limsup of a sequence of random variables (definition) Let $X_n$ be a sequence of random variables.
First, $\limsup X_n=\inf_n\{\sup_{m\ge n}X_m\}$. So,
$$\{\limsup X_n\le c\}=\bigcup_n\bigcap_{m\ge n}\{X_m\le c\}$$
Is it correct to say that if $P\{X_n\le c \text{   i.o.}\}\equiv P\{\limsup\{X_n\le c\}\}=1$ then $\limsup X_n\le c \text{ a.s.}$?
Edit 1:
By the same way
$$\{\liminf X_n\ge c\}=\bigcup_n\bigcap_{m\ge n}\{X_m\ge c\}$$
So, $P\{X_n\ge c \text{   i.o.}\}=1 \Rightarrow \liminf X_n\ge c \text{ a.s.}$
Edit 2:
Following @Did's comments, the answer follows from
$$[X_n\geqslant c\ \text{i.o.}]\subseteq[\limsup X_n\geqslant c] \text { and } [X_n\leqslant c\ \text{i.o.}]\subseteq[\liminf X_n\leqslant c]$$

No.
Let $X_{n}=\left(-1\right)^{n}$ (constant random variables) so that: $$P\left\{ X_{n}\leq0\text{ i.o.}\right\} =1=P\left\{ X_{n}\geq0\text{ i.o.}\right\}$$
However $\limsup X_{n}=1$ and $\liminf X_{n}=-1$ so that: $$P\left\{ \limsup X_{n}\leq0\right\} =0=P\left\{ \liminf X_{n}\geq0\right\}$$

addendum:
Let $\left(a_{n}\right)$ be some sequence in $\mathbb{R}$. 
Then:
$$\limsup a_{n}>c\implies a_{n}>c\text{ i.o.}\implies\limsup a_n\geq c$$ 
