I am not familiar with laws of large numbers (LLN) and I have a question on whether some LLN is applicable to the following setting:
Assumption (*): Let $i\in I\equiv \{1,...,n\}$ and $j\in J\equiv \{1,...,m\}$. Consider the following random variables defined on a probability space $(\Omega, \mathcal{F}, P)$: $X_{ij}, Z_{ij}, X_i, Z_j$ $\forall i\in I$, $\forall j \in J$ (for a total of $2nm+n+m$ random variables).
Assume $\Big(X_{ij}, Z_{ij}, X_i, Z_j\text{ }\forall i\in I \text{ } \forall j \in J\Big)$ are i.i.d. with standard normal cdf.
$\forall i\in I$, $\forall j \in J$, define the random variable $$ W_{ij}\equiv X_{ij}+Z_{ij}+X_{i}+Z_{j} $$ By Assumption (*), $W_{ij}\sim N(0,4)$ and we denote its cdf by $\Phi$.
We can see that $\Big(W_{ij}\text{ }\forall i\in I \text{ } \forall j \in J\Big)$ are identically distributed but not mutually independent (e.g., $W_{11}=X_{11}+Z_{11}+X_{1}+Z_{1}$ and $W_{12}=X_{12}+Z_{12}+X_{1}+Z_{2}$ which are clearly not independent).
Define $\hat{F}_{nm}(x)=\frac{1}{nm}\sum_{i,j\in I\times J}1\{W_{ij}\leq x\}$ for any $x\in \mathbb{R}$.
Since $\Big(W_{ij}\text{ }\forall i\in I \text{ } \forall j \in J\Big)$ are not i.i.d., we cannot use the strong LLN to say that $\hat{F}_{nm}(x)\rightarrow_{a.s.}\Phi(x)$ as $nm\rightarrow \infty$.
Still, does Assumption (*) tell us something about convergence (almost surely or in probability) of $\hat{F}_{nm}(x)$ as $nm\rightarrow \infty$?
Here are some ideas. Hopefully, you will be able develop them into a complete proof.
I will work in a simplified setup not involving $X_{ij}$ and $Z_{ij}$ (which btw can be combined into one variable). Consider sums of the form $$ S_{n,m}(f) = \frac{1}{mn} \sum_{i=1}^n \sum_{j=1}^m f(X_{i},Z_{j}), $$ where $f$ is a bounded measurable function (the boundedness assumption can be relaxed to the intgrability of $f(X_1,Z_1)$ but let's keep it simple). When $f(x,z) = \mathbf{1}_A(x) \mathbf{1}_B(z)$, we have by SLLN that $S_{n,m}(f) \to \mathbb{P}(X_1\in A)\mathbb{P}(Z_1\in B) = \mathbb{E}[f(X_1,Z_1)]$, $n,m\to \infty$, a.s. Then the same is true for finite linear combinations of such products. Now use a (functional) monotone class argument to conclude that this is true for any bounded measurable function, in particular, for $f(x,z) = \mathbf{1}_{x+z\le t}$.