If $\sigma _n\nearrow \infty $ a.s. does there is $n\in\mathbb N$ s.t. $\sigma _n>t$ a.s.?

Question

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(\sigma _n)$ a sequence of stopping time s.t. $\sigma _n\nearrow \infty $ a.s. Fix $t>0$. Does there exist $n\in\mathbb N$ s.t. $\sigma _n>t$ a.s. ?

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I don't think so, for me, $\sigma _n\nearrow \infty $ a.s. means there is a null set $N$ s.t. for all $\omega \in \Omega \setminus N$, $\sigma _n(\omega )\nearrow \infty $. Therefore, for all $\omega \in \Omega \setminus N$ there is $N(\omega )$ s.t. $\sigma _{N(\omega )}(\omega )>t$. So, if $\sup_{\omega \in \Omega }N(\omega )<\infty $, then, we can take $$m\geq \sup_{\omega \in \Omega }N(\omega ),$$ and thus $\sigma _m>t$ a.s., but I don't see any reason for $\sup_{\omega \in \Omega }N(\omega )<\infty $, so a priori, for the existence of $n$ s.t. $\sigma _n>t$ a.s. is not correct.

Question

1) What do you think ?

2) Can someone provide a counter example ?

Answer 1

You're right.

Let $(\Omega,\mathcal{F},\mathbb{P})=((0,1],\mathcal{B},\lambda)$ with the latter denoting the Borel algebra and the Lesbegue measure respectively. Define $X_n(x)=\frac{n}{x}$ and let $\sigma_k$ be the first hitting time of $[k,\infty)$. Then $\sigma_k(x)=\lceil kx\rceil,$ which goes to $\infty$ for every $x$. However, for any $k$ and any $t$, we have $\sigma_k(x)\geq t$ implies $kx+1\geq t$ and hence, $x\geq\frac{t-1}{k}$, which implies that $(\sigma_k\geq t)\subseteq [\frac{t-1}{k},1],$ which never has full measure for $t>1$.