First let $\mu$ be the induced distribution of the random variable $X$ on $(\mathbb{R},\mathcal{B})$ and denote $EX=m$.
We also define for all $A\in G_{n+1}$ and $\omega\in X^{-1}(A)$ $$E(X\mid\mathcal{G}_{n+1})(\omega)=\frac{E(1_{A}X)}{\mu(A)}$$
They now construct a binary tree as follows, the root of the tree is a node with value $m$. In the construction of $\mathcal{G}_{1}$, with each interval associated with $m$, they put an out-arrow and node at level 1. With the out-arrow they associate the corresponding interval and with the node, the corresponding value of $E(X\mid\mathcal{G}_{1})$. This construction is then repeated to find the full tree.
Suppose $A_{n}\in G_{n}$ is associated with the arrow from the vertex with value $s_{n-1}$ to the vertex with value $s_{n}$. Then either $s_{n}=s_{n-1}$, this then implies $A_{n}=A_{n-1}$ and so $\mu(A_{n})=\mu(A_{n-1})$. Or $s_{n-1}$ has a second out-arrow associated with $A_{n-1}\setminus A_{n}$ leading to a vertex with value $s_{n}'$. Since $X^{-1}(A_{n-1})\in \mathcal{G}_{n-1}\subset \mathcal{G}_{n}$ $$s_{n-1}\mu(A_{n-1})=E1_{\{X^{-1}(A_{n-1})\}}E(X\mid\mathcal{G}_{n-1})$$ $$=s_{n}\mu(A_{n})+s_{n}'\mu(A_{n-1}\setminus A_{n})$$
The question is now to derive such a binary tree for a homogeneous distribution on $[0,1]$ but frankly I have no idea as to how to go about this. Can anyone help me out with this?
