# Convergence theorem for bounded continuous martingale in $L^2$

I would like to be sure to understand the proof this theorem.

Consider the set $$\mathcal{M}^2$$ of continuous martingale bounded in $$L^2$$. If $$(X_t)_{t\geq0}\in\mathcal{M}^2$$, there exists $$X_{\infty}\in\mathcal{M}^2$$ such that $$(X_t)_{t\geq0}$$ converges almost surely and in $$L^2$$ to $$X_{\infty}$$. Moreover $$\lVert X\rVert_{L^2} = \lVert X_{\infty}\rVert_{L^2}$$ and for all $$t\geq 0$$ we have $$\mathbb{E}(X_{\infty} | \mathcal{F}_t) = X_t$$

Proof : Consider $$X_t$$ in $$L_2$$ a continuous martingale, to prove the convergence in $$L^2$$ norm it suffices to use the sequential characterization of the continuity (of the $$L^2$$ norm) and show that for all increasing sequence $$t_n$$ of positive numbers that goes to $$\infty$$ we have $$X_{t_n}\to X_{\infty}$$.

Consider such a sequence, since $$X_{t_n}$$ is a martingale and we are in $$L^2$$ we have for $$m\geq n$$

$$\lVert X_{t_m} - X_{t_n}\rVert_{L^2}^{2} = \lVert X_{t_n}\rVert_{L^2}^{2} - \lVert X_{t_m}\rVert_{L^2}^{2}$$

The $$L^2$$ norm of a martingale is increasing and here it is bounded, so as $$n, m$$ go to $$\infty$$ we have that $$\lVert X_{t_m} - X_{t_n}\rVert_{L^2}^{2}\to 0$$. $$L^2$$ being complete and $$X_{t_n}$$ being a Cauchy sequence we conclude it has a limit $$X_{\infty}$$. This limit is in $$\mathcal{M}^2$$ using the continuity of the $$L^2$$ norm. By continuity of the conditional expectation we also have

$$\mathbb{E}(X_{\infty} | \mathcal{F}) = \lim_{s\to\infty}\mathbb{E}(X_s | \mathcal{F}_t) = X_t$$

Next we want to prove the almost sure convergence : from $$X_{t_n}$$ we can extract a subsequence $$X_{\phi(t_n)}$$ that converges almost surely to $$X_{\infty}$$, we denote $$A$$ the set of probability $$1$$ on which we have this convergence.

We would like to use that for all $$\omega\in A$$ we have

$$\lvert X_t - X_{\infty} \rvert\leq \lvert X_t - X_{\phi(t_n)} \rvert + \lvert X_{\phi(t_n)} - X_{\infty} \rvert$$

However at the stage we have not upper bound for $$\lvert X_t - X_{\phi(t_n)} \rvert$$.

For this, we consider for all $$n$$ the martingale $$Y_{n,s} = X_{\phi(t_n)+s} - X_{\phi(t_n)}$$. It is in $$L^2$$ so by Doob’s inequality and the first result we prove earlier for convergence in $$L^2$$ of martingale in $$L^2$$ we have

$$\lVert \sup_{s\geq 0}\lvert Y_{n,s}\rvert\rVert_{L^2} \leq 2\lVert Y_n \rVert_{L^2}=2\lVert Y_{\infty}\rVert_{L^2} = 2\lVert X_{\infty} - X_{\phi(t_n)}\rVert_{L^2}\to 0$$

When $$t_n\to\infty$$. This proves that $$\sup_{s\geq 0}\lvert Y_{n,s}\rvert$$ converges to $$0$$ in $$L^2$$. By extracting a sub sequence $$\left(Y_{\psi(n),s}= X_{\psi(\phi(t_n))+s} - X_{\psi(\phi(t_n))}\right)$$ we get a convergence almost surely to $$0$$, denote $$B$$ the set where take place such convergence.

Now consider $$C=B\cap A$$, we have $$\mathbb{P}(C)=1$$.

Hence, consider $$\epsilon>0$$, $$\exists N\left(\frac{\epsilon}{2}\right)$$ such that $$\psi(\phi(t_n))> N\left(\frac{\epsilon}{2}\right)$$ implies

$$\sup_{s\geq 0}\lvert Y_{\psi(n),s}\rvert + \lvert X_{\psi(\phi(t_n))} - X_{\infty} \rvert\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}$$

Now take $$t$$ big enough such that $$s = t-\psi(\phi(t_n))>0$$, then we have

$$\lvert X_t - X_{\psi(\phi(t_n))} \rvert = \lvert X_{\psi(\phi(t_n)) + s} - X_{\psi(\phi(t_n))} \rvert\leq \sup_{s\geq 0}\lvert Y_{\psi(n),s}\rvert$$

Thus

$$\lvert X_t - X_{\infty}\rvert\leq \lvert X_t - X_{\psi(\phi(t_n))} \rvert + \lvert X_{\psi(\phi(t_n))} - X_{\infty} \rvert\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

This holds for all $$\omega\in C$$ so we conclude for the convergence almost surely.

Is this seems correct please ? I try to make clear every step in order to show my understanding of all notions that are used here, I would like to know if my understanding (and hence my proof) is correct please.

Thank you !