Assume a probability space $\left(\Omega, \, F, \, P \right)$ with a filtration $\left(F_t \right)_{t\geq 0}$. I am dealing with a discrete time case, that is $t = \left\{1, \, 2, \, 3 , \, \cdots \right\}$.
$\tau: \Omega \rightarrow \mathbb{R}$ is a stopping time if \begin{align} \left\{w\in \Omega \, \, | \, \, \tau\left(w \right) \leq t \right\}\in F_t \end{align}
So we already know that $\mathbb{1}_{\left\{\tau\left(w \right)\leq t \right\}}$ is $F_t$- measurable, where $\mathbb{1}_{\left\{. \right\}}$ is an indicator function.
But what about $\mathbb{1}_{\left\{\tau\left(w \right)\geq t \right\}}$? In my opinion, $\mathbb{1}_{\left\{\tau\left(w \right)\geq t \right\}}$ is $F_{t-1}- measurable$. I can say that $\left\{\tau(w) \geq t\right\} = \left\{\tau(w) \leq t-1 \right\}^{c} \in F_{t-1}$, where $c$ designates "complement".
Is what I am writing correct? If yes, how can I prove that $\left\{\tau(w) \leq t-1 \right\}^{c} \in F_{t-1}$ ?
Any help will be very appreciated!
The other answer has some superfluous (and slightly incorrect) information. (For example, the definition of a $\sigma$-algebra $\mathcal F$ is incorrect, since $\sigma$-algebras are closed under countable unions, not just finite unions.)
To answer the question regarding the measurability of $\mathbf 1_{\{\omega \in \Omega \colon \tau(\omega) \geq t\}}$, note that
\begin{align} \{\omega \in \Omega \colon \tau(\omega) \geq t\} = \{ \omega \in \Omega \colon \tau(\omega) \leq t - 1\}^{\complement} \in \mathcal F_{t - 1} \end{align}
by the definition of a $\sigma$-algebra. (To be clear, we know from the assumption that $\tau$ is a stopping time that $\{ \omega \in \Omega \colon \tau(\omega) \leq t - 1\} \in \mathcal F_{t - 1}$, which implies that the complement is in $\mathcal F_{t - 1}$ as well.)
Thus $\mathbf 1_{\{\omega \in \Omega \colon \tau(\omega) \geq t\}} \colon \Omega \to \mathbb R$ is $\mathcal F_{t - 1}$-$\mathcal B(\mathbb R)$-measurable.