I want to demonstrate the issue using an example. Let $(X_n)_{n\in\mathbb{N}}$ be iid. with $P(X_n> x)=e^{-x}$. We want to show $\forall \epsilon > 0$ $$\limsup_{n\to\infty}\frac{X_n}{\ln n}\ge 1-\epsilon \ a.s$$
To do this the Borel-Cantelli theorem is used. The argument goes like this: $$P(X_n> \ln n(1-\epsilon))=e^{-\ln n(1-\epsilon)}=n^{-(1-\epsilon)}$$ which is not summable. Therefore, for $A_n:=\{X_n>\ln n (1-\epsilon)\}$ we get $$P(\limsup_{n\to\infty} A_n)=1$$
Now in the script I am using it is concluded that $\limsup_{n\to\infty}\frac{X_n}{\ln n}\ge 1-\epsilon \ a.s$ which is what we wanted to show.
Why is $$P(\limsup_{n\to\infty} \{X_n> \ln (1-\epsilon)\})=1$$ the same as $$P(\limsup_{n\to\infty} X_n> \ln (1-\epsilon))=P(\omega\in\Omega\colon \limsup_{n\to\infty}X_n(\omega)>\ln n(1-\epsilon)) = 1$$
For a sequence of r.v.s. $\{Z_n\}$ and a constant $z$, $$ \left\{\limsup_{n\to\infty}Z_n> z\right\}\subseteq \left\{Z_n> z \text{ i.o.}\right\}\subseteq \left\{Z_n\ge z \text{ i.o.}\right\}\subseteq\left\{\limsup_{n\to\infty}Z_n\ge z\right\}, $$ which follows from the properties of sequences of real numbers.