Let
- $(\Omega,\mathcal{A})$ be a measurable space
- $X$ be $\mathcal{A}$-$\mathcal{B}(\overline{\mathbb{R}})$-measurable
Moreover, let $$X_n:=\frac{1}{2^n}\lfloor 2^nX\rfloor$$ How can we prove that $\left(X_n\right)_{n\in\mathbb{N}}$ is monotonically increasing and converges to $X$?
I'm failing to show that $X_n\le X_{n+1}$.
To prove monotonicity, observe that if $2^nX=k+d$ where $k \in \mathbb Z$ and $0\le d <1$ then you have that $$\begin{align*}\lfloor 2^{n+1}X \rfloor&=\lfloor 2\cdot2^nX \rfloor=\lfloor 2\cdot (k+d) \rfloor=\lfloor 2k+2d \rfloor\\&=2k \text{ or } 2k+1\\& \ge 2k\\&= 2\lfloor (k+d)\rfloor\\&=2\lfloor 2^nX\rfloor\end{align*}$$ This implies that $$\begin{align*}\lfloor 2^{n+1}X \rfloor \ge 2 \lfloor 2^nX \rfloor &\iff \frac{1}{2^n}\lfloor 2^{n+1}X \rfloor \ge \frac{1}{2^n}2 \lfloor 2^nX \rfloor \\&\iff \frac{1}{2^{n+1}}\lfloor 2^{n+1}X \rfloor \ge \frac{1}{2^n} \lfloor 2^nX \rfloor \\&\iff X_{n+1}\ge X_n\end{align*}$$