I have the following problem:
Let $(\Omega,F)$ be a measure space, $f:\Omega\rightarrow\mathbb{R}$, $f_n:\Omega\rightarrow\mathbb{R}$ with $x\mapsto min\{\frac{1}{2^n}\lfloor 2^n f(x)\rfloor,n\}$. Show $f_n\leq f_{n+1}$ and $\lim f_n=f$ pointwise.
I want to show $\frac{1}{2^n}\lfloor 2^n f(x)\rfloor\leq \frac{1}{2^{n+1}}\lfloor 2^{n+1} f(x)\rfloor$ since $n<n+1$. I'm not sure how to deal with the floor function. I tried to prove it with induction and proved it for $n=1$ by using different cases for $x$, but I don't know how to continue there. Can someone help me?
HINT: your inequality is equivalent to this other:
$$2\lfloor 2^n y\rfloor\le \lfloor 2^{n+1}y\rfloor\tag1$$
for any chosen $y\in\Bbb R$. Note that $y=\lfloor y\rfloor+\{y\}$ where $\{y\}\in[0,1)$. Then
$$\lfloor 2^k y\rfloor=\lfloor 2^k\lfloor y\rfloor+2^k\{y\}\rfloor=2^k\lfloor y\rfloor+\lfloor2^k\{y\}\rfloor\tag2$$
And $(1)$ becomes
$$2(2^n\lfloor y\rfloor +\lfloor 2^n\{y\}\rfloor)\le2^{n+1}\lfloor y\rfloor+\lfloor 2^{n+1}\{y\}\rfloor\implies\\ 2\lfloor 2^n\{y\}\rfloor\le\lfloor 2^{n+1}\{y\}\rfloor\tag3 $$
So you need to prove $(3)$, that setting $x:=2^n\{y\}$ is equivalent to show that $2\lfloor \{x\}\rfloor\le\lfloor 2\{x\}\rfloor$, what is immediate because $\lfloor \{x\}\rfloor=0$.