Under what conditions does $ \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n a_{i,n}= \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n a_{i,\infty}$ Suppose we have  sequence $\{a_{i,n} \}$ of non-negative numbers.
Consider the following two limits
\begin{align}
 L_1&=\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n a_{i,n}\\
 L_2&= \lim_{n \to \infty}   \frac{1}{n}  \sum_{i=1}^n a_{i,\infty}
\end{align}
where $a_{i,\infty} =\lim_{n \to \infty} a_{i,n}$.
My question: under what conditions does $L_1=L_2$?  Are there any good references for this or keywords?
 A: A sufficient condition is that $a_{i,n} \to a_{i,\infty}$ as $n \to \infty$ uniformly for all $i \in \mathbb{N}$.
Note that
$$|L_1- L_2| = \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n} +\frac{1}{n}\sum_{i=1}^n a_{i,n} - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}+\frac{1}{n}\sum_{i=1}^n a_{i,\infty}- L_2\right|\\ \leqslant  \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right|+ \left|\frac{1}{n}\sum_{i=1}^n a_{i,n} - \frac{1}{n}\sum_{i=1}^n a_{i,\infty} \right| \\ \leqslant \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right|+ \frac{1}{n}\sum_{i=1}^n |a_{i,n} - a_{i,\infty}|$$
Since $a_{i,n} \to a_{i,\infty}$ uniformly, there exists $N \in \mathbb{N}$  such that $|a_{i,n} - a_{i, \infty}|< \epsilon$ for all $i \in \mathbb{N}$ when $n > N$, and, hence,
$$ \frac{1}{n}\sum_{i=1}^n |a_{i,n} -  a_{i,\infty}| < \frac{1}{n}\sum_{i=1}^n\epsilon =\epsilon$$
Thus for all $n > N$, we have
$$|L_1- L_2|< \left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right| + \epsilon$$
and
$$|L_1- L_2| = \limsup_{n \to \infty} |L_1 - L_2| \\< \limsup_{n \to \infty}\left|L_1 - \frac{1}{n}\sum_{i=1}^n a_{i,n}\right| +  \limsup_{n \to \infty}\left|L_2 - \frac{1}{n}\sum_{i=1}^n a_{i,\infty}\right| + \epsilon = \epsilon$$
Since $\epsilon > 0$ can be arbitrarily small it follows that $L_1 = L_2.$
