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Suppose $\{X_n\}$ is a sequence of random variables with $0<X_n\leq n$ and $X_n\rightarrow c$, almost surely. Form the definition of almost sure convergence, we obtain $$P(\cap_{n=1}^\infty\cup_{m=1}^\infty |X_{n+m}-c|\geq\epsilon)=0,\ for\ all\ \epsilon>0.$$ Thus, defining $A=\{\cup_{n=1}^\infty\cap_{m=1}^\infty |X_{n+m}-c|\leq\epsilon\}$, we know that $P(A)=1.$ On the set $A$, there exists a constant $N(\epsilon)$ such that for all $n\geq N(\epsilon)$, $$c-\epsilon\leq X_{n+m}\leq c+\epsilon.$$ For $n<N(\epsilon)$, we know that $$0<X_n\leq N(\epsilon),\ a.s.$$ Therefore, for any $\epsilon>0$, there exists a constant $K(\epsilon)$ such that $$P(\sup_n|X_n|\leq K(\epsilon))=1.$$

Is the above proof correct?

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What are you trying to prove? The fact that a sequence of random variables that converges almost surely is almost surely bounded is obvious: any sequence that converges is bounded. But your last statement appears to say that there is some uniform bound $K$ such that all $|X_n| \le K$ almost surely, and that is false. For example, consider $X_n$ defined on the probability space $[0,1]$ with Lebesgue measure, with $X_n(t) = n$ for $0 < t < 1/n$ and $0$ otherwise.

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  • $\begingroup$ Sorry for the confusion. I want to study the following question. For the above sequence $\{X_n\}$, for some constant $k$, which of the following statements are true:$$\lim_nP(X_n>k)\rightarrow0,$$ $$P(\lim_nX_n>k)=0,$$ $$\lim_nP(\sup_nX_n>k)\rightarrow0,$$ $$P(\lim_n\sup_nX_n>k)=0?$$ $\endgroup$ – SHAN Apr 3 '17 at 7:41

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