Let $M\in\mathbb R$. How can I show that $$p\{\limsup_{n\to \infty }A_n\geq M\}\geq \limsup_{n\to \infty }p\{A_n\geq M\}\ \ \ ?$$ What I tried is $$p\{\limsup_{n\to\infty }A_n\geq M\}=p\left\{\bigcap_{n\in\mathbb N}\bigcup_{k=n}^\infty A_n\geq M\right\}=\lim_{n\to \infty }p\left\{\bigcup_{k=n}^\infty A_k\geq M\right\}$$ but $\bigcup_{k=n}^\infty A_n\geq M$ and $\bigcap_{n\in\mathbb N}\bigcup_{k=n}^\infty A_n\geq M$ doesn't has a sense... (they are probably not sets as I could consider).
Any idea ?
By the reverse Fatou's lemma, we have
$$ \limsup_{n\to\infty}\Bbb{P}(A_n \geq M) \leq \Bbb{E}( \limsup_{n\to\infty} \mathbf{1}_{\{ A_n \geq M \}} ) = \Bbb{P}( A_n \geq M \text{ i.o.} ) $$
But on the event $\{ A_n \geq M \text{ i.o.} \}$ we have $\limsup A_n \geq M$. Therefore the inequality follows.