# A subinvariant random variable is already invariant

Let $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space, $$\tau:\Omega\to\Omega$$ be a measurable map on $$(\Omega,\mathcal A)$$ with $$\operatorname P\circ\:\tau^{-1}=\operatorname P$$ and $$X:\Omega\to\overline{\mathbb R}$$ be $$\mathcal A$$-measurable with $$X\circ\tau\le X\;\;\;\operatorname P\text{-almost surely}\tag1.$$ I would like to conclude $$X\circ\tau=X\;\;\;\operatorname P\text{-almost surely}\tag2.$$ I assume that this is somehow almost trivial, but I can't figure out how we need to approach it. Maybe by showing that $$\{X\circ\tau\ge X\}$$ has probability $$1$$ or by showing that $$\{X\circ\tau is a null set?

• Are you assuming ergodicity? Apr 27, 2020 at 17:02
• @FelipePérez No, but do you think it's easy to prove then? Apr 27, 2020 at 18:12
• @0xbadf00d $\int X\circ \tau = \int X$, so $\int (X-X\circ \tau) = 0$, but integrand is non-negative, so it must be identically $0$ Apr 28, 2020 at 14:40

Partial answer: Obviously, $$\{X>c\}\subseteq\tau^{-1}\left(\left\{X>c\right\}\right)\tag3$$ and hence (since $$\operatorname P\circ\:\tau^{-1}=\operatorname P$$) $$\operatorname P[X\le cc\right\}\right)\right]-\operatorname P[X>c]=0\tag4$$ for all $$c\in\mathbb R$$. I'm not sure whether I'm missing something or not, but we should be able to conclude $$\operatorname P[X Since, on the other hand, $$\operatorname P[X\le X\circ\tau]=1$$, this would yield $$(2)$$.