Is the set $\limsup\limits_n\{\frac{X_n}{\log(n)}>\frac{1}{\lambda}\}$ equal to $\{\limsup\limits_n\frac{X_n}{\log(n)}>\frac{1}{\lambda}\}$?

Question

Difference between $\limsup\limits_n\{\frac{X_n}{\log(n)}>\frac{1}{\lambda}\}$ and $\{\limsup\limits_n\frac{X_n}{\log(n)}>\frac{1}{\lambda}\}$

are the sets equal ?

I think they would be not necessarily equal if we had;

$\limsup\limits_n\Pr\bigg(\{\frac{X_n}{\log(n)}>\frac{1}{\lambda}\}\bigg)$ and $\Pr\bigg(\{\limsup\limits_n\frac{X_n}{\log(n)}>\frac{1}{\lambda}\}\bigg)$

Answer 1

Consider $X_n\equiv 1+\log n$ everywhere, and $\lambda=1$. Then every element belongs to LHS but RHS is empty... so this is a counterexample.