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Suppose you have a doubly infinite sequence of random variables [itex]X_{i,n}[/itex] that obey the following almost sure convergence relations. For each [itex]i = 1,2,3,...[/itex],

[tex]X_{i,n} \xrightarrow{a.s} a_i \quad \mbox{ as } \quad n\xrightarrow{} \infty[/tex].

Further, we have that [itex]\sum_{i=1}^\infty a_i = \mu < \infty[/itex]. Since this series converges, for any [itex]\delta > 0[/itex], there is some smallest [itex]I[/itex] such that [itex]\left| \sum_{i=1}^m a_i - \mu \right | < \delta[/itex] for all [itex]m \geq I[/itex]. Consider a sequence of deltas decreasing to zero, and the increasing sequence of their corresponding [itex]I[/itex]'s.

[tex]\delta_1 > \delta_2 > ... \xrightarrow{} 0 \quad \mbox{and} \quad I_1 < I_2 < ....[/tex]

Consider some particular pair [itex](\delta_j, I_j)[/itex]. Since almost sure convergence is linear,

[tex]\sum_{i=1}^{I_j}X_{i,n} \xrightarrow{a.s} \sum_{i=1}^{I_j} a_i \quad \mbox{ as } \quad n\xrightarrow{} \infty [/tex]

This is the same thing as saying the set

[tex] \{ \omega: \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \sum_{i=1}^{I_j} a_i \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \} [/tex]

has probability zero for any choice of [itex]\epsilon > 0[/itex]. From the definition of the deltas and I's, the set

[tex] A_j = \{ \omega: \left| \sum_{i=1}^{I_j}X_{i,n}(\omega) - \mu \right| > \epsilon + \delta_j \quad i.o. \quad n\xrightarrow{} \infty \} [/tex]

also probability zero.

**My question is,**does the sequence of sets [itex]A_j[/itex] have a limit of

[tex]A = \{ \omega: \left| \sum_{i=1}^{\infty}X_{i,n}(\omega) - \mu \right| > \epsilon \quad i.o. \quad n\xrightarrow{} \infty \}[/tex]

Thanks for wading through that!