Completeness using Conditional Expectation as a Norm Background for my question:

Fix a probability space $(\Omega,\mathcal A, P)$, and let $\mathcal F$ be a sub-$\sigma$-algebra of $\mathcal A$. For any nonnnegative measurable map $X:\Omega\to [0,\infty]$ there exists a  $\mathcal F$-measurable map $Y:\Omega\to [0,\infty]$ satisfying
$$\int_A X\,dP=\int_A Y\, dP \quad \text{for all } A\in \mathcal F$$
Moreover $Y$ is $P$-almost surely unique, and we denote it by $E[X| \mathcal F]$. (Theorem 15.1 in Probability Theory by Heinz Bauer).
Define $\mathcal P^+$ to be the set of all real random variables $X$ on $(\Omega,\mathcal A, P)$ satisfying $\|X\|_\mathcal F:=E[X^2|\mathcal F]^{1/2}<\infty$ $P$-almost surely. From the conditional Minkowski's inequality and the pull-out property of conditional expectations we have that $W_1X_1+W_2X_2\in \mathcal P^+$ whenever $X_1,X_2\in \mathcal P^+$ and $W_1,W_2$ are $\mathcal F$- measurable. We say that $\mathcal P^+$ is conditionally linear.
A sequence $(X_n)\subset \mathcal P^+$ is said to converge conditionally to $X\in \mathcal P^+$ if $P[\|X_n-X\|_\mathcal F>\epsilon]\to 0$ as $n\to\infty$ for all $\epsilon>0$, and is said to be conditionally Cauchy if $P[\|X_m-X_n\|_\mathcal F>\epsilon]\to 0$ as $m,n\to\infty$ for all $\epsilon>0$. This is well-defined by the almost sure uniqueness of conditional expectations.

Lemma. If $(X_n)\subset \mathcal P^+$ is conditionally Cauchy, then it is Cauchy in probability.
Proof. Suppose $(X_n)\subset \mathcal P^+$ is conditionally Cauchy, and let $\epsilon>0$ be given. Define $\delta:=\min(\varepsilon/2,\sqrt{\varepsilon^3/2})>0$. Since $(X_n)$ is conditionally Cauchy there exists $N$ such that $P[\|X_m-X_n\|_\mathcal F>\delta]<\delta$ for all $m,n\geq N$. Let $$A_{m,n}:=\{|X_n-X_m|>\epsilon\} \quad \text{and}\quad  B_{m,n}:=\{\|X_n-X_m\|_\mathcal F>\delta\}$$ and let  $A^c_{m,n}$ and $B^c_{m,n}$ denote their complements respectively. Since $B^c_{m,n}\in\mathcal F$ we have
$$\delta^2\geq \int_{B^c_{m,n}} \|X_n-X_m\|_{\mathcal F}^2 \,dP =\int_{B^c_{m,n}} (X_n-X_m)^2 \,dP \geq \int_{A_{m,n}\cap B^c_{m,n} } (X_n-X_m)^2 \,dP \geq \epsilon^2 P[A_{m,n}\cap B^c_{m,n} ]  $$
Therefore if $m,n\geq N$ then
$$P[A_{m,n}]=P[A_{m,n}\cap B_{m,n} ]+P[A_{m,n}\cap B^c_{m,n} ]\leq P[ B_{m,n} ]+P[A_{m,n}\cap B^c_{m,n} ] < \varepsilon/2+\varepsilon/2=\varepsilon $$
We have shown that for any $\epsilon>0$ there exists $N$ such that $m,n\geq N$ implies $P[|X_n-X_m|>\epsilon]< \epsilon$. This is equivalent to $(X_n)$ being Cauchy in probability.

Theorem. $\mathcal P^+$ is conditionally complete.
Proof. Let $(X_n)$ be a conditionally Cauchy sequence in $\mathcal P^+$. From the above lemma and the definition of conditionally Cauchy we can extract a subsequence $(X_{n_k})$ of  $(X_n)$  such that
$$ P[A_k]:=P[|X_{n_{k+1}}-X_{n_{k}}|>1/2^k]<1/2^k$$ $$ P[B_k]:= P[\|X_{n_{k+1}}-X_{n_{k}}\|_{\mathcal F}>1/2^k]<1/2^k $$
for all $k$. Define $Y_k:=X_{n_{k+1}}-X_{n_{k}}$ for each $k$. Since $\sum_{k=1}^\infty P[A_k]<\infty$ and $\sum_{k=1}^\infty P[B_k]<\infty$, the Borel-Cantelli lemma implies that
$$\sum_{k=1}^\infty |Y_k|<\infty \quad \text{and} \quad  \sum_{k=1}^\infty \|Y_k\|_{\mathcal F}<\infty  \quad P\text{-almost surely}$$
Let $N\in \mathcal A$ with $P[N]=0$ be such that the above holds on $\Omega\setminus N$. Define the real random variable $X$ on $(\Omega,\mathcal A)$ by
$$X(\omega)=\begin{cases} 
      \sum_{k=1}^\infty Y_k(\omega) +X_{n_1}(\omega)& \text{if } \omega\in \Omega\setminus N \\
      0 & \text{if } \omega\in N  
   \end{cases}$$
Note that $X$ is $\mathcal A$-measurable and that we have $|X|\leq  \sum_{k=1}^\infty |Y_k|+|X_{n_1}|$ pointwise. Hence
$$\|X\|_{\mathcal F} \leq \Big\|  \sum_{k=1}^\infty |Y_k|+|X_{n_1}| \Big\|_{\mathcal F} \leq \Big\|  \sum_{k=1}^\infty |Y_k| \Big\|_{\mathcal F}+ \Big\||X_{n_1}|\Big\|_{\mathcal F}\leq \sum_{k=1}^\infty \|Y_k\|_{\mathcal F} + \|X_{n_1}\|_{\mathcal F}<\infty $$
$P$-almost surely, where the third inequality is from the conditional MCT and the conditional Minkowski's inequality. This shows that $X\in\mathcal P^+$. We will show that $(X_{n_k})$ converges conditionally to $X$.
Since $|X-X_{n_k}|=\Big|\sum_{i=k}^\infty Y_i\Big|\leq \sum_{i=k}^\infty |Y_i|$  on $\in \Omega\setminus N$ for each $k$, we have
$$\|X-X_{n_k}\|_{\mathcal F}\leq \Big \| \sum_{i=k}^\infty |Y_i| \Big \|_{\mathcal F} \leq \sum_{i=k}^\infty \|Y_i\|_{\mathcal F} \quad P\text{-almost surely}$$ for each $k$. Using the fact that a countable union of null sets is null and that $\sum_{k=1}^\infty \|Y_k\|<\infty$ $P$-almost surely we get $\|X-X_{n_k}\|_{\mathcal F}\to 0$ as $k\to \infty$, $P$-almost surely. Since almost sure convergence implies convergence in probability we get that $(X_{n_k})$ converges conditionally to $X$.
Finally we show that  $(X_{n})$ converges conditionally to $X$. Since $(X_n)$ is conditionally Cauchy and $n_k\geq k$ for each $k$ we have $P[\|X_{n_k}-X_{k}\|_{\mathcal F}>\epsilon]\to 0$ as $k\to \infty$ for any $\epsilon>0$. Moreover we have already shown that $P[\|X_{n_k}-X\|_{\mathcal F}>\epsilon]\to 0$ as $k\to \infty$ for any $\epsilon>0$. Therefore
$$P[\|X_{k}-X\|_{\mathcal F}>\epsilon]\leq P[\|X_{n_k}-X_{k}\|_{\mathcal F}>\epsilon/2]+P[\|X_{n_k}-X\|_{\mathcal F}>\epsilon/2]\to 0 \quad \text{as}\quad  k\to \infty$$ for any $\epsilon>0$.

