There is a claim from a paper which I do not understand:



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> Let $D$ be a domain in $\mathbb{R}^d$. Let $(p^{\eta})\_{\eta >0}$ be a family of densities for random variables on $(C[0,T], \mathbb{R}^d)$. Suppose that 
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> (i) For any bounded open domain $\mathcal{O}$ in $D$, the family of densities $(p^{\eta}|\_{\mathcal{O}})\_{\eta >0}$ is relatively compact in $L^2 ( [0,T], L^2 (\mathcal{O}))$, i.e. with the norm defined by 
>  $$ \| f \| = \bigg( \int\_0^T \| f\_t \|^2\_{L^2 (\mathcal{O})} \,dt \bigg)^{1/2}, \quad \quad \forall f \in L^2 ( [0,T], L^2 (\mathcal{O})).$$
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> (ii) The set $(P^{\eta})\_{\eta >0}$ of laws for the random variables in $(C[0,T], \mathbb{R}^d)$ corresponding to the above densities is tight.
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> Then it claims that we can select a subsequence $\eta\_k \rightarrow 0$ such that
>  the sequence of functions $p^{\eta\_k}$ converges a.e. on $[0,T] \times D$ and in $L^2([0,T], L^2 ( \mathcal{O}))$, for any bounded domain $\mathcal{O}$ of $D$, as $k \to \infty$.
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My main concern/problem about proving this is that the subsequences of densities converging in the $L^2([0,T], L^2 ( \mathcal{O}))$-norm depend on choice of $\mathcal{O}$ and there might be uncountably many $\mathcal{O}$ in general. How can we go about proving such a claim? Any ideas?



A diagonalisation argument applied to density functions
Suppose $\mathcal O\_1 \subseteq \mathcal O\_2$. If a sequence $\eta\_k$ converges in $L^2([0,T], L^2(\mathcal O\_2))$ to $\eta$, then the restrictions 
$R(\eta\_k) = \left.\eta\_k(t)\right|\_{\mathcal O\_1}$ also converge to $R(\eta) = \left. \eta(t)\right|\_{\mathcal O\_1}$ in $L^2([0,T], L^2(\mathcal O\_`))$, with $\|R(\eta\_k) - R(\eta)\| \le \|\eta\_k - \eta\|$. So it suffices to take a sequence of bounded open subsets of $D$ such that every bounded open subset of $D$ is contained in some member of the sequence.


