Let $\{ X_{n} \}_{n=1}^{ \infty}$ denote sequence of random variables with the same distribution on $ \Omega $ and also: $$ A_{n} = \{ \omega \in \Omega : \ X_{n} ( \omega) \leq \varepsilon \log (n) \}, \ A = \limsup A_{n} \text{ and} $$ $$ \ Q = \{ \omega \in \Omega : \ \liminf \frac{X_{n}(\omega)}{\log (n)} \leq \varepsilon \} $$
Then I want to prove (as partial result in more complex proof) that $ A \subset Q$. We know that $A$ means that infinitely many of $A_{n}$ happen, so $A$ is a subset of $ \Omega$ for which it is true that for infinitely many $n$ it holds: $ \frac{X_{n}(\omega)}{\log (n)} \leq \varepsilon$ and $Q$ is a subset of $ \Omega$ for which it is true that for almost all of $n$ (excluding finitely many $n$) it is true that: $ \frac{X_{n}(\omega)}{\log (n)} \leq \varepsilon$. However I don't know how to put it formally.
Fix any $\omega \in A$. Then $X_n(\omega) \leq \epsilon \log n$ for infintely many values of $n$. Hence $\frac {X_n(\omega)} { \log n} \leq \epsilon$ for infinitely many values of $n$. This implies that $\lim \inf_n \frac {X_n(\omega)} { \log n} \leq \epsilon$ so $\omega \in Q$.
The last step follows from the fact that if $\lim \inf a_n >\epsilon$ then $a_n >\epsilon$ for all $n$ sufficiently large so the inequality $a_n \leq \epsilon$ can only hold finitely many values of $n$.