Consider the Lebesgue measure on $[0,1]$.
We define $I_k^n=[k2^{-n},(k+1)2^{-n})$, for $0\leq k\leq 2^n-1$, and $F_n=\sigma(\{I^n_k:k\})$.
Finally $M_n=1_{I_0^n}2^n$.
How do we show that $M_n$ is a martingale? And what is its limit $M$?
For martingale we must show $$\mathbb{E}[1_{I_0^n}2^n|F_{n-1}]=1_{I_0^{n-1}}2^{n-1}.$$ I know that $I_0^{n-1}=I_0^{n}\sqcup I_1^n$, but how does one then show the identity?
Further I have a feeling that $M=0$ since $I_0^n\to\{0\}$, but again, how is this shown?