Let $W(t)$ be a standard Brownian motion, and $\Delta_n(t)=W(\frac{T(m-1)}{n})$ for $t\in[\frac{T(m-1)}{n}, \frac{Tm}{n})$ for $T>0$ given. I'd like to show that $E\lim_{n\to\infty}\int_0^T |\Delta_n(t)-W(t)|^2\,dt$=0.
Let $\Omega_{h}=\{\omega:\forall t\in[0,T],\ B(t,w)$ is Holder continuous with Holder exponent $\gamma<\frac{1}{2}$}; we know $P(\Omega_{h})=1$. So the desired quantity is $E\lim_{n\to\infty}\int_0^T |\Delta_n(t)-W(t)|^2\,dt=\int_{\Omega_{h}}\lim_{n\to\infty}\int_0^T |\Delta_n(t)-W(t)|^2\,dt$.
For $\omega\in\Omega_{h}$, $m\ge1$, and $t\in[\frac{T(m-1)}{n},\frac{Tm}{n})$, Holder continuity (using $\gamma=\frac{1}{4}$) yields, for some $C>0$,
$|W(\frac{T(m-1)}{n}-W(t)|^2\le C^2|t-\frac{T(m-1)}{n}|^{1/2}=\frac{C^2}{n^{1/2}}.$
So $\int_{\frac{T(m-1)}{n}}^\frac{Tm}{n} |W(\frac{T(m-1)}{n})-W(t)|^2\,dt\le \frac{C^2}{n^{3/2}}$, and
$\int_0^T|W(\frac{T(m-1)}{n})-W(t)|^2\,dt=\lim_{n\to\infty}\sum_{m=1}^n\int_{\frac{T(m-1)}{n}}^\frac{Tm}{n} |W(\frac{T(m-1)}{n})-W(t)|^2\,dt\le$
$\lim_{n\to\infty}\sum_{m=0}^n \frac{C^2}{n^{3/2}}=\lim_{n\to\infty} \frac{C^2}{n^{1/2}}=0.$
I would greatly appreciate some confirmation that the proof is correct. Thanks in advance!