Counter-example of Subsequence Criterion?

The last argument shows that if $$X_n\to X_\infty$$ a.s. and $$N(n)\to\infty$$ a.s., then $$X_{N(n)}\to X_\infty$$. We have written this out with care because the analogous result for convergence in probability is false. If $$X_n\in\{0,1\}$$ are independent with $$P(X=1)=a_n\to0$$ and $$\sum_n a_n=\infty$$, then $$X_n\to0$$ in probability, but if we let $$N(n)=\inf\{m\ge n; X_m=1\}$$ then $$X_{N(n)}=1$$ a.s.

It is from Durrett 5th Edition To be consistent to the Subsequence Criterion, N(n) should have further subsequence that converges to 0 almost surely. But it is not possible. So, what's wrong with my understanding to subsequence criterion?

The sequence $$(X_{N(n)})_{n=1}^\infty$$ is not a subsequence of $$(X_n)_{n=1}^\infty$$, which I believe is how you are reading it, since you refer to a "further" subsequence. $$(X_{N(n)})_{n=1}^\infty$$ is not a subsequence because the $$N(n)$$ are random variables. So $$X_{N(n)}$$ is a Frankenstein monster cobbled together from different terms of the sequence, depending on the various values $$N(n)$$ takes over the probability space.
With the understanding that $$(X_{N(n)})_{n=1}^\infty$$ is not a subsequence of $$(X_n)_{n=1}^\infty$$, but a sequence of Frankensteins, I think this should clear things up. Any subsequence of $$(X_n)_{n=1}^\infty$$ has a further subsequence which converges to $$0$$ a.s., but since $$(X_{N(n)})_{n=1}^\infty$$ is not a subsequence, this does not apply to $$(X_{N(n)})_{n=1}^\infty$$. $$(X_{N(n)})_{n=1}^\infty$$ does not converge to zero in probability or a.s.
• Of course, $X_{N(n)}$ does converge to 0 a.s. under the stated hypotheses. – ofer zeitouni Mar 22 '19 at 18:46
• @ofer "Of course, $X_{N(n)}$ does converge to $0$ a.s. under the stated hypotheses." The specific choice of $X_{N(n)}$ from the end of the text is $1$ a.s, and does not converge to $0$ a.s. – bangs Mar 24 '19 at 13:11