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Let $(A_i)$ be a sequence of subsets of $\Omega$. Let $A = \bigcap_{n=0}^\infty \bigcup_{m=n}^\infty A_m.$ Prove that $A = \{\omega \in \Omega \mid \omega \in A_n\text{ for infinitely many $n$}\}$.

I kinda only understand the Union of An part, where it means sum of probability of the event $A_n$, $n$ element of natural numbers..

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    Please don't make us follow links around and **translate**! If you need help marking up your equations, copy from [this site](http://codecogs.com/latex/eqneditor.php) and wrap your equations in dollar signs (\$)2012-04-23

3 Answers 3

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If $\omega_1 \in \Omega$ has the property $\omega_1 \in A_n\text{ for infinitely many $n$}$ then $\omega_1 \in \bigcup_{m=N}^\infty A_m$ for all $N$ and so $\omega_1 \in \bigcap_{N=0}^\infty \bigcup_{m=N}^\infty A_m = A.$

If $\omega_2 \in \Omega$ has the property $\omega_2 \in A_n\text{ for finitely many $n$ (possibly zero)}$ then there exists $N_2$ such that $\omega_2 \not \in \bigcup_{m=N}^\infty A_m$ for all $N \ge N_2$ and so $\omega_2 \not \in \bigcap_{N=0}^\infty \bigcup_{m=N}^\infty A_m = A.$

So $A = \{\omega \in \Omega \mid \omega \in A_n\text{ for infinitely many $n$}\}.$

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Denote for $n \in \mathbb N$ by $B_n$ the set $\bigcup_{m=n}^\infty A_m$. Let us first think about what it means for an $\omega\in \Omega$ to be an element of $B_n$. $\omega$ is an element of the union iff it is an element of one of the sets $A_m$, $m\ge n$. So $\omega \in B_n \iff \exists m\ge n\, \omega \in A_m$. Now $\omega \in A = \bigcap_{n=0}^\infty B_n$ iff $\omega$ is an element of all sets $B_n$, that is \[ \omega \in A \iff \forall n\, \omega \in B_n \iff \forall n\,\exists m \ge n\,\omega \in A_m. \] So given $\omega \in A$ there is a $n_1 \ge 1$ such that $\omega \in A_{n_1}$, and a $n_2\ge n_1 + 1$ with $\omega \in A_{n_2}$, and so on (inductively). So there are infinitely many $n$ with $\omega \in A_n$. If on the other hand $\omega \in A_n$ infinitely often, then given any $n$ we have $\omega \in B_n$ as there are only finitely many $n'\in \mathbb N$ with n' < n so $\omega$ has to be contained in some $A_m$, $m \ge n$. Hence $\omega \in B_n$ and as $n$ was arbitrary, $\omega \in A$.

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Essentially you are to prove that the two sets are equal. To prove that two sets are the same is to prove that every element in one set is in the other and vice versa.

If $x\in A$, then for every $n$: $x\in\cup_{m=n}^{\infty}A_m$, i.e. for every $n$ there exists some $m\ge n$ such that $x\in A_m$. Now, for $n=1$, this means that there is $m\ge 1$ such that $x\in A_m$, hence, $x$ is contained in at least one $A_m$. If there were only finite number of $A_m$'s containing $x$, then there were the maximum index $m=M$ such that $x\in A_m$. But then for $n$ equal to $M+1$ there is another $m\ge M+1>M$ such that $x\in A_m$, and $M$ is not the maximum index. Contradiction. Hence, there are infinitely many sets containing $x$.

If $x$ is contained in infinitely many sets, then for every $n$ there is some $m\ge n$ such that $x\in A_m$. Hence, for every $n$: $x\in\cup_{m=n}^{\infty}A_{m}$, and $x\in A$.