Let $(X_n)_{n\in \mathbb{N}}$ be a sequence of iid random variables. Given that $X_n$ converge almost surely to $x_0\in [m, M]$, where $0<m<M<1$, I try to find the upper bounds of $\mathbb{E}\left(\dfrac{1}{X_n}\right)$ and $\mathbb{E}\left(\dfrac{1}{1-X_n}\right)$.
My calculations:
Since $X_n \underset{n\to +\infty}{\longrightarrow} x_0$ a.s, there exists $\epsilon$ such that $X_n \ge x_0 -\epsilon \ge m - \epsilon$. If we choose $\epsilon = \dfrac{m}{2}$, we obtain $X_n \ge \dfrac{m}{2}$. We get $\dfrac{1}{X_n} \le \dfrac{2}{m}$ a.s.
Similarly, using again the property $X_n \underset{n\to +\infty}{\longrightarrow} x_0$ a.s, there exists $\delta$ such that $X_n \le x_0 + \delta \le 1 - m + \delta$, this implies $1-X_n \ge m - \delta$. Therefore if we choose $\delta = m/2$, we obtain $\dfrac{1}{1-X_n} \le \dfrac{2}{m}$ a.s.
Then we can conclude that for $n$ large enough $$ \mathbb{E}\left(\dfrac{1}{X_n}\right) \le \frac{2}{m}\quad\text{ and }\quad \mathbb{E}\left(\dfrac{1}{1-X_n}\right) \le \frac{2}{m}. $$
However, I have a doubt about my results. Indeed, the upper bounds of $1/X_n$ and $1/(1-X_n)$ are only true for $n \ge N(\omega)$ (where $\omega\in\Omega$ of the probability space), so we have also a dependence on $\omega$ and taking the expectations over $\Omega$ makes a problem!! Furthermore, when we say the upper bounds of the expectations are valid for "$n$ large enough", this seems vague. But I'm not sure. So I would be appreciated for any comment of suggestion that helps to make clear this point. Thank you very much.