We wish to show $\lim_{n \rightarrow \infty} X_n = X$.
Does that follow from $\limsup_{n \rightarrow \infty} |X_n - X| = 0$, and if so, why?
$X_n$and $X$ are RVs.
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I don't know what RVs are, but surely this condition implies the limsups of both $Y - X_n$ and $X_n - Y$ are zero, and hence both the limsup and liminf of $X_n - Y$ is zero too. – Theo Bendit Oct 11 '17 at 01:02
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Assuming that $X$ is finite, this has nothing to do with probability. It follows from the equivalent $$ \lim_{n\to\infty} a_n = \alpha \in \mathbb{R} \quad \Leftrightarrow \quad \limsup_{n\to\infty} |a_n - \alpha| = 0,$$ which is quite straightforward to prove from standard $\epsilon$-$N$ definition. – Sangchul Lee Oct 11 '17 at 01:08