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.
Best Answer
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 $m \in \mathbb N$:
Conclude that $x \in \limsup A_n$.