Suppose we have a general continuous stochastic process $X_t$ which satisfies $$\lim_{m \rightarrow \infty} \sup_{t \ge 0} P \left( X_t < \frac{1}{m} \right) = 0 \quad \quad \quad \textbf{(1)}$$
I want to use this to show a long run lower bound for $X$, i.e. $P-$almost surely, there exists some $m$ and $T$ such that $X_t \ge 1/m$ for every $t \ge T$. I am not sure if $\textbf{(1)}$ implies some sort of result like this. Formally, this is equivalent to showing $$P \left( \bigcap_{n=1}^\infty \bigcup_{t \in [n, \infty) \cap\mathbb{Q}} \{ X_t < 1/m_n \}\right) = 0$$ for some appropriately chosen $m_n$ but I cannot show this. I tried thinking about it in terms of Borel-Cantelli, but the $P(X_t < 1/m)$ tending to $0$ uniformly in $t$ does not suffice as far as I can gather.
Maybe I am taking the wrong approach. Any help would be massively appreciated!
No, this is not the case. Let $Y \sim U(0,1)$. For $t \in [n,n+1]$ and $j=0,...,n-2$, if $Y \in [\frac jn, \frac{j+1}n)$, define $X_{n + \frac{j+1}n} = \frac 1n$, with $X_t = 1$ for $t \in [n,n+\frac jn]$ or $t \in [n+\frac{j+2}{n},n+1]$, and linearly interpolate between. Then for $t \ge n$, $P(X_t \ne 1) \le \frac 3n$, and for $t < n$ we have $P(X_t < \frac 1n) = 0$. Hence for all $t \ge 0$ and $n \in \mathbb{N}$ we have $P(X_t < \frac 1n) \le \frac 3n$, so $\lim_{n \rightarrow \infty} \sup_{t \ge 0} P(X_t < \frac 1n) \le \lim_{n \rightarrow \infty} \frac 3n = 0$. But for all $m$ and $T$ there exists $t \ge 0$ such that $X_t < \frac 1m$.