 Solution

Summing the exact cubic increment identity givesThe first sum tends to zero in L2 space. Indeed, odd moments of centered normal random variables vanish, a centered normal random variable of variance $v$ has sixth moment $15v^3$, and independence of the Brownian increments givesThe second sum converges in L2 space to $3{\int }_0^t{B}_{s}ds$ by (i).

For the last sum, first note that $\mathbb{E}{\int }_0^t{B}_s^{4}ds={\int }_0^{t}3s^{2}ds=t^3$, so $B_s^2$ is a square-integrable stochastic integrand. If $s=t_j+u$ and $D=B_s-B_j$, then $D$ is independent of $B_j$, with $D\sim N(0,u)$. HenceIt follows thatThe Itô isometry therefore makes ${\sum }_j{B}_j^2\Delta B_j$ converge in L2 space to ${\int }_0^t{B}_s^{2}d{B}_s$, with bounded truncations supplying the initial simple predictable processes if needed. Taking limits in the cubic identity provesThe proof uses only increment algebra, moments of normal random variables and the definition of the Itô integral.