Boundedness in probability Suppose we have a set $M$ of random variables on a probability space. Then we defined boundedness of $M$ as,
$M$ is bounded if $\sup_{X\in M}P(|X|>N) \to 0$ as $ N \to \infty$. 
This definition means that the measure of the set where the elements of $ M $ are big is very small, in fact tends to zero.
I have three questions to this definition and some further conclusions. 
Now suppose we have a unbounded set. My questions are: 


*

*Unbounded would mean, that for every $N>0$ there exists an $ \epsilon >0 $ and a $ X \in M $ such that $ P(|X| > N) \ge \epsilon $. Is this conclusion right?

*If theres a sequence $ (X_n) $ of random variables, unbounded and positive, then there's a subsequence $ (X_{n_k}) $ and a $\lambda>0 $ such that $ P(|(X_{n_k})| > k)\ge \lambda $ for every $ k \in \mathbb{N}$.


My observations so far to 2. 
After the comment of Srivatsan (see below), we therefore have:$ \exists \epsilon > 0 $ sucht that for all $ N>0 $ exists a $ X_n $ such that $ P(X_n(\omega) > N) \ge \epsilon $ Put $ N=1 $, hence there is a $ X_{n_1} $ sucht that $ P(X_{n_1} > 1 ) \ge \epsilon$. Now put $ N=2 $, hence there is a $ X_i $ such that $ P(X_i > 2) \ge \epsilon$. Now the problem is, why do I know that $ i> n_1 $ ? Otherwise it isn't a subsequence. 
Thanks for your help
hulik
 A: (1) is not correct.  Let $Y$ be a single random variable such that, for every $N$, we have $P(|Y| > N) > 0$; for example, a normal random variable.  (I would call such a random variable "unbounded", but that conflicts with your terminology.)  Then let $M = \{Y\}$.  $M$ is bounded in your sense (exercise: verify this, using continuity of probability) but it is still true that for every $N$ there exists $X \in M$ (namely $X=Y$) and $\epsilon > 0$ (namely $\epsilon = \frac{1}{2}P(|Y|>N)$) such that $P(|X| > N) > \epsilon$.
Srivatsan's comment above gives a corrected statement; I just wanted to show explicitly that the statement in the original question (for every $N>0$ there exists an $\epsilon >0$ and $X \in M$ such that $P(|X|>N) \ge \epsilon$) is not correct.
Regarding (2): You know that there is an $\epsilon$ such that for any $N$ there is an $X_n$ with $P(|X_n|>N) > \epsilon$.  In fact, for any $N$ there are infinitely many such $X_n$; thus you can always choose one that occurs later in the sequence than all the ones chosen so far.  To see there must be infinitely many such $X_n$, suppose there were only finitely many and show that in fact we would have to have $\sup_n P(|X|>N) \to 0$ as $N \to \infty$.  (Warmup: what if there were only one such $X_n$?  Now what if there were only two?)
Hint: Fix a random variable $X$.  This is a measurable function $X : \Omega \to \mathbb{R}$; for every $\omega \in \Omega$, $|X(\omega)|$ is some real number, and so there is an integer (depending on $\omega$) which is greater than it.  It follows that $\bigcap_{N \in \mathbb{N}} \{|X| \ge N\} = \emptyset$.  Now "continuity from above" (which follows from countable additivity) implies that $\lim_{N \to \infty} P(|X| \ge N) = 0$.  This is the key step you need.
