L^2 convergence with support in closed subset Suppose that $X_n$ is a sequence of random-variables in $L^2(\Omega,\mathcal{F},\mathbb{P})$, where $(\Omega,\mathcal{F},\mathbb{P})$ is a complete probability space.
Further, suppose that there exists a proper closed subset $K$ of $\mathbb{R}$, such that for every $n \in \mathbb{N}$
$$
\mathbb{P}\left(
X_n \in K
\right)=1
$$
and that there exists $X\in L^2(\Omega,\mathcal{F},\mathbb{P})$ which is the limit of $X_n$, then does this imply that $$
\mathbb{P}\left(
X \in K
\right)=1
?
$$
 A: Many thanks to @астон вілла олоф мэллбэрг for their guidance:
Since $X_n\overset{L^2}{\mapsto} X$ then $X_n\overset{\mathbb{P}}{\mapsto} X$. Since the Ky-Fan topology is a metrization of the topology describing convergence in probability, then it is Hausdorff; hence the limiting random-variable $X$ is uniquely defined.  
Fix $k \in K$ (this is possible because $K$ is non-empty) and consider the modifications $\tilde{X}_n$ and $\tilde{X}$ defined by
$$
\begin{aligned}
\tilde{X}_n(\omega)\triangleq &\begin{cases}
X_n(\omega) & : (\forall k \in \mathbb{N}) X_k(\omega)\in K \mbox{ and } X(\omega)\in K\\
k & : else.
\end{cases}
\\
\tilde{X}(\omega)\triangleq &\begin{cases}
X(\omega) & : (\forall k \in \mathbb{N}) X_k(\omega)\in K \mbox{ and } X(\omega)\in K\\
k & : else.
\end{cases}
\end{aligned}
.
$$
Since $X$ is uniquely well-defined, then $\bigcup_{n \in \mathbb{N}}\{\omega \in \Omega: X_n(\omega) \not\in K\}\cup \{\omega \in \Omega: X(\omega) \not\in K\}$ is a countable union of probability $0$ subsets of $\Omega$; hence is itself of probability $0$.  Therefore the modifications are indeed well-defined.  
Since $X_n\overset{\mathbb{P}}{\mapsto} X$, then there exists a sub-sequence $\left\{\tilde{X}_{n_k}\right\}_{k \in \mathbb{N}}$ converging to $\tilde{X}$ $\mathbb{P}$-a.s.  Therefore the subset defined by
$$
\mathcal{X}\triangleq \bigcup_{k \in \mathbb{N},\omega\in\Omega} \tilde{X}_{n_k}(\omega),
$$
must contain $X(\omega) \in \overline(\mathcal{X})$, for every $\omega \in \Omega$.  Since $K$ is closed in $\mathbb{R}$, and $\mathcal{X}\subseteq K$, then $\overline(\mathcal{X})\subseteq K$.  
Therefore, $\tilde{X}$ is $\mathbb{P}$ in $K$ for every $\omega \in \Omega$.  Since $\tilde{X}$ is a modification of $X$, then $\mathbb{P}\left(X \in K\right)=1$!
