Edit I think I solved my problem. Thinking about the application of LLN conditionally on R as:
$$\mathbb{P}\left(\frac{1}{n}\sum\limits_{i=1}^n g_{l,i}g_{l',i}\longrightarrow R_{l,l'}\bigg|R\right)=1,$$
then taking the expectation on both sides, this gives us the unconditioned probability equal to 1.
original question: I want to understand the proof of the Dovbysh-Sudakov theorem (this pdf in case someone is interested, but I think my description of the problem should suffice)
I am given some infinite symmetric, positive definite random matrix $R$ (actually it is weakly exchangeable but lets keep that aside). Then I can generate Gaussian vectors $g \in N(0, R)$ conditioned on R. Let $g_i$ be such a sequence of conditionally independent Gaussian vectors in $\mathbb{R}^\mathbb{N}$.
Some other theorem (Hoovers representation theorem) now tells us that we are able to represent these Gaussian vectors as
$(1)\hspace{60mm}(g_{li})_{li}\overset{d}{=}f(W,U_l,V_{l'},X_{l,l'}).$
where $$f:[0,1]^4\rightarrow\mathbb{R}\hspace{4mm}\text{ and }\hspace{4mm}W,(U_{l}),(V_{l'}),(X_{l,l'})\sim\mbox{Unif}[0,1]\hspace{3mm}\text{ i.i.d. }$$
Ok we are nearly done: Now we take two limits. For one we take the limit "conditionally on $R$" (LLN):
$$\frac{1}{n}\sum\limits_{i=1}^n g_{l,i}g_{l',i}\longrightarrow R_{l,l'}.$$
For the other we take the limit "conditionally on $W$ and $(U_l)$" (LLN): $$\frac{1}{n}\sum\limits_{i=1}^n f(W,U_l,V_i,X_{l,i})f(W,U_{l'},V_i,X_{l',i})\longrightarrow\mathbb{E}_{v,x}[f(W,U_l,V_1,X_{l,1})f(W,U_{l'},V_1,X_{l',1})],$$
where the expectation is only over $V_1$ and $(X_{l,1})_{l}$, since we conditioned on the other two elements. Till here I'm fine with the proof, but now it says: (1) implies that
$$R_{l,l'}\overset{d}{=}\mathbb{E}_{v,x}[f(W,U_l,V_1,X_{l,1})f(W,U_{l'},V_1,X_{l',1})],$$
i.e. the limits are distributional equal, although we conditioned the sum on different random elements.
I am very thankful for any idea what I should think about or what I should look up.