How to show the following is a martingale Show that $X_n = \{f((k+1)2^{-n})-f(k2^{-n})\}/2^{-n}$ on $I_{k,n} = [k2^{-n},(k+1)2^{-n}]$ defines a martingale where $\mathscr{F}_n = \sigma(I_{k,n})$
I'm not sure what $E(X_{n+1}|\mathscr{F}_n)$ means in this context. I can write $I_{k,n} = I_{m,n+1} \cup I_{m+1,n+1}$ for some $m$ but would be great to know the physical meaning. Thanks.
 A: I am assuming that the underlying probability filtered space is $([0,1],\{\mathscr{F}_n:n\in\mathbb{Z}_+\},\lambda)$, where $\lambda$ is Lebesgue measure. To check that $(X_n,\mathscr{F}_n)$ is indeed a Martingale, it is enough to see whether for $I_{n,k}:=\Big(\frac{k}{2^n},\frac{k+1}{2^n}\Big]=I_{n+1,2k}\cup I_{n+1,2k+1}$ you have that $\mathbb{E}[X_{n+1};I_{n,k}]=\mathbb{E}[X_n;I_{n,k}]$.
\begin{aligned}
\int_{I_{n,k}}X_{n+1}(t)\,dt &= \int_{I_{n+1,2k}}X_{n+1}(t)\,dt + \int_{I_{n+1,2k+1}}X_{n+1}(t)\,dt\\
&=\Big(f\big(\frac{2k+1}{2^{n+1}}\big)-f\big(\frac{2k}{2^{n+1}}\big)\Big)\frac{1}{2^{n+1}} + \Big(f\big(\frac{2k+2}{2^{n+1}}\big)-f\big(\frac{2k+1}{2^{n+1}}\big)\Big)\frac{1}{2^{n+1}}\\
&=\Big(f\big(\frac{2k+2}{2^{n+1}}\big)-f\big(\frac{2k}{2^{n+1}}\big)\Big)\frac{1}{2^{n+1}}\\
&=\Big(f\big(\frac{k+1}{2^n}\big)-f\big(\frac{k}{2^n}\big)\Big)\frac{1}{2^{n+1}}\\
&=\frac{1}{2}\int_{I_{n,k}}X_n(t)\,dt
\end{aligned}
Unless I am missing something, it does not seem that we have a martingale. $Y_n=2^nX_n$ on the other hand does seem to be a Martingale.
