I have some difficulties understanding the following:
Let $(X_n)$ be a sequence of independent random variables s.t.
$P[X_n=1]=1−P[X_n=0]=\frac{1}{n}$
After using the Borell Cantelli lemma, I could show that:
$P(\limsup [X_n=0])=1$
$P(\limsup [X_n=1])=1$
Now using the fact that $\operatorname{[\limsup F_n]}^c = [\liminf \operatorname{(F_n)}^c]$ I concluded that
$P(\liminf [X_n=1])=0$
$P(\liminf [X_n=0])=0$ as $P(A)+P(A^C)=1$.
This implies that as $X_n$ is either $0$ or $1$, $\liminf X_n$ is never constant with probability $1$, contradicting some of the implications of the Kolmogorov $0$-$1$ law.
Could someone please clarify this? Much appreciated.
Consider the events $A_n=[X_n=0]$. By definition, the event $\limsup A_n$ happens if and only if $A_n$ happens for infinitely many $n$, hence the general formula $$ \limsup A_n=\bigcap\limits_n\bigcup\limits_{k\geqslant n}A_k. $$ In our case, consider $N=\{n\mid X_n=0\}$, then $$ \limsup A_n=[X_n=0\ \text{for infinitely many}\ n]=[N\ \text{is infinite}]. $$ Consider $Y=\liminf X_n$. This is a random variable $Y$ characterized by the fact that $Y=1$ if $X_n=1$ for every $n$ large enough, that is, $$ [Y=1]=[\exists n,\ \forall k\geqslant n,\ X_k=1]. $$ and that $[Y=0]=[N\ \text{is infinite}]$.
Finally, for $\{0,1\}$-valued random variables $(X_n)$, $$ \limsup[X_n=0]=[\liminf X_n=0]. $$ Likewise, again for $\{0,1\}$-valued random variables $(X_n)$, $$ \liminf[X_n=0]=[(X_n)\ \text{converges and}\ \lim X_n=0]=[\limsup X_n=0], $$ $$ \liminf[X_n=1]=[(X_n)\ \text{converges and}\ \lim X_n=1]=[\liminf X_n=1], $$ and $$ \limsup[X_n=1]=[X_n=1\ \text{for infinitely many}\ n]=[\limsup X_n=1]. $$