Why does adding $\sup$ render the following probability measures not equal I cannot wrap my head around why the following is not equivalent:
$\lim\limits_{n \to \infty} P(\sup\limits_{m \geq n}\vert X_{m}- X\vert > \epsilon )=0\iff \lim\limits_{n \to \infty} P(\vert X_{n}- X\vert > \epsilon )=0$
Obviously $\Rightarrow$ is trivial as $P(\sup\limits_{m \geq n}\vert X_{m}- X\vert > \epsilon )\geq P(\vert X_{n}- X\vert > \epsilon )$
and I cannot understand why the $\Leftarrow$ is not necessarily true. Let's say we get the following relationship from the fact that  $\lim\limits_{n \to \infty} P(\vert X_{n}- X\vert > \epsilon )=0$: say $P(\vert X_{n}- X\vert > \epsilon )\leq \frac{1}{n}(*)$ for any $n \in \mathbb N$. Surely this means that for any $m \geq n$ that $P(\vert X_{m}- X\vert > \epsilon )\leq \frac{1}{m}\leq \frac{1}{n}$ and hence $P(\sup\limits_{m \geq n}\vert X_{m}- X\vert > \epsilon )\leq \frac{1}{n}$
Fair enough, $(*)$ may not always hold, however, my intuition tells me that: when $n$ gets larger in terms of the RV's $(X_{n})_{n}$ and only a certain number (decreasing) of outcomes $\omega$ so that: $\vert X_{n}(\omega)- X(\omega)\vert > \epsilon$ is satisfied then surely the exact same would hold for $\sup\limits_{m \geq n}\vert X_{m}(\omega)- X(\omega)\vert > \epsilon$. Someone please explain why I am wrong
 A: Think of what the expressions mean in words.
The event $\{|X_n - X| \le \epsilon\}$ is demanding that $X_n$ is close to $X$. $P(|X_n - X| \le \epsilon) = 1 - o(1)$ means that for large $n$, the individual $X_n$ must each be close to $X$ with high probability.
But, the event $\{\sup_{m \ge n} |X_m - X| \le \epsilon\}$ is very different - it is demanding that all of the $X_m$ (for large $m$) are simultaneously close to $X$ - any one being different will screw things up. Formally, the event is the same as $ \bigcap_{m \ge n} \{ |X_m - X| \le \epsilon\}$. If we have many trials (countably infinite in this case), then even if any individual event is very likely, it can become quite probable that that at least one fails.

Here's a simple counterexample that may be instructive:
Pick any $\epsilon \in (0,1),$ let $X \equiv 0$ and consider the random variables $$X_n = \begin{cases} 0 & \textrm{with probability } 1 - 1/n \\ 1 & \textrm{with probability } 1/n\end{cases},$$ with all $X_n$ independent.
Then clearly $P(|X_n - X| \le \epsilon) = 1 - 1/n \to 1.$
But, for every $n$, $$P( \sup_{m \ge n} |X_n - X| \le \epsilon) = P(\forall m \ge n: X_m = 0) = \prod_{i = n}^\infty (1 - 1/i) = 0.$$
(I'll leave the proof of the last equality above to you.)
A: I give a counterexample: Let us consider the Lebesgue measure $\mu$ on $[0,1]$. This is a nice simple probability space. Let $X=0$ and 
$$X_n=\mathbb1_{[(n-2^m)1/2^{m},(n-2^m)1/2^{m}+1/2^{m}]},$$
where $m$ is the greatest integer such that $2^m\leq n$. Thus $X_1=1$, $X_2=\mathbb 1_{[0,1/2]}$, $X_3=\mathbb 1_{[1/2,1]}$, $X_4=\mathbb 1_{[0,1/4]}$, etc. Clearly $\lim_{n\to\infty}\mu(|X_n-X|>1/2)=0$, but we have that $\lim_{n\to\infty}\mu(x:\sup_{k\geq n}|X_n(x)-X(x)|>1/2)=1$. 
To see this, note that for every $n$ and every $x$ there exists a $k\geq n$ such that $X_k(x)=1$, because all we are doing is basically, at every step $2^m$ dividing $[0,1]$ into $2^m$ slices, and looking at indicator functions of these successive slices for the next $2^m$ steps. These slices get thinner and thinner with successive powers of $2$, but you will always be able to find a thinner slice containing $x$, which is precisely what we are looking for when taking the supremum. However when we are not taking the supremum we are then just looking at the probability that $x$ is in the $n$-th slice, which decreases to $0$ as the slices get smaller as $n$ gets larger.
