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 $(\Omega, \mathcal{F}, P)$: $X_{ij}, Y_{ij}, X_i, Z_j$ $\forall i\in I$, $\forall j \in J$ (for a total of $2nm+n+m$ random variables).

Assume $\Big(X_{ij}, Y_{ij}, X_i, Z_j\text{ }\forall i\in I \text{ } \forall j \in J\Big)$ 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 $nm\rightarrow \infty$?