Notation: $\forall n \in \mathbb{N}$, consider the real-valued random variables $$ X_{n,1}, X_{n,2}, X_{n,3}, ..., X_{n,n} $$ We can picture them as a sort of "infinite triangle'' $$ A\equiv \begin{cases} X_{1,1}\\ X_{2,1} & X_{2.2}\\ X_{3,1} & X_{3,2} & X_{3,3}\\ X_{4,1} & X_{4,2} & X_{4,3} & X_{4,4}\\ ...\\ \end{cases} $$
$\forall n \in \mathbb{N}$, consider the real-valued random variables $$ Y_{n,1}, Y_{n,2}, Y_{n,3}, ..., Y_{n,n} $$ We can picture them as a sort of "infinite triangle'' $$ B\equiv \begin{cases} Y_{1,1}\\ Y_{2,1} & Y_{2.2}\\ Y_{3,1} & Y_{3,2} & Y_{3,3}\\ Y_{4,1} & Y_{4,2} & Y_{4,3} & Y_{4,4}\\ ...\\ \end{cases} $$
Assumption 1: all the random variables in $A$ are i.i.d.; all the random variables in $B$ are i.i.d.
Assumption 2: $\forall i\in \mathbb{N}$ $$ X_{n,i}-Y_{n,i}\rightarrow_{a.s.} 0 \hspace{1cm}\text{ as $n\rightarrow \infty$} $$
Question: are Assumptions 1,2 sufficient to claim $$ \frac{1}{n}\sum_{i=1}^nX_{n,i}- \frac{1}{n}\sum_{i=1}^n Y_{n,i}\rightarrow_{a.s.}0 \text{ as $n\rightarrow \infty$} $$ If not, which other conditions would be sufficient?
I am assuming that expectations are finite. If all the $X_{n,i}$ are all i.i.d. (even when $n$ increases), then $\frac{1}{n}\sum_{i=1}^n X_{n,i}\to \mathbb{E}[X_{1,1}]$ by the strong law of large numbers. Similarly, $\frac{1}{n}\sum_{i=1}^n Y_{n,i}\to \mathbb{E}[Y_{1,1}]$ and therefore $$ \frac{1}{n}\sum_{i=1}^n X_{n,i} - \frac{1}{n}\sum_{i=1}^n Y_{n,i} \to \mathbb{E}[X_{1,1}] - \mathbb{E}[Y_{1,1}] $$ almost surely. It remains to understand whether the hypothesis "$X_{n,i}-Y_{n,i}\to 0$ almost surely" implies that $\mathbb{E}[X_{1,1}] = \mathbb{E}[Y_{1,1}]$ but this is true because the $X_{n,i}$'s are i.i.d. (to see this, for instance, observe that almost sure convergence implies converge in distribution and since the $X_{n,i}$'s are i.i.d. $ \mathbf{1}_{\{t\ge 0\}} = \lim_{n} P(X_{n,1} -Y_{n,1} \le t) = P(X_{1,1} -Y_{1,1} \le t) $, which gives $\mathbb{E}[X_{1,1} - Y_{1,1}] = 0$).