For an $L^1([0,1])$ function $f$, if we partition $[0,1]$ into intervals by setting $I_j = [x_j,x_{j+1}]$ with $0 =: x_0 < x_1 < \cdots < x_N := 1$ and let $\mathcal{F}$ be the sigma algebra generated by this partition, then $$ \mathbb{E}[f ~\big| ~\mathcal{F} ] := \sum_{j=1}^N \chi_{I_j}(x) \frac{1}{m(I_j)} \int_{I_j} f(y) dy $$ is a conditional expectation.
Furthermore, if we have now a filtration of sigma algebras generated by refining the partition above as $n$ increases (such that $\cdots \subseteq \mathcal{F}_{n} \subseteq \mathcal{F}_{n+1} \subseteq \cdots$, then we set $$X_n := \mathbb{E}[ f~\big| ~ \mathcal{F}_n ]$$ then $X_n$ is a Martingale sequence. We can show this using the Tower property somewhat trivally by writing $$\mathbb{E}[X_{n+1} \big| \mathcal{F}_n] = \mathbb{E}[ \mathbb{E} [ f \big| \mathcal{F}_{n+1} ] \big| \mathcal{F}_n ] = \mathbb{E}[f\big| \mathcal{F}_n] = X_n$$.
I want to be able to show this in a less abstracted way though using the definition that $X_n = \mathbb{E}[ f~\big| ~ \mathcal{F}_n ] :=\sum_{j=1}^N \chi_{I_j}(x) \frac{1}{m(I_j)} \int_{I_j} f(y) dy$ , to in particular compute that $$ \mathbb{E}[X_{n+1} \big| \mathcal{F}_n] = \sum_{k=0 }^{N_{n+1}} \frac{\chi_{I_j \cap I_k}}{m(I_j)m(I_k)} \int_{I_k} f(y) dy = \cdots = X_n $$ but I have been stuck trying to manipulate the sums and characteristic functions. Any help would be appreciated!
[From Muscalu and Schlag's "Classical and Multilinear Harmonic Analysis" Vol I, Ch 5]
I will work on $([0,1),\mathscr{B}[0,1),m)$ and on interval partitions of the form $([x_{k}^n,x_{k+1}^n))_{0\leq k < M_n}$ where $0=x_0^n<...<x_{M_n}^n=1$. You may want to prove that for all $n$ we have $E[\mathbf{1}_{A}X_{n+1}]=E[\mathbf{1}_AX_{n}], A \in \mathscr{F}_{n}$. We have that $\mathscr{F}_{n}$ is generated by the intervals of the $n$-th partition and these partitions are refining. Fix $n$, which fixes a partition. Let $A=[x^n_k,x_{k+1}^n)$ for some $k\in \{0,1,...,M_n-1\}$. Now:
$$\begin{aligned} E[\mathbf{1}_{[x^n_k,x_{k+1}^n)}X_{n+1}] &=\int_{[x^n_k,x_{k+1}^n)}\bigg(\sum_{0\leq \ell< M_{n+1}}\mathbf{1}_{[x_\ell^{n+1},x_{\ell+1}^{n+1})}(x)\frac{1}{m([x_\ell^{n+1},x_{\ell+1}^{n+1}))}\int_{[x_\ell^{n+1},x_{\ell+1}^{n+1})}f(y)dy\bigg)dx\\ &=\sum_{0\leq \ell< M_{n+1}}\frac{m([x^n_k,x_{k+1}^n)\cap [x_\ell^{n+1},x_{\ell+1}^{n+1}))}{m([x_\ell^{n+1},x_{\ell+1}^{n+1}))}\int_{[x_\ell^{n+1},x_{\ell+1}^{n+1})}f(y)dy \end{aligned}$$ Since the partitions are refining, for each $k$ there exists only one $\ell_k$ s.t. $[x^n_k,x_{k+1}^n)=\cup_{0\leq h <N_k}[x_{\ell_k+h}^{n+1},x_{\ell_k+h+1}^{n+1})$ for some $N_k$. So we get, since these are disjoint sets $$E[\mathbf{1}_{[x^n_k,x_{k+1}^n)}X_{n+1}]=\sum_{0\leq h <N_k}\int_{[x_{\ell_k+h}^{n+1},x_{\ell_k+h+1}^{n+1})}f(y)dy=\int_{[x^n_k,x_{k+1}^n)}f(y)dy$$ But now note: $$\begin{aligned}E[\mathbf{1}_{[x^n_k,x_{k+1}^n)}X_{n}]&=\sum_{0\leq u< M_n}\frac{m([x^n_k,x_{k+1}^n)\cap [x_u^{n},x_{u+1}^{n}))}{m([x_u^{n},x_{u+1}^{n}))}\int_{[x_u^{n},x_{u+1}^{n})}f(y)dy\\ &=\int_{[x^n_k,x_{k+1}^n)}f(y)dy\end{aligned}$$ So we conclude that $E[\mathbf{1}_{[x^n_k,x_{k+1}^n)}X_{n+1}]=E[\mathbf{1}_{[x^n_k,x_{k+1}^n)}X_{n}]$. Since each set $A\in \mathscr{F}_n$ is a union of some (or all) of the intervals defined by the $n$-th partition, we conclude.