A vectorial subspace $\mathcal{H} \subset L^2(\Omega,\mathcal{F},\mathcal{P})$ is called Gaussian space if every element is a gaussian distributed random variable.
Already proven :
Taken $X_1,\ldots,X_n \in \mathcal H \Rightarrow (X_1,\cdots,X_n)$ is gaussian.
Prove that the closure of a Gaussian space is a Gaussian space.
I'd like to proof the following : If $\mathcal{H}$ is a Gaussian space the closure of $\mathcal{H}$ with respect to the convergence in probability and $L^2$ are the same.
My proof goes and interrupts as follows : We already know that if $X_n \overset{L^2}{\longmapsto} X$ then $X_n \overset{P}{\longmapsto} X$ which implies thanks to the previous points that $\overline{\mathcal{H}}^{L^2} \subset \overline{\mathcal{H}}^{P}$.
Conversely suppose that $X_n \overset{P}{\longmapsto} X$, using that $E[\lvert X \rvert ^p] = \int_0^{+\infty} P(X > t)pt^{p-1} dt$
We have that $$\lvert \lvert {X_n - X}\rvert \rvert_{2}^{2} = E[\lvert {X_n - X}\rvert^2] = 2 \int_0^{+\infty} t P(\lvert X_n - X \rvert > t)dt$$
Noting that the integrand goes pointly to $0$ I'd like to pass the limit under the integral sign.
From here I don't how to find inequalities to bound $t P(\lvert X_n - X \rvert > t)$ and applying dominated convergence theorem (or other limit under integral sign-theorems), to use the fact that $X_n \overset{P}{\longmapsto} X$.
I could also use a strengthening of dominated convergence theorem which says : $X_n \overset{P}{\longmapsto} X$ and $Y \in L^p (p \in [1,+\infty)) : \lvert X_n \rvert \leq Y a.e \Rightarrow X_n \overset{L^p}{\longmapsto} X$ and $X \in L^P$ but the problem remains how to bound $X_n$ for me.
I'm noting that the hypothesis of $X_n,X$ being gaussian, i.e. $X_n, X \in \mathcal{H}$ hasn't been used yet.
Any help or hint would be appreciated, thanks.