Assume that we have the probability field $([0,1],\mathcal{B}([0,1]),\lambda)$ where $\lambda$ is the lebesgue-measure. The goal is to explicitly develop the binomial tree, defined by the stochastic process, $X_n:[0,1] \rightarrow\mathbb{R}$ , given by $$X_n(\omega)=U\cdot\mathbf{1}_{A_n}(\omega)+D\cdot\mathbf{1}_{A_n^c}(\omega), \quad D<U$$
To get $P(X_n)=p\in(0,1)$ for all $n$, the way we construct the sets $A_n$, is by first splitting $[0,1]$ up in two: $[0,p)$ and $[p,1]$ and set $A_1=[0,p)$. Next we divide the to former in two with same proportion as before: $[0,p^2)$, $[p^2,p)$, $[p,2p-p^2)$ and $[2p-p^2,1]$ and now $A_2=[0,p^2)\cup[p,2p-p^2)$ and so on...
If $p=1/2$, we fairly simple get $$ A_n=\bigcup_{i=1}^{2^{n-1}}\left[\frac{2(i-1)}{2^n},\frac{2i-1}{2^n}\right)$$ But is there a general formula for the sets $A_n$ when we want it to be a general $p$, i.e. $P(X_n=U)=p\in(0,1)$?