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How do I conceptualise this expression : Let {$A_n$}$^{n=\infty}_{n=1}$ belong to sigma algebra $A$. Define, $\limsup\{A_{n}\}=\bigcap_{n=1}^{\infty}\{\bigcup_{m=n}^{\infty}A_{n}\} $ and similarly $\liminf$ with intersection and union exchanged. I know what it means to say, like the first expression means to say is that $\limsup A_n$ is the set of points which are in infinitely many $A_n$ and similarly $\liminf A_n$ is the set of points which fails to be in at most finitely many $A_n$ . How do I see this explanation from the expression.

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Let us describe it for limsup: as Michael has mentioned $$ A = \limsup A_n = \bigcap\limits_{n=0}^\infty\bigcup\limits_{k=n}^\infty A_k. $$ Now, let us focus on the two-layer nature of limsup and consider the deepest layer: denote $$ B_n = \bigcup\limits_{k=n}^\infty A_k. $$ Note that $B_n$ contains points from all $A_k$ starting from $k = n$. Also, $B_{n+1} \subseteq B_n$ because in general $B_{n+1}$ contains "less" points: $A_n$ is not included in that union, only $A_{n+1}.$ So, $(B_n)_{n\geq0}$ is a non-increasing sequence and hence has a limit - namely and intersection of all $B_n$ $$ A =\limsup A_n = \bigcap_{n=0}^\infty B_n. $$

If $x$ appears as an element in the sequence of sets $(A_n)_{n\geq 0}$ infinitely often, then it will belong to all $B_n$'s and hence to their intersection $A$. Conversely, if $x$ is in $A$ then it is in all $B_n$'s hence for any $n\geq 0$ there is $k\geq n$ s.t. $x\in A_k$ - so $x$ appears as an element in the sequence $(A_n)_{n\geq 0}$ infinitely often.

The similar goes for the liminf where instead you have a non-decreasing sequence of sets. Please tell me, if you want me to clarify this case separately.

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well explained :) –  Theorem May 4 '12 at 12:45
    
@Vedananda: nice to know :) –  Ilya May 4 '12 at 12:45

An element $x$ is in infitely many of the $A_n$ if for each $n$, there is an $m\geq n$ such that $x\in A_m$.

Formally: $$\forall n\exists m: m\geq n:x\in A_m.$$ We can rewrite this as $$\forall n: x\in\{y\in A_m:\text{ for some } m\geq n\}$$ $$\forall n: x\in\bigcup_{m=n}^\infty A_m$$ $$x\in\{y\in\bigcup_{m=n}^\infty:\text{ for all }n\}$$ $$x\in\bigcap_{n=0}^\infty\bigcup_{m=n}^\infty A_m.$$

Quite generally, one can translate infinite intersections to "for all"-statements and infinite unions to "there exists"-statements. This way, one can learn to interpret such as expression more easily.

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$\bigcup_{m=n}^\infty A_m$ is the set for which $A_m$ occurs for some $m \ge n$. Taking the intersection of these over $n$ gives you the points such that for arbitrarily large $M \ge 0 $, we can find an $m\ge M$ such that $A_m$ that occurs. You can check that this equivalent to $A_n$ occurring infinitely often.

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