a question about random variable I am thinking whether it is possible to have a random variable $X$ defined on the probability space $(\Omega, \mathscr{F}, \mathbb{P})$ such that $\mathbb{P}(|X|>N)=1$ as $N\to +\infty$. To me, Borel measurability of $X$ alone does not seem to prevent this from being true. The definition of $X$ also allows it to take on values of $\pm \infty$. Please correct me if I am wrong, or give an example if this can be true.
Thank you.
 A: The confusion, if any, stems from not specifying the target space $E$ of the arrow $X:\Omega\to E$. 
If $E$ is the real line, obviously $P(|X|\ge N)\to0$ when $N\to+\infty$, but if $E$ is the extended real ine, this need not be true anymore since $P(X=+\infty)$ and $P(X=-\infty)$ may be any nonnegative real numbers whose sum does not exceed $1$. In the second case, the sum $P(X=+\infty)+P(X=-\infty)$ might even be $1$, this would happen if $P(|X| \mbox{finite})=0$ and $P(|X|=+\infty)=1$. 
Both topological spaces $E$ are endowed with a natural Borel sigma-algebra. Some people call random variable any random variable with values in the (non extended) real line, some consider the extended real line, and, although I prefer one choice over the other, both seem legitimate to me. 
A common situation where the extended real line enters the picture is when a sequence $(\xi_t)$ of random variables indexed by $t\in T$ is given, for example with $T=\mathbb{N}$, and one defines $X$ as the smallest index $t$ such that an event like $[\xi_t\in B]$ happens, for a given measurable subset $B$ of the common target space of the random variables $\xi_t$. Then the natural target space of $X$ is the extended line because $X=+\infty$ on the event $[\forall t\in T, \xi_t\notin B]$, and it may well happen that $P(X=+\infty)>0$. 
However, I can think of no natural situation where a random variable $X$ such that $P(|X|=+\infty)=1$ is involved.
The relevant wikipedia paragraph explains this concisely and competently.
A: If ${\rm P}(|X| > N) = 1$, then $1-{\rm P}(|X| > N) = 0$; that is, ${\rm P}(|X| \leq N) = 0$. But the distribution function of $|X|$ necessarily tends to $1$ as $N \to \infty$, that is ${\rm P}(|X| \leq N) \to 1$ as $N \to \infty$, a contradiction. While a random variable is often allowed to take the value $+\infty$ or $-\infty$ with probability $0$, it is not allowed to take these values with positive probability.
EDIT: I considered real-valued random variables, but in general random variables need not be real-valued, hence Jonas' answer may be good as well.
A: Of course this is true if $\mathbb P(X=\pm\infty)=1$.  That is the only case.  When you have a decreasing sequence of events, the limit of the probabilities is the probability of the limit.  Therefore $\displaystyle{\lim_{N\to\infty}\mathbb{P}(|X|>N)=\mathbb{P}(X=\pm\infty)}$.
