I read Schilling's "Brownian Motion"(2nd edition). This book say that $\mathcal{M}^{2,c}_T$ (the family of continuous $L^2$-martingale $(M_t)_{0\leq t\leq T}$) is complete under a norm $\|M\|_{\mathcal{M}^{2,c}}:=(\mathbb{E}[\sup_{s\leq T}|M_s|^2])^{1/2}$. Lemma $15.10$ in this book already proved $\|M\|_{\mathcal{M}^{2,c}}$ is norm on $\mathcal{M}^{2,c}_T$ and $\mathcal{M}^{2,c}_T$ is closed under this norm. But, proof of completeness is nothing. My proof is correct?
Continuous $L^2$-martingale $\mathcal{M}^{2,c}_T$ is complete under $\|M\|_{\mathcal{M}^{2,c}}=(\mathbb{E}[\sup_{s\leq T}|M_s|^2])^{1/2}$.
my proof
Let $M^{(n)}$ be a Cauchy sequence in $\mathcal{M}^{2,c}_T$. \begin{equation}\lim_{n,m\to\infty}\mathbb{E}[\sup_{s\leq T}|M^{(n)}_s-M^{(m)}_s|^2]=0.\end{equation} For any $0\leq t\leq T$, $\mathbb{E}[|M^{(n)}_t-M^{(m)}_t|^2]\leq\mathbb{E}[\sup_{s\leq T}|M^{(n)}_s-M^{(m)}_s|^2]$ is clear. So, $M^{(n)}_t$ is a Cauchy sequene in $L^2(\mathbb{P})$. By completeness of $L^2(\mathbb{P})$, $M_t:=L^2(\mathbb{P})$-$\lim_{n\to\infty}M^{(n)}_t$ exists. This $(M_t)_{0\leq t\leq T}$ is martingale because following property of conditional expectation. "$X_n\to X$ in $L^1$ $\Longrightarrow$ $\mathbb{E}[X_n|\mathcal{G}]\to\mathbb{E}[X|\mathcal{G}]$ in $L^1$." Then, by Doob's inequality, \begin{equation} \mathbb{E}[\sup_{s\leq T}|M^{(n)}_s-M_s|^2]\leq 4\mathbb{E}[|M^{(n)}_T-M_T|^2]. \end{equation} Taking limit in this inequality and the fact "$\mathcal{M}^{2,c}_T$ is closed (by the book lemma)" shows that any Cauchy sequence is convergence sequence. Therefore, $\mathcal{M}^{2,c}_T$ is complete in this norm.
If not correct, please give me some clues. Thanks.