Oké, so this question was one we had with a course of Stochastic Integration, it is however part of bigger proof, but I'll formulate the part I am uncertain about. The question is as follows:
$T$ is a stopping time.
Define for all $n \in \mathbb{N}$: $T_n := 2^{-n} \lfloor2^n T +1 \rfloor$. Show that $T_n$ is non-increasing in $n$.
Now, i want to check if my proof is right.
So let $m,n \in \mathbb{N}$, fixed and arbitrary.
Then: look at $T_m - T_n = (2^{-m} \lfloor2^m T +1 \rfloor) - (-2^{-n} \lfloor2^n T +1 \rfloor)$. Now denote by $\{T\}$ the decimal part of $T$. Then we have:
\begin{eqnarray} (2^{-m} \lfloor2^m T +1 \rfloor) - (-2^{-n} \lfloor2^n T +1 \rfloor)\\ =(2^{-m} \lfloor2^m T \rfloor) - (-2^{-n} \lfloor2^n T\rfloor)\\ = (2^{-m} \lfloor2^m (\lfloor T \rfloor + \{T\}) \rfloor) - (-2^{-n} \lfloor2^n (\lfloor T \rfloor + \{T\}) \rfloor) \\ = \lfloor T \rfloor + 2^{-m}\lfloor 2^m\{T\} \rfloor -( \lfloor T \rfloor + 2^{-n}\lfloor 2^n\{T\} \rfloor \\ = 2^{-m}\lfloor 2^m\{T\} \rfloor - 2^{-n}\lfloor 2^n\{T\} \rfloor. \end{eqnarray}
So now, we only have to prove the statement for $T$ in $[0,1)$.
My argument now is, that $2^{-m}\lfloor 2^m\{T\} \rfloor$ only has finitely many values, namely $\{0, 1, 2, \dots, 2^m-1\}$. Due to the increasingness of $2^m\{T\}$ and $2^n\{T\}$ with a lower derivative for $2^n\{T\}$, we have that $2^{-m}\lfloor 2^m\{T\} \rfloor$ -$2^{-n}\lfloor 2^n\{T\} \rfloor \leq 2^{-m}(2^m)-2^{-n}(2^n)=0$. With this the proof is given.
I would like to know if this is correct and understandable. Also i found it a whole lot of work for such a (sort of) easy question. Is there a easier solution?