# Weaker version of the martingale convergence theorem

Let $\mathcal{A}_n$ be a sequence of finite sigma-algebras, let $\mathcal{B}_{q,p}= \sigma(\mathcal{A}_n, q \geq n \geq p )$. Moreover, we suppose $\mathcal{A}_k \subset \mathcal{B}_{\infty,p}$ for any $k <p$.

Let $X$ be a random variable that is $\mathcal{B}_{\infty,p}$ measurable for any integer $p$.

Let $u_{q,p}= E[ |E[X|\mathcal{B}_{q,p}]-E[X|\mathcal{A}_q] |^2]$.

We suppose for any $p \geq 1$, $u_{q+1,p}=u_{q,p-1}$ (this is a weaker equivalent of the "filtration" condition in the martingale convergence theorem) , and that $u_{n+1,0}-u_{n,0} \rightarrow_{n \rightarrow + \infty} 0$.

Is it true that almost surely, $E[X|\mathcal{A}_n] \rightarrow_{n \rightarrow + \infty} X$.

If not, can you give a counterexample?

Let $\Omega=\{-1,1\} \times \{-1,1\}$ with the discrete $\sigma$-algebra and uniform measure, let $X(\omega_1,\omega_2)=\omega_2$, and let $$\mathcal{A}_n \ = \ \left\{ \begin{array}{l l} \sigma(\{\omega_1=\omega_2\}) & n \textrm{ odd} \\ \sigma(\{\omega_1=1\}) & n \textrm{ even}. \end{array} \right.$$ So $\mathcal{B}_{q,p}$ is the discrete $\sigma$-algebra for all $p<q$, but $E[X|\mathcal{A}_n]=0$ for all $n$.