# Approximating stopping times from the right (related to Bass exercise 3.5)

I am trying to show something that I found in Bass' book on stochastic processes that would give me an alternative way to solve a previous question of mine: Cadlag and adapted imply progressively measurable (related to Protter theorem 6)

Given a stopping time $T:\Omega \rightarrow [0,\infty)$ for a filtration $(\mathcal{F}_t)_{t\geq0}$, define $T_n(\omega) = \frac{k+1}{2^n}$ where $k$ is the unique integer such that $\frac{k}{2^n} \leq T(\omega) < \frac{k+1}{2^n}$. Show that $T_n$ is a stopping time for $(\mathcal{F}_t)_{t\geq0}$. Again this is pretty much exercise 3.5 in Bass'book on stochastic processes.

I'm pretty much stuck. We want to show that for $t \geq 0$, $T_n^{-1}([0,t]) \in \mathcal{F}_t$ and somehow have to relate to the fact that $T$ satisfies such a condition.

Could someone give me a hint? I think just an illustrative hint would be nice so that I could get the idea and finish myself. I have been stuck on the linked problem for a long time and solving this problem would let me solve the linked problem. Thanks for your time.

Note that $T_n$ takes values only in the countable set $\{k 2^{-n}; k \geq 1\}$, and therefore $T_n$ is a stopping time if we can show that
$$\{T_n = k 2^{-n}\} \in \mathcal{F}_{k2^{-n}} \quad \text{for all k \geq 1}. \tag{1}$$
By the very definition of $T_n$, we have
$$\{T_n = k 2^{-n}\} = \{(k-1) 2^{-n} \leq T < k 2^{-n}\} = \{T < k 2^{-n}\} \backslash \{T < (k-1) 2^{-n}\},$$
and since $T$ is by assumption a stopping time, this shows $(1)$.