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.


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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