I have encountered an $L^2$ limit, which I am not sure of:
Let $B_t$ be a Brownian motion and $a=t_0<t_1<\dots <t_n=b$. Show that
$$\lim_{max(t_{i+1}-t_i)\mapsto 0} \sum_{0\leq i \leq n-1} (B_{t_{i+1}} - B_{t_i})^2 = b-a \hspace{1cm}\text{ in } L^2.$$
Is there an elegant way to show this?
Let $\Pi=\{a=t_0<t_1 < \ldots < t_n=b\}$ a partition of $[a,b]$ and $$S_2^{\Pi}(B) := \sum_{0 \leq i \leq n-1} (B_{t_{i+1}}-B_{t_i})^2$$ Then $$\mathbb{E}S_2^{\Pi}(B) = \sum_{0 \leq i \leq n-1} (t_{i+1}-t_i) = b-a$$
since $B_{t_{i+1}}-B_{t_i} \sim N(0,t_{i+1}-t_i)$. Thus
$$\begin{align} \mathbb{E}\big((S_2^{\Pi}(B)-(b-a))^2 \big) &= \text{var}(S_2^{\Pi}) = \sum_{j=0}^{n-1} \underbrace{\text{var}((B(t_{j+1})-B(t_j)^2)}_{\mathbb{E} \bigg( \big( (B(t_{j+1})-B(t_j))^2-(t_{j+1}-t_j) \big)^2 \bigg)} \tag{1} \\ &= \sum_{j=0}^{n-1} \mathbb{E} \bigg( \big( B(t_{j+1}-t_j)^2-(t_{j+1}-t_j) \big)^2 \bigg) \\ &\stackrel{B_t \sim \sqrt{t} \cdot B_1}{=} \sum_{j=0}^{n-1} (t_{j+1}-t_j)^2 \cdot \underbrace{\mathbb{E}((B_1^2-1)^2)}_{2} \\ &\leq 2 \cdot \max_j (t_{j+1}-t_j) \cdot (b-a) \to 0 \qquad \text{as} \, \max_i (t_{j+1}-t_j) \to 0 \end{align}$$
where we used the independence of the increments in $(1)$.