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Cud someone please explain the proof of $ P(X_n \to X)=1 $ iff $$ \lim_{n \to \infty}P(\sup_{m \ge n} |X_m -X|>\epsilon) \to 0 $$. Im not able to understand the meaning of the various sets they take during the course of the proof.

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Intuitively, the result means that to converge almost everywhere is equivalent to "bound the probability of the $\omega$'s for which $|X_n-X|$ is infinitely often larger than a positive number". Here is a more formal argument.

Assume that $X_n\to X$ almost surely and fix $\varepsilon\gt 0$. Define $A_m:=\{|X_m-X|\gt \varepsilon\}$ and $B_n:=\bigcup_{m\geqslant n}A_m$. The sequence $(B_n)$ is non-increasing and $\bigcap_{n\geqslant 1}B_n$ is the set of $\omega$'s for which $\omega\notin A_m$ for infinitely many $m$'s. Since this set is contained in $\{\omega,X_n(\omega)\mbox{ doesn't converge to }X(\omega)\}$, as set of measure $0$, we are done.

Conversely, assume that for each $\varepsilon\gt 0$, $\mathbb P(\sup_{m\geqslant n}|X_m-X|\gt \varepsilon)\to 0$. Take $\varepsilon=2^{-k}$ for a fixed $k$, and $n_k$ such that $\mathbb P(\sup_{m\geqslant n_k}|X_m-X|\gt 2^{-k})\leqslant 2^{-k}$ (we can assume $(n_k)_k$ increasing). Then by Borel-Cantelli's lemma, $\mathbb P(\limsup_k\{\sup_{m\geqslant n_k}|X_m-X|\gt 2^{-k}\})=0$. This means that there exists $\Omega'\subset\Omega$ of probability $1$ for which given $\omega\in\Omega'$, there is $k(\omega)$ such that $\sup_{m\geqslant n_k}|X_m-X|\leqslant 2^{-k}$ for $k\geqslant k(\omega)$.

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