When $X_s<Y_t$ almost surely, for every $s<t$, implies that $X_s<Y_t$ for every $s<t$, almost surely?

Question

When I have shown, for $s\le t$ and for two continuous stochastic process an inequality:

$$ X_s \le Y_t$$ P-a.s.

How can I deduce that this P-a.s. simultaneously for all rational $s\le t$ ? Thank you for your help

EDIT: According to Ilya's answer, I see that we have $$P(X_s\le Y_t\text{ simultaneously for all rationals }s\le t) = 1.$$ How could we use continuity of $X,Y$ to deduce $P(X_s\le Y_t,s\le t)=1$. Of course we take sequences of rational, however I mess up the details. So a detailed answer how to do this, would be appreciated.

Answer 1

If I got you correct, you have $P(X_s\leq Y_t) = 1$ for all $s\leq t$. Recall that if $P(A_n) $ for all $n\in \mathbb N$ then also you have $P(\bigcap_n A_n) = 1$ since $$ P(\bigcap_n A_n) = 1-P(\bigcup_n A^c_n)\geq 1-\sum\limits_n P(A^c_n) =1. $$ Now take $s_n$ be the $n$-th rational number that is less or equal to $t$ and $A_n=\{X_{s_n}\leq Y_t\}$. You can even use continuity of $X,Y$ to show that $P(X_s\leq Y_t,s\leq t) = 1$.

To show the latter, consider two sets: $$ C = \{\omega\in \Omega: X_t(\omega),Y_t(\omega)\text{ are continuous }\} $$ $$ D = \{\omega\in \Omega:X_s(\omega)\leq Y_t(\omega) \text{ simultaneously for all rational }s\leq t\}. $$

It is given to us that $\mathsf P(C) = 1$ and we have proved above that $\mathsf P(D) = 1$. As a result, $$ \mathsf P(C\cap D) = 1-\mathsf P(C^c\cup D^c)\geq 1-\mathsf P(C^c)-\mathsf P(D^c) = 1. $$

Consider now any $\omega\in C\cap D$.

1. It holds that for this $\omega$: $X_s(\omega)\leq Y_t(\omega)$ for all rational $s\leq t$. Since both $X,Y$ are continuous on $\omega$, it follows that $X_s(\omega)\leq Y_t(\omega)$ for all $s\leq t$. Indeed, if that would not be true, i.e. for some $s'\leq t$ you have $X_{s'}(\omega)>Y_{t}(\omega)$ then $X_s(\omega)>Y_t(\omega)$ in the neighborhood of $s'$ which cannot happen since there will be at least one rational number $s$.

2. Thus, for any $\omega\in C\cap D$ the desired relation holds: $C\cap D \subseteq \{X_s\leq Y_t,s\leq t\}$, so that $$ 1 =\mathsf P(C\cap D) \leq \mathsf P\{X_s\leq Y_t,s\leq t\}. $$

Answer 2

This follows from the fact that the complement of the event $[\forall s\leqslant t,\,X_s\leqslant Y_t]$ is the event $$ \left[\exists s\leqslant t,\,X_s\gt Y_t\right]\ =\left[\exists n\in\mathbb N,\,\exists s\in\mathbb Q,\,\exists t\in\mathbb Q,\,s\leqslant t,\,X_s\geqslant Y_t+\frac1n\right], $$ hence $$ \left[\exists s\leqslant t,\,X_s\gt Y_t\right]\ =\bigcup\limits_{s\leqslant t,\, s\in\mathbb Q,\,t\in \mathbb Q}\ \bigcup_{n\geqslant1}\ \left[X_s\geqslant Y_t+\frac1n\right]. $$ Since $\mathrm P(X_s\leqslant Y_t)=1$ for every $s\leqslant t$, $\mathrm P(X_s\geqslant Y_t+\frac1n)=0$ for every $n\geqslant1$. The union on the RHS of the displayed identity above is countable hence $\mathrm P(\exists s\leqslant t,\,X_s\gt Y_t)=0$. Considering the complement, one gets $$ \mathrm P(\forall s\leqslant t,\,X_s\leqslant Y_t)=1. $$