TITLE: Filtrations and Sigma-Algebras and Stopping Times
QUESTION [7 upvotes]: 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

REPLY [2 votes]: 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.