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I'm working on the problem: $\liminf\limits_{n\to\infty} A_{n} \subset \limsup \limits_{n\to\infty} A_{n}$ where $\left( A_{n}\right) _{n\geq 1}$ a sequence of events in the $\sigma$-algebra $\mathcal{A}$.

I 'think' I understand the definitions, \begin{align*} \omega\in \limsup \limits_{n\to\infty} A_{n} &\Leftrightarrow \omega \in \bigcap\limits_{n=1}^\infty\bigcup\limits_{k=n}^\infty A_k\\ &\Leftrightarrow \forall n\exists k\geq n: \omega\in A_{k}\\ &\Leftrightarrow \omega\in A_{n} \text{ for infinitely many n,}\\ \omega\in \liminf \limits_{n\to\infty} A_{n} &\Leftrightarrow \omega \in \bigcup\limits_{n=1}^\infty\bigcap\limits_{k=n}^\infty A_k\\ &\Leftrightarrow \exists n\forall k\geq n: \omega\in A_{k}\\ &\Leftrightarrow \omega\in A_{n} \text{ for all but finitely many n.} \end{align*}

By definitions, the problem can be solved easily. But I want to proceed like proving deMorgan's Law, by two ways inclusion. I tried in this way:

\begin{align*} &\text{Let } \omega \in \bigcup\limits_{n=1}^\infty\bigcap\limits_{k=n}^\infty A_k.\\ &\exists \text{ at least one } n, \text{ denoted by } \hat{n},\ \omega\in\bigcap\limits_{k=\hat{n}}^\infty A_k.\quad \Leftrightarrow \omega\in A_{k}\ \forall k\geq \hat{n}.\\ &\omega\in \bigcup\limits_{k=\hat{n}}^\infty A_k. \end{align*}

I CANNOT conclude $\omega\in\bigcap\limits_{n=1}^\infty\bigcup\limits_{k=n}^\infty A_k$ since $\bigcap\limits_{n=1}^\infty\bigcup\limits_{k=n}^\infty A_k \subset \bigcup\limits_{k=\hat{n}}^\infty A_k$.

What goes wrong? Thanks.

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    $\begingroup$ $\omega \in \bigcap\limits_{k= \hat{n}}^\infty A_k \Rightarrow \omega \in \bigcup\limits_{i=m}^\infty A_i$ for all $m$. $\endgroup$ – Daniel Fischer Sep 25 '13 at 17:30
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We want to show that $$ \bigcup_{n = 1}^\infty \bigcap_{k = n}^\infty A_k = \liminf A_n \subseteq \limsup A_n = \bigcap_{n = 1}^\infty \bigcup_{k = n}^\infty A_k.$$

Let $x \in \liminf A_n$, then there exists $N \in \mathbb N$ such that $x \in \displaystyle \bigcap_{k = N}^\infty A_k$.

Observe that $x \in \limsup A_n$ if and only if $x \in \displaystyle \bigcup_{k = m}^\infty A_k$ for all $m \in \mathbb N$.

So let $m \in \mathbb N$:

  • If $m \geq N$, then $x \in A_m$ since $x \in \displaystyle \bigcap_{k = N}^\infty A_k \subseteq A_m$. Hence $x \in \displaystyle \bigcup_{k = m}^\infty A_k$.
  • If $m < N$, then $x \in A_N$ since $x \in \displaystyle \bigcap_{k = N}^\infty A_k$ and $A_N \subseteq \displaystyle\bigcup_{k = m}^\infty A_m$. Hence $x \in \displaystyle \bigcup_{k = m}^\infty A_k$.

Conclude that $x \in \limsup A_n$.

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