Let $X$ be a measurable space with $\sigma$-algebra $\Sigma$. Let $A_{1},A_{2},\dots$ be a countably infinite sequence of sets in $\Sigma$. Write $A_{i.o.}$ for the set of points $x \in X$ in infinitely many of the $A_{n}$s. Show that $A_{i.o.} \in \Sigma$. I'm unsure how to do this, so any help would be greatly appreciated.
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3Hint: $x$ is in infinitely many of the $A_n$'s means $x\in \bigcap\limits^\infty_{k=1}\bigcup\limits^\infty_{j=k}A_j$. – Oct 09 '16 at 18:17
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3This set is usually called the "limit superior" of $A_j$, and in fact denoted by $\limsup A_j$ (by analogy with the case of real sequences, for example: $\limsup a_j=\inf_{k\geq 1}\sup_{j\geq k} a_j$ - supremumss become unions and infimums become intersections). – Oct 09 '16 at 18:25
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To make @Bye_World's comment an answer:
To say that a point $x$ is in infinitely many of the $A_n$s is the same as saying that $x \in \bigcap_{k=1}^{\infty}\bigcup_{j=k}^{\infty} A_j$. See here for a discussion of why this is so. Hence $A_{i.o.} = \bigcap_{k=1}^{\infty}\bigcup_{k=1}^{\infty} A_j$, and since a $\sigma$-algebra is closed under countable unions and intersections, it follows that $\bigcap_{k=1}^{\infty}\bigcup_{j=k}^{\infty} A_j \in \Sigma$, so $A_{i.o.} \in \Sigma$.
