A diagonalisation argument applied to density functions

Question

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

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

(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 $$ \| f \| = \bigg( \int_0^T \| f_t \|^2_{L^2 (\mathcal{O})} \,dt \bigg)^{1/2}, \quad \quad \forall f \in L^2 ( [0,T], L^2 (\mathcal{O})).$$

(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.

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 $k \to \infty$.

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?

Answer 1

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.