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Let $(X, \mathcal A, \mathcal A_n, \mu)$ be a $\sigma$-finite filtered measure space and let $X_n$, $n\in\mathbb N$ be a sequence of $\mathcal A_n$ measurable real valued functions. For a fixed Borel set $B\subset\mathbb R$ define $$ T(x)=\inf\{n\in\mathbb N:X_n(x)\in B\}, $$ and $T(x)=+\infty$ if $\{n\in\mathbb N:X_n(x)\in B\}=\emptyset$. Show that $T$ is a stopping time.

Define next $$ L(x)=0\lor\sup\{n\leq 5:X_n(x)\in B\}. $$ Show that $L$ is not a stopping time.

My attempt. Intuitively, I know that, using the definition A random variable $T$ is a stopping time (w.r.t. a filtration $\{\mathcal F_n\}$) if $\{T\leq n\}$ for every $n$ (in other words $\{T\leq n\}\in \mathcal F_n$ means asking if, from the information we have at time $n$, we know whether $T$ is less or equal to $n$ or not), $T$ is a stopping time and $L$ is not. But I don't know how to write it down formally. Can someone help me? Thank you

Redeldio
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1 Answers1

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A key thing here is that $\sup \emptyset = -\infty$.

First, $\{T> n\}=\bigcap_{k\leq n}\{X_k \notin B\}\in \mathscr{F}_n,\,\forall n$ so $\{T\leq n\}=\{T>n\}^c \in \mathscr{F}_n,\,\forall n$.

Second, consider the measurable space $([0,1],\mathscr{B}([0,1]))$, suppose $B=\{1\}$, $X_1=...X_4=X_6=...=0$ and $X_5=\mathbf{1}_{[1/2,1]}$. Let $\mathscr{F}_n=\sigma(X_k,k\leq n)$. Then $$\{L>1\}=\{L=5\}=\{X_5=1\}=[1/2,1]\notin \mathscr{F}_1=\{\emptyset,[0,1]\}$$

Snoop
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