**Definitions:**


Two real valued random variables $X\_0$ and $X\_1$ are called a *one step martingale* if $E[X\_1| X\_0] = X\_0$.


We say the one step martingale is in $L^2$ if both $X\_0$ and $X\_1$ are in $L^2(P)$.


**Question:**


Given an $L^2$ one step martingale $(X\_0, X\_1)$ does there exist a sequence $(Y\_0^n, Y\_1^n)$ of $L^2$ one step martingales satisfying the following two conditions?


* $Y\_0, Y\_1$ are simple random variables, i.e. variables taking only a finite number of values.
* $Y\_0^n, Y\_1^n$ converge to $X\_0, X\_1$ respectively in $L^2$.



Discrete approximation of one step martingale
Let $\mathcal F\_i^n$ be a sequence of $\sigma$-algebras increasing to $\mathcal F(X\_i)$, the $\sigma$-algebra generated by $X\_i$ and let $Y\_1^n=\mathbb E(X\_1|\mathcal F^n\_1\vee \mathcal F^n\_0)$ and $Y\_0^n=\mathbb E(X\_1|\mathcal F^n\_0)$.


Then $Y^n\_1$ and $Y^n\_0$ converge pointwise a.s. to $X\_1$ and $X\_0$ by Doob's martingale convergence theorem. They therefore converge in $L^2$ to $X\_1$ and $X\_0$ also.


From the tower law, we see that $\mathbb E(Y^n\_1|\mathcal F^n\_0)=Y^n\_0$.


