If $X$ and $Y$ have same distributions and $X \leq Y$ almost surely, does $X=Y$ almost surely? Here is a specific problem I met.
Assume $f(\omega)$ be a real-valued random variable which is defined on a metric dynamical system $(\mathcal{\Omega},\mathcal{F},\theta_{t},\mathbb{P})$ and $f(\theta_{t}\omega)\leq f(\omega)$ $\mathbb{P}$-a.s. Can we cliam that $f(\theta_{t}\omega)=f(\omega)$ $\mathbb{P}$-a.s. ?
Recall that a metric dynamical system $(\mathcal{\Omega},\mathcal{F},\theta_{t},\mathbb{P})$ means that $(\mathcal{\Omega},\mathcal{F},\mathbb{P})$ is a probability space. $\theta_t\circ\theta_s=\theta_{t+s}$, $(t,\omega)\mapsto\theta_{t}\omega$ is measurable. And $\mathbb{P}$ is measure-preserving, that is, $\mathbb{P}(B)=\mathbb{P}(\theta_{t}B)$ for all $B\in\mathcal{F}$.
Here is my option. Clearly, $f(\theta_{t}\omega)$ and $f(\omega)$ has same distribution since $\mathbb{P}$ is measure-preserving. Thus we have $\mathbb{P}\{\omega:f(\theta_{t}\omega)\in[0,f(\theta_{t}\omega]\}=\mathbb{P}\{\omega:f(\omega)\in[0,f(\theta_{t}\omega]\}$.
Since $\mathbb{P}\{\omega:f(\theta_{t}\omega)\in[0,f(\theta_{t}\omega]\}=1$, we have $\mathbb{P}\{\omega:f(\omega)\in[0,f(\theta_{t}\omega]\}=1$, that is $f(\omega)\leq f(\theta_{t}\omega)$ $\mathbb{P}$-a.s.
Similarly, $\mathbb{P}\{\omega:f(\omega)\in[0,f(\omega]\}=\mathbb{P}\{\omega:f(\theta_t\omega)\in[0,f(\omega]\}$. And $\mathbb{P}\{\omega:f(\omega)\in[0,f(\omega]\}=1$. Hence $\mathbb{P}\{\omega:f(\theta_t\omega)\in[0,f(\omega]\}=1$, that is $f(\theta_{t}\omega)\leq f(\omega)$ $\mathbb{P}$-a.s.
However, We don't use the assumption $f(\theta_{t}\omega)\leq f(\omega)$ almost surely, we directly follow the conclusion $f(\theta_{t}\omega)=f(\omega)$ almost surely from the validity of the same distribution of $f(\theta_{t}\omega)$ and $f(\omega)$. It is not ture in general. Could you tell me where I deduced wrong? Thanks!
Yes, $\arctan X\le \arctan Y$ (because $\arctan$ is increasing) and $E\arctan X=E \arctan Y$ (because $\arctan X$ and $\arctan Y$ have the same distribution). Hence, $\arctan X=\arctan Y$ a.s. which implies $X=Y$ as. [Use of $\arctan$ is to make expectations exist].