Garrett Black

Answered

2022-07-01

Let ${I}_{1},{I}_{2}...$ be an arrangement of the intervals $[\frac{i-1}{{2}^{n}},\frac{i}{{2}^{n}})$ in a sequence. If ${X}_{n}=n$ on ${I}_{n}$ and 0 elsewhere then $su{p}_{n}E{X}_{n}<\mathrm{\infty}$ but $P(\underset{n}{sup}{X}_{n}<\mathrm{\infty})=0$. My basic space is [0,1] with Lebesgue measure.

$A:=\{\underset{n}{sup}{X}_{n}<\mathrm{\infty}\}=\{\omega \in \mathrm{\Omega}:\mathrm{\exists}M>0,\mathrm{\forall}n\in \mathbb{N},{X}_{n}(\omega )\le M\}$

hence the complement is

$\begin{array}{rl}\overline{A}& =\{\omega \in \mathrm{\Omega}:\mathrm{\forall}M\in \mathbb{N},\mathrm{\exists}n\in \mathbb{N},{X}_{n}(\omega )>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\in \mathbb{N}}\{{X}_{n}>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\ge M}\{{X}_{n}>M\}\end{array}$

How do i go on from here?

$A:=\{\underset{n}{sup}{X}_{n}<\mathrm{\infty}\}=\{\omega \in \mathrm{\Omega}:\mathrm{\exists}M>0,\mathrm{\forall}n\in \mathbb{N},{X}_{n}(\omega )\le M\}$

hence the complement is

$\begin{array}{rl}\overline{A}& =\{\omega \in \mathrm{\Omega}:\mathrm{\forall}M\in \mathbb{N},\mathrm{\exists}n\in \mathbb{N},{X}_{n}(\omega )>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\in \mathbb{N}}\{{X}_{n}>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\ge M}\{{X}_{n}>M\}\end{array}$

How do i go on from here?

Answer & Explanation

nuvolor8

Expert

2022-07-02Added 32 answers

The example provided there is as follows:

${X}_{n}(\omega )=\{\begin{array}{ll}m& \omega \in [{\displaystyle \frac{k}{{2}^{m}}},{\displaystyle \frac{k+1}{{2}^{m}}}]\\ 0& \text{otherwise}\end{array}$

where $m=\lfloor \mathrm{log}(n)\rfloor $ and $k=n-{2}^{m}$.

To prove $\underset{n}{sup}{X}_{n}=\mathrm{\infty}$ almost surely, consider a single point $\omega \in [0,1]$. You can easily verify that ${X}_{n}(\omega )$ will go to infinity. Therefore, the set of $\omega $ such that $\underset{n}{sup}{X}_{n}=\mathrm{\infty}$ is [0,1]. This simple example really does not need more advanced technique like Borel–Cantelli lemma.

${X}_{n}(\omega )=\{\begin{array}{ll}m& \omega \in [{\displaystyle \frac{k}{{2}^{m}}},{\displaystyle \frac{k+1}{{2}^{m}}}]\\ 0& \text{otherwise}\end{array}$

where $m=\lfloor \mathrm{log}(n)\rfloor $ and $k=n-{2}^{m}$.

To prove $\underset{n}{sup}{X}_{n}=\mathrm{\infty}$ almost surely, consider a single point $\omega \in [0,1]$. You can easily verify that ${X}_{n}(\omega )$ will go to infinity. Therefore, the set of $\omega $ such that $\underset{n}{sup}{X}_{n}=\mathrm{\infty}$ is [0,1]. This simple example really does not need more advanced technique like Borel–Cantelli lemma.

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