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It is known that

A sequence $(X_n)$ of random variables converges in probability to $X$ if and only if every subsequence $(X_{f(n)})$ has a sub-subsequence ($X_{g(f(n))}$) that converges almost surely to $X$.

Now, assume that for all $\omega\in\Omega$, every subsequence $(X_{f(n)}(\omega))$ has a sub-subsequence ($X_{g(f(n))}(\omega)$) that converges to $X(\omega)$. This should be different from the assertion in the second "if" of the result above, and it should imply that $(X_n)$ converges almost surely to $X$ because $\lim_n X_n(\omega)=X(\omega)$ for each $\omega$. Am I right or doing some mistake?

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  • $\begingroup$ Yes, It's different. In your case $X_n\to X$ everywhere. $\endgroup$ – d.k.o. Apr 28 '18 at 13:46

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