# Given a stopping time $\tau,$ there is a decreasing discrete stopping times sequence which converges to $\tau$

I'm reading the proof of the next proposition:

Let $(\mathcal{F}_{t})_{t\geq 0}$ be aright-continuous filtration. Given $\tau$ stopping time, there is a decreasing discrete stopping times sequence $(\tau_{j})_{j\in\mathbb{N}}$ which converges to $\tau.$

To prove this, we define $$\tau_{j}(\omega) = \begin{cases} \tau(\omega) & \text{if}\space\tau(\omega)=\infty; \\ k2^{-j} & \text{if}\space(k-1)2^{-j}\leq\tau(\omega)<k2^{-j}.\end{cases}$$

Fix $t\geq 0$ and $j\geq 1$ and pick $k=k(t,j)$ such that $(k-1)2^{-j}\leq t<k2^{-j}.$

Then $$\{\tau_{j}\leq t\}=\{\tau_{j}\leq (k-1)2^{-j}\}=\{\tau<(k-1)2^{-j}\}.$$

My doubt is on the equalities above. I don't get them; They seem contradictory to me because of the construction of the sequence and $t.$

Any kind of help is thanked in advanced.

I will show that $\{\tau_j \leq t\} \subset \{\tau_j \leq (k-1)2^{-j}\} \subset \{\tau < (k-1)2^{-j}\}$. I believe you can verify the reverse inclusions. Let $\tau_j \leq t$. Then $\tau_j <k2^{-j}$. Now you have to observe that (by definition) $\tau_j$ is of the form $\frac l {2^{j}}$ for some integer $l$. This gives $l<k$. Since $l$ and $k$ are integers this implies $l \leq k-1$. Hence $\tau_j=\frac l {2^{j}} \leq \frac {k-1} {2^{j}}$. We have proved the first inclussion. Now suppose $\tau_j \leq \frac {k-1} {2^{j}}$. Then $\tau_j =\frac l {2^{j}}$ and $\frac {l-1} {2^{j}} \leq \tau < \frac l {2^{j}}$ for some $l \leq k-1$. Hence $\tau < \frac {k-1} {2^{j}}$.