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Let $X^n=(X^n_t)_{t\ge 0}$ and $X=(X_t)_{t\ge 0}$ be RCLL (right-continuous with left limits) processes such that

$$\lim_{n\to\infty}X^n=X,\quad \quad \mbox{almost surely},$$

where this convergence holds under "suitable topology" (e.g. Skorokhod topology). Define the hitting times $\tau_n:=\inf\{t\ge 0: X^n_t\le 0\}$ and $\tau:=\inf\{t\ge 0: X_t\le 0\}$. My question is, under which additional conditions, one may ensure $\lim_{n\to\infty}\mathbb P(\tau_n=\infty)=\mathbb P(\tau=\infty)$? The above topology can be changed to any suitable topology.

Any answers, references and comments are highly appreciated!

PS : If we take $X^n_t = 1 + \alpha^n t + W_t$ and $X_t = 1 + \alpha t + W_t$ with $\lim_{n\to\infty} \alpha^n=\alpha>0$, where $(W_t)_{t\ge 0}$ is a standard Brownian motion. Then the desired convergence holds, but both $\lim_{t\to\infty}X^n_t$ and $\lim_{t\to\infty}X_t$ do not exist. Therefore, I am very curious what kind of properties at $t=\infty$ may ensure the above convergence.

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    $\begingroup$ Maybe an even worse example is just to let $X_n = 1/n$ independent of both $t$ and $\omega$, with $X=0$. Then $X_n \to X$ uniformly and almost surely, which is about the strongest mode one could imagine, but $\mathbb{P}(\tau_n = \infty)=1$ and $\mathbb{P}(\tau = \infty)=0$. So I don't think you will get your desired result without knowing something more specific about the processes themselves. $\endgroup$ Commented Oct 27, 2020 at 23:44

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