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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

$$[X_n\geqslant c\ \text{i.o.}]\subseteq[\limsup X_n\geqslant c] \text { and } [X_n\leqslant c\ \text{i.o.}]\subseteq[\liminf X_n\leqslant c]$$

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  • $\begingroup$ your question needs repairs. $\endgroup$ – drhab Oct 4 '14 at 9:40
  • $\begingroup$ @drhab pls explain... $\endgroup$ – d.k.o. Oct 4 '14 at 9:55
  • $\begingroup$ $P\left\{ X_{n}\geq c\: i.o\right\} \Rightarrow$... How can a value on its own imply something? $\endgroup$ – drhab Oct 4 '14 at 9:59
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    $\begingroup$ The third line is wrong. In general, $$\{\limsup X_n\le c\}\ne\bigcup_n\bigcap_{m\ge n}\{X_m\le c\}=\liminf\{X_n\le c\}.$$ (I corrected the typo $X_n$ for $X_m$.) It is not even always true that $$\{\limsup X_n\le c\}=\limsup\{X_n\le c\}.$$ For example, if $X_n(\omega)=c+1/n$ for every $n$, then $\limsup X_n(\omega)=\lim X_n(\omega)=c$ hence $\omega$ is in the event $\{\limsup X_n\le c\}$ but $X_n(\omega)\gt c$ for every $n$ hence $\omega$ is neither in the event $\limsup\{X_n\le c\}$ nor in the event $\liminf\{X_n\le c\}$. $\endgroup$ – Did Oct 4 '14 at 10:36
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    $\begingroup$ $[X_n\geqslant c\ \text{i.o.}]\subseteq[\limsup X_n\geqslant c]$ and $[X_n\leqslant c\ \text{i.o.}]\subseteq[\liminf X_n\leqslant c]$ $. $\endgroup$ – Did Oct 4 '14 at 20:06
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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: $$\limsup a_{n}>c\implies a_{n}>c\text{ i.o.}\implies\limsup a_n\geq c$$

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  • $\begingroup$ I think you made a mistake in the addendum. We have $1/n >0$ infinitely often, yet $\limsup 1/n = 0$. $\endgroup$ – Bart Michels Mar 16 '19 at 21:09
  • $\begingroup$ @rabota You are correct. Thank you for attending me. I repaired. $\endgroup$ – drhab Mar 16 '19 at 22:05

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