martingale convergences wrt dyadic intervals Consider $[0,1]$ and thereon $D_n = \{ [k2^{-n}, (k{+}1) 2^{-n} ) : 0\le k \le 2^n-1\}$ and $\mathcal F_n$ the generated sigma-algebras by $D_n$. We do know that uniformly bounded (in $L_p$-norm) martingales converge. 
In this particular situation, $X_n$ is a linear combination of dyadic indicator functions, and so an element of a ball (with uniform radius) of a finite-dimensional subspace $E_n$ of dimension $2^n$ of $L_p(0,1)$. So I wonder whether the martingale convergence theorem can be proved in this setting via a compactness argument, using the linear projections from $E_{n+1}$ to $E_n$ that "average two averages". I admit the question is not very precise, but I still hope for some interesting comments.
 A: Yes, the Martingale convergence theorem can be proven by a weak compactness argument. Let me restrict to the case of a Hilbert space $H$ for simplicity. Any convergence in $H$ that is proven using completeness can be obtained by a weak compactness argument.
The $L^2$ Martingale convergence theorem is deduced from the following result. 
Theorem
Let $H$ be a Hilbert space, $E_i$ an increasing sequence of closed subspaces of $H$ such that the union is dense in $H$ and $\pi_i$ the orthogonal projections on $E_i$. Then for all $v\in H$, the following convergence holds with respect to the Hilbert norm on $H$:
$$ \pi_i(v) \rightarrow v.$$
This is usually proven by showing that the sequence $\pi_i(v)$ is a Cauchy sequence. Here is an argument that uses instead the weak compactness of the unit ball. 
We consider a weak limit $v_\infty$ of $\pi_i(v)$. It is not hard to show that $v-v_\infty$ is orthogonal to all the $E_i$ hence is zero. As a result, $v = v_\infty$, there is a single possible limit point wrt the weak topology. By compactness, $\pi_n(v)$ converges weakly to $v$.
But this implies strong convergence because the $\pi_i$ are orthogonal projections. More precisely, we have
$$\langle v - \pi_i v , v \rangle \rightarrow 0$$
This quantity is the square of the norm of the quantity we are interested in:
$$\parallel \pi_i(v) - v\parallel^2 = \langle (id - \pi_i)^*(id-\pi_i)(v), v \rangle = \langle (id-\pi_i)(v),v\rangle
$$
since $id-\pi_i$ is an orthogonal projection.