Question: If $X_1,\dots, X_p$ are fixed random variables in $L^2(\Omega,\mathcal A,P)$ and $\mathcal S$ is the subspace of $\mathcal P^+$ consisiting of all linear combinations of the form
$$W_1X_1+\dots +W_pX_p$$
with $W_i$ $\mathcal F$-measurable for each $i$, then is $\mathcal S$ conditionally complete?
Thanks a lot for your help.
 A: Write $X:=[X_1, \dots, X_p]^\top$. Then $\mathcal S=\{W^\top X: W \text{ is }\mathcal F \text{-measurable}\}$. I will make the following assumption:
Assumption. $E[XX^\top |\mathcal F]$ is positive definite $P$-almost surely.
Let $(Y_n)=(W_n^\top X)$ be a conditional Cauchy sequence in $\mathcal S$. Since $\mathcal S\subset \mathcal P^+$ and $\mathcal P^+$ is conditionally complete,  there exists $Y \in \mathcal P^+$ such that $(Y_n)$ converges conditionally to $Y$. Moreover, from the proof of the Theorem in the post, there exists a subsequence $(Y_{n_k})$ of $(Y_n)$ such that  $\|Y-Y_{n_k}\|_{\mathcal F}\to 0 $ as $k\to \infty$ $P$-almost surely. From the conditional Minkowski's inequality we get
$$ \|Y_{n_k}-Y_{n_l}\|_{\mathcal F}\leq\|Y-Y_{n_k}\|_{\mathcal F} +\|Y-Y_{n_l}\|_{\mathcal F}\to 0  \quad \text{as } k,l\to \infty $$
$P$-almost surely. But
$$\|Y_{n_k}-Y_{n_l}\|^2_{\mathcal F}=E[(W_{n_k}-W_{n_l})^\top XX^\top(W_{n_k}-W_{n_l})|\mathcal F]= (W_{n_k}-W_{n_l})^\top E[XX^\top|\mathcal F](W_{n_k}-W_{n_l}) $$ $P$-almost surely so we get
$$(W_{n_k}-W_{n_l})^\top E[XX^\top|\mathcal F](W_{n_k}-W_{n_l}) \to 0 \quad \text{as } k,l\to \infty $$ $P$-almost surely. Since $E[XX^\top |\mathcal F]$ is positive definite $P$-almost surely by assumption, this implies that
$$W_{n_k}-W_{n_l}\to 0 \quad \text{as } k,l\to \infty $$
$P$-almost surely. In other words $(W_{n_k})$ is $P$-almost surely Cauchy. Using completeness of $\mathbb R^n$ define
$$ W(\omega)=\begin{cases} 
      \lim_{k\to \infty} W_{n_k}(\omega) & \text{if }   (W_{n_k}(\omega)) \text{ is Cauchy} \\
      0 & \text{otherwise}
   \end{cases}$$
Note that $W$ is $\mathcal F$-measurable and that
$$\|Y_{n_k}-W^\top X\|^2_{\mathcal F}=(W_{n_k}-W)^\top E[XX^\top|\mathcal F](W_{n_k}-W) \to 0 \quad \text{as } k\to \infty $$ $P$-almost surely. Hence
$$\|Y-W^\top X\|_{\mathcal F}\leq\|Y-Y_{n_k}\|_{\mathcal F} +\|Y_{n_k}-W^\top X\|_{\mathcal F}\to 0  \quad \text{as } k\to \infty $$
$P$-almost surely, which yields $\|Y-W^\top X\|_{\mathcal F}=0$ $P$-almost-surely. This implies $Y=W^\top X$ $P$-almost surely, and so $(Y_n)$ converges conditionally to $W^\top X \in \mathcal S$.
