Brownian Motion $dW_t \, dt=0$ proof! I am facing a bit weird issue here. I am going through Shreeve book on stochastic calculus and faced the following theorem, while proving $dWdt=0$.
$\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_j))(t_{j+1}-t_j)$
$\le$ $\max_{0 \le k \lt n-1}  |(W(t_{j+1})-W(t_j))| \cdot \sum_{j=0}^{n-1}(t_{j+1}-t_j)$
Now he argued that the because $W$ is continuous , the first term in the above equation goes to zero as the time partition goes to infinity. Now the doubt I am having is that,
$(W(t_{j+1})-W(t_j))=y \cdot \sqrt {t_{j+1}-t_j}$, where $y$ is a standard normal random variable. Then the maximum of this term can be anything from $-\infty$ to $\infty$. Now you may argue that as time partition goes to infinity, the time difference goes to zero (so the square root term goes to zero), but even in this case it can be $0 \cdot \infty$ scenario as the random variable has positive probability, which would again be undefined.
Then how the author argued that the maximum of difference in Brownian terms is zero! Please help.
Best Regards,
 A: *

*Mind that for each fixed $\omega \in \Omega$, the mapping $ t\mapsto W_t(\omega)$ is continuous and therefore $$\sup_{s,t \in [0,T]} |W(t,\omega)-W(s,\omega)|$$ is always finite. Now, the author uses that the mapping is even uniformly continuous on compact intervals and therefore $$\sup_{s,t \in [0,T],|s-t| \leq \delta} |W(t,\omega)-W(s,\omega)| \to 0 \qquad \text{as} \, \, \delta \to 0.$$ 

*You mentioned the identity $$W(t_{j+1}) - W(t_j) = Y \sqrt{t_{j+1}-t_j}.$$ Mind that this equality holds in distribution. Since we are interested in the behavior for each fixed $\omega$ (i.e. in the path behavior of Brownian motion), this identity is of no use. (A similar problem: $W_t = \sqrt{t} W_1$ (in distribution) does not imply $W_t = \sqrt{t} W_1$ (almost surely).)

A: The idea here is that you first fix an $\omega$ such that $t\mapsto W_t(\omega)$ is continuous on a compact interval hence uniformly continuous in $t$. 
I think what's confusing you is that the convergence is not uniform in $\omega$, and this is indeed the case, but this is not what the writer is claiming. 
