If $T$ is measure preserving and $X \circ T \le X \ a.s$, does $X = X \circ T \ a.s$? If $T$ is measure preserving and $X \circ T \le X \ a.s$, does $X = X \circ T \ a.s$?
I presume the proof would involve working splitting integrals up on null sets and such, but I am not sure how the details are.
 A: Let $(\Omega,\mathbb{F},P)$ be a probability space and let $T$ be $P$-measure preserving. Let
    $X$ be an integrable random variable and assume that $X \circ T \leq  X $ almost surely. Then
    $X = X \circ T$ almost surely. 
Proof: Since $X$ is $P$-integrable and $T$ is measure preserving it is especially $T(P)=P$-integrable. By the abstract change of variable theorem we now get that
$$
\int X dP = \int X dT(P)  =\int X\circ T dP \iff \int X-X\circ T dP = 0
$$
Now note that on the full measure set $(X\circ T \leq X) $  we have that  $X-X\circ T = |X-X\circ T|$. Thus 
$$
0= \int X-X\circ T dP = \int_{(X\circ T \leq X)} |X-X\circ T | dP+\int_{(X\circ T \leq X)^c} X-X\circ T dP 
$$
$$= \int_{(X\circ T \leq X)} |X-X\circ T | dP +\int_{(X\circ T \leq X)^c} |X-X\circ T| dP =\int |X-X\circ T | dP
$$
where we changed the integrand in the last integral, but this does not change the value since we integrate over a zero measure set (i.e. both integrals are zero). Hence we know that (Theorem 10.9 [Schilling])
$$|X-X\circ T|=0 \textit{ $P$-a.s.} \iff X=X\circ T \textit{ $P$-a.s.}
$$ 
 qed.
