Actually $\frac{1}{\int_{0}^{\infty}f(X_s)ds}=0$ almost surely does not imply $\lim_{t\rightarrow\infty}\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{t}f(X_s)ds}\right)^2\right\}=0$, but what about $\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{\infty}f(X_s)ds}\right)^2\right\}$?

@S.Surace $X_\infty$ is $\lim_{t\rightarrow\infty}X_t$.

Sorry but can you explain the last step a bit more? (3) implies that $\frac{1}{\int_{0}^{\infty}f(X_s)ds}=0$ almost surely, and that does not necessarily mean $\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{\infty}f(X_s)ds}\right)^2\right\}=0$, right?

Wow Thank you!!! I didn't finish reading yet but a first glance at your paper already shows how powerful and relevant it is! Thank you for pointing out such a relevant reference!