$\sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2\xrightarrow{L^2}\frac{T}{2}$? I was reading this post that at some point explains why  $\sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2\xrightarrow{L^2}\frac{T}{2}$. Since the post is old, I thought it may be easier to ask the question here.
First, if we define $S_n:= \sum_{j=1}^n (B_{t_j}-B_{t_j^*})^2$, I am struggling to understand the equality $$\text{Var} \, (S_n)  = \sum_{j=1}^n \text{Var} \, \left[ (B_{t_j}-B_{t_j^{\ast}})^2 - (t_j-t_j^{\ast})\right]$$
Indeed, if $X$ and $Y$ are independent we have $Var(X+Y)=Var(X)+Var(Y)$, so wouldn't we get $\text{Var} \, (S_n)  = \sum_{j=1}^n \text{Var} \, \left[ (B_{t_j}-B_{t_j^{\ast}})^2\right]$?
Also I am struggling to understand ($|\Pi|$ is the size of the mesh) the bound $$E \left[ \left( \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 - \frac{T}{2} \right)^2 \right] =Var(S_n)\leq C|\Pi| T$$
Is it because $B_{t_j}-B_{t_j^*}\overset{d}{=}\sqrt{t_j-t_j^*}Z$ where $Z\sim\mathcal{N}(0,1)$ and so
\begin{align}
Var(S_n)&=\sum_{j=1}^n Var[(B_{t_j}-B_{t_j^*})^2]\\&=\sum_{j=1}^n Var[(t_j-t_j^*)Z^2]\\&=\sum_{j=1}^n (t_j-t_j^*)^2\underbrace{Var(Z^2)}_{=2}\le\sum_{j=1}^n |\Pi|(t_j-t_j^*)\cdot2 =2|\Pi|\underbrace{\sum_{j=1}^n t_j-t_j^*}_{\le T}
\end{align}
 A: Let be $ (\Omega, \mathcal{F}, \mathbb{P})$ a probability space on which we define a Brownian motion $ B $. Let $\{t_i\}_{i\in \{0,..,n\}}$ be a partition of $[0,T]$ and we note the middle point $t^*_i = \frac{t_i + t_{i-1}}{2}$.
We want to show that $\sum_i (B_{t_i} - B_{t^*_i})^2 \to \frac{T}{2}$ in $L^2(\Omega, \mathcal{F}, \mathbb{P})$ when the mesh size $\Pi \to 0$.
The idea is to first "center" the r.v. $\sum_i (B_{t_i} - B_{t^*_i})^2$: $E[\sum_i(B_{t_i} - B_{t^*_i})^2] = \sum_i E[(B_{t_i} - B_{t^*_i})^2]=\sum_i(t_i-t^*_i)=\frac{T}{2}$.
As such, we have:
\begin{align}
Var\left(\sum_i(B_{t_i} - B_{t^*_i})^2 - \frac{T}{2}\right) &= E\left[\left(\sum_i(B_{t_i} - B_{t^*_i})^2 - \frac{T}{2}\right)^2\right] - \underbrace{E\left[\left(\sum_i(B_{t_i} - B_{t^*_i})^2 - \frac{T}{2}\right)\right]^2}_{=0}\\
&=||\sum_i(B_{t_i} - B_{t^*_i})^2 - \frac{T}{2}||^2_{L^2}
\end{align}
We know that the Brownian increments are independent and stationnary. Thus,
\begin{align}
Var\left(\sum_i(B_{t_i} - B_{t^*_i})^2 - \frac{T}{2}\right) &= \sum_i Var((B_{t_i} - B_{t^*_i})^2) \\
&= \sum_i 2(t_i - t^*_i)^2 \\
&\leq 2 \sup_i |t_i - t_{i-1}| \sum_i (t_i - t^*_i)= \Pi T \xrightarrow[\Pi\to0]{}0  
\end{align}
In second equality we used that if $X\sim \mathcal{N} (0,\sigma^2)$, then $E[X^2] = \sigma^2$ and $E[X^4] = 3\sigma^4$.
To conclude, $||\sum_i(B_{t_i} - B_{t^*_i})^2 - \frac{T}{2}||^2_{L^2}  \to 0$ when $\Pi \to 0$.
