Assume that for a sequence of random variables $(X_n)$ one has
$\lim_{x\to -\infty}\lim_{a\to 0}\lim_{n\to\infty} P[X_n \ge x_0(1-a)+a\cdot x]=1.$
Does this imply $\liminf_{n\to\infty} X_n\ge x_0$ almost surely?
Edit: I've tried to prove $P[\liminf \{M_n \ge x_0\}]=1$ directly via $\sigma$-continuity and estimates but failed. I think that
$\lim_{a\to0} P[ \liminf \{X_n \ge x_0(1-a)+ax\}] = P[\bigcap_{a>0} \liminf \{X_n \ge x_0(1-a)+ax\}] =P[\liminf \{X_n \ge x_0\}]$
and similarly for the limit of $x$, but I don't see how to use the limit of $n$ to infinity. Is there a weaker condition for $\liminf X_n\ge x_0$ a.s. than $P[\liminf \{X_n \ge x_0\}]=1$ one could use in this case?