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I want to show that $\omega\in \limsup A_n\implies\sum_{n\geq 0}\Bbb{1}_{A_n}(\omega)=+\infty$ where $\{A_n\}_{n\geq 0}$ is a sequence of subsets of $\Omega$ and \begin{align} \Bbb{1}_{A_n}(\omega)=\begin{cases}1& \text{if}\;\omega\in A,\\0 &\text{if}\;\omega\not\in A\end{cases} \end{align}

HERE IS A PROOF

Assume that $\omega\in \limsup A_n$. The idea is to construct a subsequence $n_k,\;\forall k\in \Bbb{N}$ such that $\omega\in A_{n_k}$ for infinitely many $k.$ \begin{align} \omega\in \limsup A_n &\Leftrightarrow \omega\in\bigcap_{n\geq0}\left(\bigcup_{m\geq n}A_m\right)\\&=\forall n\in \Bbb{N},\,\omega\in\bigcup_{m\geq n}A_m=B_n\;\;\;(1)\end{align} So, for $n=0,\;\omega\in B_0=\bigcup_{m\geq 0}A_m.$ Then, there exists $n_0$ such that \begin{align} \omega\in A_{n_0}\subseteq \bigcup_{m\geq {n_0}}A_m=B_{n_0}.\end{align} Also, from $(1)$, $\omega\in B_{n_0+1}=\bigcup_{m\geq {n_0+1}}A_m.$ Then, there exists $n_1>n_0+1$ such that $\omega\in A_{n_1}.$ Again, from $(1)$, $\omega\in B_{n_1+1}=\bigcup_{m\geq {n_1+1}}A_m.$ Then, there exists $n_2>n_1+1$ such that $\omega\in A_{n_2}.$ $$\vdots$$ Continuing in the same approach, there exists $(n_k)_{k\in\Bbb{N}}$ such that $\omega\in A_{n_k},\forall k\in \Bbb{N}.$ This implies that $\omega$ belongs to an infinite number of $A_n$ and

\begin{align} \sum_{n\geq 0}\Bbb{1}_{A_n}(\omega)\geq \sum_{k\geq 0}\Bbb{1}_{A_{n_k}}(\omega)=+\infty\end{align}

MY QUESTION

I don't really get this last line \begin{align} \sum_{n\geq 0}\Bbb{1}_{A_n}(\omega)\geq \sum_{k\geq 0}\Bbb{1}_{A_{n_k}}(\omega)=+\infty\end{align} Any explanation for this, please? Alternative proofs are welcome!

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  • $\begingroup$ Since $(n_k)_{k\in\mathbb{N}}$ is a subsequence, the sum on the right contains less terms that the one on the right. Does that answer your question? $\endgroup$ – Ernie060 Sep 20 '18 at 20:13
  • $\begingroup$ @Ernie060: Why is it $=\infty?$ $\endgroup$ – Omojola Micheal Sep 20 '18 at 20:14
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    $\begingroup$ Since the subsequence is constructed such that $\omega \in A_{n_k}$ for infinitely many $k$. $\endgroup$ – Ernie060 Sep 20 '18 at 20:16
  • $\begingroup$ @Ernie060: Oh, I see! $\endgroup$ – Omojola Micheal Sep 20 '18 at 20:17
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It is well-known that for a sequence of sets $A_n\subseteq \Omega$: $$\limsup_{n \to \infty} A_n = \{\omega \in \Omega: \{n: \omega \in A_n\} \text{ is infinite}\}$$ whereas $$\liminf_{n \to \infty} A_n = \{ \omega \in \Omega: \{n: \omega \in A_n\} \text{ has finite complement }\}$$

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