How to show that these spaces are Banach spaces I want to show, that the following spaces are Banach spaces:
$X_1:=\{M=(M_t)_{0\le t \le T} ;\mbox{ M is an adapted RCLL process }\}$ with the norm $\|M\|_{X_1}:=\|\sup_{0\le t\le T}|M_t|\|_{L^2(P)}$ where $P$ is a probability measure on a probability space. And $ X_2:=\{M=(M_t)_{0\le t \le T}; \mbox{M is a optional process}\}$ with the norm $\|M\|_{X_2} :=(E(\int_0^T|M_s|^2ds))^{\frac{1}{2}}$
For $X_1$ any Cauchy sequence would be a Cauchy sequence uniformly in $t$ in $L^2$. Hence there is a limit. But how do I show that this limit is again in $X_1$? Also I have no idea how to prove this for $X_2$
 A: Let $\{M^{(n)}\}$ a Cauchy sequence in $X_1$. Then $\sup_{0\leq t\leq T}|M_t^{(n)}-M_t^{(m)}|\to 0$ when $m,n\to +\infty$ in $L^2(P)$. In particular, for each $t$ fixed in $[0,T]$, the sequence $\{M_t^{(n)}\}$ is Cauchy in $L^2(P)$. Since this space is complete, we can define $M_t$ as the limit in $L^2(P)$ of this sequence. Now, have have several things to check: 


*

*We have that for all $\varepsilon >0$, we can find $n_0$ such that if $m,n\geq n_0$ and $t\in [0,T]$ then $\lVert M_t^{(n)}-M_t^{(m)}\rVert_{L^2(P)}\leq \varepsilon$ (since $|M_t^{(n)}-M_t^{(m)}|\leq \sup_{0\leq s\leq T}|M_s^{(n)}-M_s^{(m)}|$). Taking the limit $m\to +\infty$, we get that $\lVert M-M^{(n)}\rVert_{X_1}\to 0$. Indeed, you can check, using the fact that converge in $L^2$ implies convergence of a subsequence almost everywhere, that for each subsequence it's true for a further subsequence.  

*We check that $M\in X_1$. We look at right continuity. We can find an increasing sequence $\{n_k\}$ of integers such that $\sup_{0\leq t\leq T}|M^{(n_k)}_t-M_t|\to 0$ almost everywhere (say in $\Omega'$. Fix $t_0\in [0,T]$. For $s\geq t_0$ and $\omega\in\Omega'$, we have 
\begin{align}
|M_s(\omega)-M_{t_0}(\omega)|&\leq |M_s(\omega)-M_s^{(n_k)}(\omega)|\\
&+|M_s^{(n_k)}(\omega)-M_{t_0}^{(n_k)}(\omega)|+|M_{t_0}^{(n_k)}(\omega)-M_{t_0}(\omega)|\\
&\leq 2\sup_{0\leq t\leq T}|M_t(\omega)-M_t^{(n_k)}(\omega)|+|M_s^{(n_k)}(\omega)-M_{t_0}^{(n_k)}(\omega)|.
\end{align}
Now, we fix $\varepsilon>0$, and pick $k$ such that $2\sup_{0\leq t\leq T}|M_t(\omega)-M_t^{(n_k)}(\omega)|\leq \varepsilon$. Then we conclude using the right continuity of $M^{(n_k)}$. 
We have now to see left limit. It follows a similar argument, plus the Cauchy criterion: fix $t_0\in [0,T]$. For $s_1,s_2\leq t$ we have 
$$|M_{s_1}(\omega)-M_{s_2}(\omega)|\leq 2\sup_{0\leq t\leq T}|M_t(\omega)-M_t^{(n_k)}(\omega)|+|M_{s_1}^{(n_k)}(\omega)-M_{s_2}^{(n_k)}(\omega)|.$$

*We have to check that $\{M_t\}$ is adapted. To see that, note that the set $\{\omega,\{M_t^{(n_k)}(\omega)\}\mbox{ doesn't converge}\}$ can we written as a countable intersection of countable union of elements of $\mathcal F_t$. 

