Pointwise Convergence of $Z_n \rightarrow 0$

Question

Let $\{W_t\}_{t\ge 0}$ be a standard Brownian motion. How can I show that

$Z_n=\sum_{j=0}^{n-1} (t_{j+1}-t_j)(W_{j+1} - W_j)$ converges to $0$ pointwise? [i.e., $Z_n \rightarrow 0$ pointwise],

which was defined from each partition $\pi_n=\{t_0,t_1,\dots,t_n\}$ of the interval [$0,t$] where $0=t_0<t_1<\dots<t_n=t$. Denoted by $\|\Delta_n\|=\max(t_{j+1}-t_j)$.

Assuming that in the sequence $\pi_1,\pi_2,...$ of partitions, the number of sub-intervals increases to $+\infty$ and that $\|\Delta_n\| \to 0$.

Answer 1

Note that $$|Z_n| \leq \sup_{\substack{|u-v| \leq \|\Delta_n\|\\ u,v \in [0,t]}} |W_u-W_v| \sum_{j=0}^{n-1} = t \sup_{\substack{|u-v| \leq \|\Delta_n\|\\ u,v \in [0,t]}} |W_u-W_v|.$$ Since the Brownian motion has continuous sample paths, we know that $s \mapsto W_s(\omega)$ is uniformly continuous on compact sets for any $\omega \in \Omega$. Thus,

$$|Z_n| \leq t t \sup_{\substack{|u-v| \leq \|\Delta_n\|\\ u,v \in [0,t]}} |W_u-W_v| \xrightarrow[]{n \to \infty} 0.$$