I am currently reading through the proof of the Doob-Meyer deocomposition and came across this statement:
Assume that $(A_t)_{t \in [0,1]}$ is a right continuous stochastic process. If $D$ is a countable dense subset of $[0,1]$, e.g. $D=\mathbb{Q} \cap [0,1]$, and for all $t,t' \in D$ with $t \leq t'$ we have $$A_t \leq A_t' \text{, P-a.s. }, $$ Then the trajectories of $A$ are almost surely increasing, i.e. the set $$\{\omega: t \mapsto A_t(\omega) \text{ is increasing}\}$$ contains a set of probability one.
I do not know how to prove this statement, since it was not proved inside the proof, so I assume it should be easy. But I still cannot see why it holds. Does anyone know how to prove this?
Thanks a lot in advance!
Since the set of rationals is countable, the inequality you have holds almost surely for all rational $t, t'$. Note the difference in the position of the quantifier "almost surely" between your inequality and what I stated above.
Then fix some $\omega$ in this full measure set and two arbitrary time points $t<s$. Since the paths are right-continuous, you can express $A_t(\omega)$ as $\lim A_{t_n}(\omega)$ with $t_n$ rational for every $n$. You can do the same for $A_s(\omega)$. Since $A_{t_n}(\omega) \leq A_{s_n}(\omega)$ (you can choose your sequences to make sure this is true), it follows that $A_{t}(\omega) \leq A_{s}(\omega)$.