Garrett Black

2022-07-01

Let ${I}_{1},{I}_{2}...$ be an arrangement of the intervals $\left[\frac{i-1}{{2}^{n}},\frac{i}{{2}^{n}}\right)$ in a sequence. If ${X}_{n}=n$ on ${I}_{n}$ and 0 elsewhere then $su{p}_{n}E{X}_{n}<\mathrm{\infty }$ but $P\left(\underset{n}{sup}{X}_{n}<\mathrm{\infty }\right)=0$. My basic space is [0,1] with Lebesgue measure.
$A:=\left\{\underset{n}{sup}{X}_{n}<\mathrm{\infty }\right\}=\left\{\omega \in \mathrm{\Omega }:\mathrm{\exists }M>0,\mathrm{\forall }n\in \mathbb{N},{X}_{n}\left(\omega \right)\le M\right\}$
hence the complement is
$\begin{array}{rl}\overline{A}& =\left\{\omega \in \mathrm{\Omega }:\mathrm{\forall }M\in \mathbb{N},\mathrm{\exists }n\in \mathbb{N},{X}_{n}\left(\omega \right)>M\right\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\in \mathbb{N}}\left\{{X}_{n}>M\right\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\ge M}\left\{{X}_{n}>M\right\}\end{array}$
How do i go on from here?

nuvolor8

Expert

The example provided there is as follows:
${X}_{n}\left(\omega \right)=\left\{\begin{array}{ll}m& \omega \in \left[\frac{k}{{2}^{m}},\frac{k+1}{{2}^{m}}\right]\\ 0& \text{otherwise}\end{array}$
where $m=⌊\mathrm{log}\left(n\right)⌋$ and $k=n-{2}^{m}$.
To prove $\underset{n}{sup}{X}_{n}=\mathrm{\infty }$ almost surely, consider a single point $\omega \in \left[0,1\right]$. You can easily verify that ${X}_{n}\left(\omega \right)$ 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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