Piecewise Linear approximation of Brownian Motion I'm looking to show that if you have a partition $0 = t_0 < \cdots < t_n = T$, then for any $t \in [0,T]$, $\lim_{n \to \infty} W^{(n)}(t) \to W(t)$ where $W(t)$ is standard Brownian motion. $W^{(n)}(t)$ is a piecewise continuous linear approximation such that
$$
W^{(n)}(t) = W(\mathrm{t}_i) + \left(W(\mathrm{t}_{i+1})-W(\mathrm{t}_i)\right) \frac{t-\mathrm{t}_i}{\mathrm{t}_{i+1}-\mathrm{t}_i}, \quad t \in \left[\mathrm{t}_i, \mathrm{t}_{i+1}\right)
$$
I'm struggling a bit here, if $t$ is taken on each $t_i$, the boundary of the partition, then everything cancels; however, for the $t \in (t_i, t_{i+1})$, I don't know how to proceed here.
Any help is appreciated, thanks.
 A: The function $W:[0,+\infty)\to\mathbb R$ is continuous.
Continuous functions on compact sets are uniformly continuous. The domain of this function is not compact, but consider $[0,a]$ when $a>0.$
For $a>0$ and $\varepsilon>0,$ there exists $\delta>0$ such that for every $s_1,s_2\in[0,a],$ if $|s_1-s_2|<\delta$ then $|W(s_1)-W(s_2)|<\varepsilon.$
(The reason for bounding this interval $[0,a]$ at a finite number $a$ rather than at $+\infty$ is that a value of $\delta$ that is small enough on such a bounded interval may not be small enough on the whole line. Do values of $\delta$ that are small enough go down to $0$ as $a$ grows? I.e. with Brownian motion does that happen? At this moment I don't know and my mind is on other things than answering that.)
But the statement above that begins with "For $a>0$ and..." is also true of $W^{(n)},$ the piecewise linear approximation. So let $\delta$ be small enough for $W$ and also small enough for $W^{(n)}.$ Then make $n$ big enough so that $\mathrm{t}_{i+1}-\mathrm{t}_i<\delta.$ Then $W^{(n)}(t)$ and $W(t)$ both differ from the values of $W$ at either of the two endpoints by less than $\varepsilon;$ hence from each other by less than $2\varepsilon.$ That should do it.
