A subinvariant random variable is already invariant

Question

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 $\{X\circ\tau<X\}$ is a null set?

Answer 1

Partial answer: Obviously, $$\{X>c\}\subseteq\tau^{-1}\left(\left\{X>c\right\}\right)\tag3$$ and hence (since $\operatorname P\circ\:\tau^{-1}=\operatorname P$) $$\operatorname P[X\le c<X\circ\tau]=\operatorname P\left[\tau^{-1}\left(\left\{X>c\right\}\right)\right]-\operatorname P[X>c]=0\tag4$$ for all $c\in\mathbb R$. I'm not sure whether I'm missing something or not, but we should be able to conclude $$\operatorname P[X<X\circ\tau]=\bigcup_{c\in\mathbb Q}\operatorname P[X\le c<X\circ\tau]=0\tag5.$$ Since, on the other hand, $\operatorname P[X\le X\circ\tau]=1$, this would yield $(2)$.