# Filtrations and Sigma-Algebras and Stopping Times

In a previous post Filtrations and Sigma-Algebras I asked the question:

$\textbf{Previous Question:}$ Let $\Omega=\{1,2,3\}, \mathcal{A}=\mathcal{P}(\Omega)$ and $P(\{\omega\})=\tfrac{1}{3}$ for each $\omega \in \Omega$. Define a stochastic process $(X(t):t\ge 0)$ by $X(t)(\omega) = \max\{t-\omega,0\}$.Then the filtration generated by the stochastic process $X$ computes as \begin{align} \mathcal{F} = \begin{cases} \{0,\Omega\}, \qquad \qquad \qquad \text{if $t\in[0,1],$} \\ \{0,\Omega,\{1\},\{2,3\}, \phantom{xx}\text{if $t \in (1,2]$,}\\ \mathcal{P}(\Omega), \qquad \qquad \qquad \phantom{.}\text{if $t>2$.} \end{cases} \end{align}

The user V.C. explained most comprehensively how the filtration was obtained for which I am most grateful. \begin{align} \end{align} I would now like to extend the question to $\textbf{Stopping Times:}$

$\textbf{Extension:}$ In light of the previous question define \begin{align} \tau : \Omega \rightarrow [0,\infty), \quad \tau(\omega) := \inf\{t \ge 0: X(t)(\omega) > 0\} \end{align} Then it may be seen that $\tau$ is not a stopping time as $\{\tau \le 1\} = \{1\}$ but $\{1\} \not \in \mathcal{F}_1$. \begin{align} \end{align} $\textbf{Question:}$ Could someone explain what $\{\tau \le 1\} = \{1\}$ represents please?

I can see that the stochastic process $X_t(\omega) = t-1$ when $\omega = \{1\}$ and $t\in(1,2]$ and thus $0 < t-1 \le 1$.

But I'm confused because in the definition of $\tau$ it is stated being a function of $\omega$, i.e. $\tau(\omega)$, thus I'm not sure what $\{\tau \le 1\}$ is supposed to represent.

All help is appreciated.

Many thanks,

John

I believe that $\{\tau\leq 1\}$ is a shorthand for \begin{align*} \{\omega\in\Omega\,|\,\tau(\omega)\leq 1\}. \end{align*}
To see that this is consistent with the statement that $\{\tau\leq 1\}=\{1\}$, note that \begin{align*} \tau(1)=&\,\inf\{t\geq 0\,|\,\max\{t-1,0\}>0\}=\inf\{(1,\infty)\}=1,\\ \tau(2)=&\,\inf\{t\geq 0\,|\,\max\{t-2,0\}>0\}=2,\\ \tau(3)=&\,\inf\{t\geq 0\,|\,\max\{t-3,0\}>0\}=3. \end{align*}
In general, if $X$ is any space and $f:X\to\mathbb R$ is a real-valued function, then $$\{f\leq c\}\equiv f^{-1}((-\infty,c])\equiv\{x\in X\,|\,f(x)\leq c\}$$ for a given $c\in\mathbb R$. To be honest, I dislike this notation—you're living proof that it's a hotbed of confusion.