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

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


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

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 !



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