Doubt about eqivalence .

Question

Regarding the second line of the first proof by @JDL in Altenate definitions of almost sure convergence

Are the following equivalent?

$P(\omega:\exists n\in\mathbf{N}, \forall m\in\mathbf{N}, \exists i>m \,\,\, \text{s.t.} \,\, |X_i(\omega) - X(\omega)| < 1/n)$

AND

$P(\omega:\forall n\in\mathbf{N}, \exists m\in\mathbf{N},\ \text{s.t.}\ \forall i>m \,\, |X_i(\omega) - X(\omega)| \ge 1/n)$

Answer 1

The original question asks to prove the equivalence: $$P[\omega:\lim_{n\to\infty}X_n(\omega) = X(\omega)] = 1\iff\lim_{n\to\infty}P[\omega:\sup_{k>n}|X_k(\omega) - X(\omega)|>\epsilon] = 0$$

Adding the missing "for every positive $\epsilon$" on the RHS and correcting the notations on both sides, one should show the equivalence of the properties:

• $P(A)=1$ where $A=[\lim\limits_{n\to\infty}X_n=X]$
• $\lim\limits_{n\to\infty}P(A_n^\epsilon)=1$ for every positive $\epsilon$, where $A_n^\epsilon=[Y_n\leqslant\epsilon]$ and $Y_n=\sup\limits_{k>n}|X_k-X|$

To show this equivalence, first note that the sequence of random variables $(Y_n)$ is nonincreasing, hence $$A=\bigcap_{\epsilon>0}[\lim\limits_{n\to\infty}Y_n\leqslant\epsilon]=\bigcap_{\epsilon>0}[\exists n,Y_n\leqslant\epsilon]=\bigcap_{\epsilon>0}\bigcup_nA_n^\epsilon$$ Now, for every positive $\epsilon$, the sequence of events $(A_n^\epsilon)$ is nondecreasing, hence $$P(A)=\inf_{\epsilon>0}P\left(\bigcup_nA_n^\epsilon\right)=\inf_{\epsilon>0}\left(\lim_{n\to\infty}P(A_n^\epsilon)\right)$$ which proves the desired equivalence. Note that the last identity above nowhere uses the hypothesis that $P(A)=1$, thus, for every $p$ in $[0,1]$, $$P(A)\geqslant p\iff\forall\epsilon>0,\lim\limits_{n\to\infty}P(A_n^\epsilon)\geqslant p$$

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Nota bene: The step $$\bigcap_{\epsilon>0}[\lim\limits_{n\to\infty}Y_n\leqslant\epsilon]=\bigcap_{\epsilon>0}[\exists n,Y_n\leqslant\epsilon]$$ is trickier than may appear at first sight, since the identity $$[\lim\limits_{n\to\infty}Y_n\leqslant\epsilon]=[\exists n,Y_n\leqslant\epsilon]$$ is false for some fixed $\epsilon$ (consider $Y_n=\epsilon+\frac1n$). But the following inclusions save the day: $$[\lim\limits_{n\to\infty}Y_n\leqslant\epsilon]\subseteq[\exists n,Y_n\leqslant2\epsilon]\subseteq[\lim\limits_{n\to\infty}Y_n\leqslant2\epsilon]$$ since indeed they imply that $$\bigcap_{\epsilon>0}[\lim\limits_{n\to\infty}Y_n\leqslant\epsilon]\subseteq\bigcap_{\epsilon>0}[\exists n,Y_n\leqslant2\epsilon]\subseteq\bigcap_{\epsilon>0}[\lim\limits_{n\to\infty}Y_n\leqslant2\epsilon]=\bigcap_{\epsilon>0}[\lim\limits_{n\to\infty}Y_n\leqslant\epsilon]$$ Another option is to use strict inequalities, since $$[\lim\limits_{n\to\infty}Y_n<\epsilon]=[\exists n,Y_n<\epsilon]$$