Why does P-a-s Convergence Equal Limm→∞ P({Supn≥m|Xn-X| ≥ε })?

[stukbv]

So I have a definition;
X_n n=1,2... is a sequence of random variables on ( Ω,F,P) a probability space, and let X be another random variable.
We say X_n converges to X almost surely (P-a-s) iff P({lim_{n →∞} X_n=X}^C) = 0

It then goes on to say that checking this is the same as checking
lim_{m →∞} P({Sup_n≥m|X_n-X| ≥ε }) = 0
Can somebody please explain why this is true, I don't understand at all how to get from one to the other properly.

Thanks!

 

[viraltux]

They're just two ways to express the same concept; that you can get X_n as close to X as you want with an n sufficiently large.

 

[Stephen Tashi]

It then goes on to say that checking this is the same as checking
lim_{m →∞} P({Sup_n≥m|X_n-X| ≥ε }) = 0

"It" isn't being very precise. I suppose mathematical tradition tells us that \epsilon is a number greater than zero. Tradition also tells us that the quantifier associated with \epsilon is "for each", so that's a hint about what it means. The notation it is using for sets is very abbreviated. The usual notation would tell us that a set is "the set of all... such that ...".

If you want to understand the assertions precisely, the first thing you must do is to understand precisely what they assert. I don't know if that is your goal. If it is, try to write out exactly what each of those statements claims using better notation.