\|v\circ T\|_p=\|v\|_p$.
Any help would be appreciated!
Here's a counterexample:
You let $X=[0,3]$ with the usual Borel $\sigma$-algebra and normalized measure (we can use the interval $[0,1]$ instead but it's less convenient).
$T:X\rightarrow X$ is the function defined by
It sends the interval $[0,1]$ to $[0,2]$ ($x\mapsto 2x$), Now it starts shrinking
It sends $[1,2]$ to $[2,2.5]$ ($x\mapsto \frac{x+3}{2}$).
It sends $[2,2.5]$ to $[2.5,2.75]$
and so on...
As it only shrink the interval by at most twice it is non-singular. But define a function $f:X\rightarrow \mathbb{R}$ by the formula $f(x)=2-x$ on $[0,2]$ and zero otherwise. You have a counterexample.