Let
- $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
- $X,X^n:\Omega\times[0,\infty)\to\mathbb R$ be continuous stochastic processes on $(\Omega,\mathcal A,\operatorname P)$ such that
- $(X-X^n)_0=0$
- $(X-X^n)_t\ge(X-X^n)_s$ for all $t\ge s\ge0$ and $n\in\mathbb N$
- $X^n\le X^{n+1}$ for all $n\in\mathbb N$
Assume $$\operatorname P\left[\sup_{s\in[0,\:t]}(X_s-X^n_s)>\varepsilon\right]\xrightarrow{n\to\infty}0\;\;\;\text{for all }t\ge0\text{ and }\varepsilon>0\tag1\;.$$
I want to conclude that $$X^n_t\xrightarrow{n\to\infty}X_t\;\;\;\text{for all }t\ge0\text{ almost surely}\;.\tag2$$
By 2., $(1)$ is equivalent to $$\operatorname P\left[X_t-X^n_t>\varepsilon\right]\xrightarrow{n\to\infty}0\;\;\;\text{for all }t\ge0\text{ and }\varepsilon>0\tag3$$ and by 3. and this question, we obtain $$X^n_t\xrightarrow{n\to\infty}X_t\;\;\;\text{almost surely for all }t\ge0\tag4\;.$$ By continuity and denseness of $[0,\infty)\cap\mathbb Q$ in $[0,\infty)$, we finally obtain $(2)$.
Is there anything I'm missing? The described scenario is an abstraction of a special problem I've encountered in a book. Therein, the author claims that $(2)$ follows from $(1)$. While, as I've shown, this seems to be correct, I wonder why he's explicitly taking the supremum in $(1)$. So, my question is, is there any more general result which allows us to conclude $(2)$ from $(1)$ (and which the author probably got in mind)?
By 3., the pointwise limit $Y_t(\omega) :=\lim_n X^n_t(\omega)$ exists for each $t$ and $\omega$. By (1) there is a subsequence $\Bbb K\subset\Bbb N$ and a measurable set $\Omega_0$ with $\Bbb P(\Omega_0)=1$ such that $\lim_{n\in\Bbb K}\sup_{0\le s\le t}|X_s(\omega)-X^n_s(\omega)|=0$ for all $s\ge 0$ and all $\omega\in\Omega_0$. It follows that $X_s(\omega)=Y_s(\omega)$ for all $s\ge 0$ and all $\omega\in\Omega_0$. That is, $\lim_n X^n_s(\omega)=X_s(\omega)$ for all $s\ge 0$ and all $\omega\in\Omega_0$. [Condition 2. is unnecessary unless you want to weaken (1) to (3).]