= Solution
To establish the <Brownian weighted quadratic variation in L2> for the deterministic <partition of an interval>, decompose
$$
\sum_jB_j(\Delta B_j)^2-\int_0^tB_s\,ds
=U_\pi+V_\pi,
$$
where $U_\pi=\sum_jB_j((\Delta B_j)^2-\Delta t_j)$ and $V_\pi=\sum_jB_j\Delta t_j-\int_0^tB_sds$.
The summands in $U_\pi$ are <martingale differences>: $B_j$ is $\mathcal F_{t_j}$-<measurable>, and the following <Brownian increment> is <independent> of $\mathcal F_{t_j}$. Their conditional <means> are zero, so distinct summands are <orthogonal> in <L2 space>. Since $\mathbb EB_j^2=t_j$ and $\mathbb E[((\Delta B_j)^2-\Delta t_j)^2]=2(\Delta t_j)^2$,
$$
\mathbb EU_\pi^2=2\sum_jt_j(\Delta t_j)^2
\leq2t^2|\pi|\longrightarrow0.
$$
For the <Riemann sum> error, the <Cauchy-Schwarz inequality> gives
$$
\mathbb EV_\pi^2
\leq t\sum_j\int_{t_j}^{t_{j+1}}\mathbb E|B_j-B_s|^2ds
=\frac t2\sum_j(\Delta t_j)^2
\leq\frac{t^2|\pi|}{2}\longrightarrow0.
$$
Thus the triangle inequality in <L2 space> proves
$$
\boxed{\sum_jB_j(\Delta B_j)^2
\longrightarrow\int_0^tB_s\,ds\quad\text{in }L^2(\mathbb P).}
$$
The time integration variable is $s$.
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