limsup and liminf of sequence of random variables and Borel Cantelli For a sequence of independent random variables $X_n$, if we have $P( \limsup [X_n > a] ) = 1$, then $P( [ \limsup X_n ] > a ) = 1$. Does this automatically mean $P( \liminf [X_n > a] ) = 1$ as well?
Since $P(([\limsup X_n] > a)^c)=P( \liminf(X_n > a)^c)=P([\liminf X_n]\le a) = 0$, we have $P( \liminf [X_n > a] ) = 1$. 
Please let me know if I have made stupid mistakes. Thank you in advance.
 A: Important inequalities
Williams - Probability with Martingales



Deduced similarly:
(iii) If $\liminf x_n > z$, then
$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)
(iv) If $\liminf x_n < z $, then
$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)


$$P( \limsup [X_n > a] ) = 1 \to P( [ \limsup X_n ] > a) = 1$$

No. Consider $X_n = a + \frac{1}{n}$.
The converse, which is,
$$P( \limsup [X_n > a] ) = 1 \leftarrow P( [ \limsup X_n ] > a) = 1$$
is true (see above).


$$P( [ \limsup X_n ] > a) = 1 \to P( \liminf [X_n > a] ) = 1$$

No.
$$P( [ \limsup X_n ] > a) = 1 \to P( \limsup (X_n  > a)) = 1$$
$$P( \limsup (X_n  > a)) = 1 \nrightarrow P( \liminf [X_n > a] ) = 1$$
The converse of the latter, which is,
$$P( \liminf [X_n > a] ) = 1 \to P( \limsup (X_n  > a)) = 1$$
is true (obviously?).


$$P(([\limsup X_n] > a)^c)=P( \liminf(X_n > a)^c)$$

No.
$$P(([\limsup X_n] > a)^c) = P([\limsup X_n] \le a)$$
$$P( \liminf[(X_n > a)^c]) = P( \liminf(X_n \le a))$$
$$P( [\liminf(X_n > a)]^c) = P( \limsup(X_n \le a))$$


$$P( \liminf(X_n > a)^c)=P([\liminf X_n]\le a)$$

No.
$$P( \liminf[(X_n > a)^c]) = P( \liminf(X_n \le a))$$
$$P( [\liminf(X_n > a)]^c) = P( \limsup(X_n \le a))$$
$$P([\liminf X_n]\le a) = P(([\liminf X_n] > a)^C)$$


$$P([\liminf X_n]\le a) = 0 \to P( \liminf [X_n > a] ) = 1$$

Yes. Contrapositive of 3
