For a $n$-dim Brownian motion $B_t=(B_t^1,\dots, B_t^n)$, I would like to ask how to derive the following result from the uniformly law of large number: as $n\to \infty$, the following converges to zero almost surely on $[0,T]^2$, $$ \sup_{(t,s)\in [0,T]^2}\left|\frac{1}{n}\sum B_t^iB_s^i-E[\frac{1}{n}\sum B_t^iB_s^i]\right|\to 0. $$
This means $$ \sup_{(t,s)\in [0,T]^2}\left|\frac{1}{n}\sum B_t^iB_s^i-E[B_t^1B_s^1]\right|\to 0 $$ and we take sup-norm. So $\frac{1}{n}\sum B_t^iB_s^i\to E[B_t^1B_s^1]$.
This question is enough to verify the condition of ULLN and I find one lecture note of this theorem: http://www.stat.cmu.edu/~arinaldo/Teaching/36709/S19/Scribed_Lectures/Apr23_Addison.pdf It seems that we need to check the Rademacher complexity is $o(1)$? How to check that?
Let us regard the Brownian motions $B^i$ as i.i.d. random elements in the space $\mathcal X = C([0,T])$ of continuous functions. Consider the function $f\colon \mathcal X \times [0,T]^2\to \mathbb{R}$ given by $$ f(x; t,s) = x(t) x(s), x\in \mathcal X, (t,s)\in [0,T]^2. $$ Then:
For all $x\in \mathcal X$, $f$ is continuous on $[0,T]^2$, and for all $(t,s)$, $f$ is measurable (moreover continuous) in $x$.
For all $x\in \mathcal X$, $(t,s)\in[0,T]^2$, $|f(x; t,s)|\le g(x) := \sup_{t\in [0,T]} |x(t)|^2$, and $$\mathbb{E}[g(B^1)] = \mathbb{E}\sup_{t\in [0,T]} |B_t^1|^2 <\infty.$$
Therefore, by the usual uniform law of large numbers,
$$ \sup_{(t,s)\in[0,T]^2}\left|\frac1n \sum_{i=1}^n f(B^i;t,s) - \mathbb{E}f(B^1;t,s)\right| = \sup_{(t,s)\in[0,T]^2}\left|\frac1n \sum_{i=1}^n B^i_t B^i_s - \min(t,s)\right|\to 0,\quad n\to\infty. $$