Given $A_n\rightarrow \infty$ almost surely (a.s). Show that $\forall\ N > 0$, $P\left\{A_n<N\ \text{infinitely often}\right\} = 0$.
My thought: By the sake of contradiction, assume there exists an $\infty> N >0$ such that $P\left\{A_n<N\ \text{infinitely often}\right\} > 0$. This is equivalent to: $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} (A_k-N) < 0\right\} > 0$. But this means there exists a $k_1\geq n$ such that $A_{k_1} < N$ (1)
On the other hand, since $A_n\rightarrow \infty$ a.s, for every $\epsilon > 0$, $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} |A_k- \infty| \geq \epsilon\right\} = 0$. This means $|A_k-\infty|<\epsilon$ for every $k\geq n$. Since $k_1\geq n$, $\infty - A_{k_1} < |A_k- \infty| < \epsilon$ for any $\epsilon > 0$. Let $\epsilon = N$, we get $\infty < N+A_{k_1} < 2N$ (due to (1)). This is a contradiction, since $\infty > N$.
My question: I don't find my mathematical notation above legit, because I am doing subtraction with $\infty$. But I would like to know if my solution above is correct, so could someone please help verify? Also, I think there should be a shorter/slicker way to solve it, as my solution above is quite complicated and use more of real analysis. Anyone wants to give it a try? Any thoughts about my solution or this problem would be really appreciated anyway.
Important inequalities (Probability w/ Martingales):
1, 2
If $$\liminf x_n > z$$ then $$(x_n > z)$$ eventually
If $$\liminf x_n < z$$ then $$(x_n < z)$$ infinitely often
$$P([\lim X_n] = \infty) = 1$$
$$\to P([\lim X_n] > N) = 1$$
$$\to P([\liminf X_n] > N) = 1$$
$$\to P(\liminf [X_n > N]) = 1$$
$$\to P(\limsup[X_n \le N]) = 0$$
QED