Let $\Omega= [0,1)$, $I_{k,n}=[k2^{-n},(k+1){2^{-n}})$ and $\mathcal{F}_n = \sigma(I_{k,n}:0 \leq k \leq 2^n)$. Show that $X_n = (f((k+1){2^{-n}})-f(k2^{-n}))/2^{-n}$ on $I_{k,n}$, where $f$ is Lipschitz continuous function, defines a martingale, $X_n \rightarrow X_{\infty}$ $a.s.$ and in $L^1$, and $f(b)-f(a) = \int_{a}^{b}X_{\infty}(w)dw$.
I know that if I prove that $X_n$ is uniformly integrable the $X_n$ will converges to $X_{\infty}$ a.s. and in $L^1$.
I am stuck in proving the third property of martingale i.e. $E(X_{n+1}|\mathcal{F}_n)=X_n$.
Can anyone help me with this part of the question?
Note that all the sets in $\mathscr{F}_n$ are made of unions of the disjoint sets $\{I_{k,n},0\leq k < 2^n\}$. Furthermore, note that $$I_{k,n}=[k2^{-n},(k+1)2^{-n})=[2k2^{-(n+1)},2(k+1)2^{-(n+1)})=I_{2k,n+1}\cup I_{2k+1,n+1}$$ and $\lambda(I_{n,k})=2^{-n}$. wlog we can choose some $I_{k,n}\in \mathscr{F}_n$. Then $$\begin{aligned}\int_{I_{n,k}}X_{n+1}d\lambda&=\int_{I_{2k,n+1}\cup I_{2k+1,n+1}}X_{n+1}d\lambda=\\ &=\int_{I_{2k,n+1}}X_{n+1}d\lambda+\int_{I_{2k+1,n+1}}X_{n+1}d\lambda=\\ &=\frac{f(2(k+1)2^{-(n+1)})-f(2k 2^{-(n+1)})}{2^{-(n+1)}}2^{-(n+1)}=\\ &=\frac{f((k+1)2^{-n})-f(k 2^{-n})}{2^{-n}}2^{-n}=\\ &=\int_{I_{n,k}}X_n d\lambda\end{aligned}$$ But this means that $E[X_{n+1}|\mathscr{F}_n]=X_n$.
Hint for the rest of the question: the function $X$ s.t. $X=f'$ wherever $f'$ exists (i.e. a.e.) might give you an idea about what we're actually trying to prove.