I am working through the exercises in Oksendal's SDE book. I am working on Exercise 3.2 which asks to show that $\int_0^t B_s^2 \,dB_s = \frac{1}{3}B_t^3 - \int_0^t B_s \,ds$ using the definition of the Ito integral.
I was able to do it but I am looking at other solutions to see if there are shorter approaches. In this solution manual, it says $$I\le \left[\sum_{j=1}^n (B_{j/n}-B_{(j-1)/n})^2\right]\max_{1\le j\le n}\left|B_{j/n}-B_{(j-1)/n}\right|\to 0\quad \text{a.s.}\,,$$ where $I = \sum\limits_{j=1}^n (B_{j/n} - B_{(j-1)/n})^3$. I find the mode of convergence here confusing.
In Oksendal, the Ito integral is defined as the $L^2$ limit and while I know it can be defined as limit in probability, I'd prefer to stick to limit in $L^2$ for now. The goal is to show that $I \rightarrow 0$ in $L^2$. I am able to do this using the definition of $L^2$ norm.
In the solutions manual, it is not clear to me why the 1) above quantity converges to $0$ a.s. and even if that's the case, 2) what does this imply for $L^2$ convergence?
Here are my thoughts:
I know that $\sum\limits_{j=1}^n (B_{j/n} - B_{(j-1)/n})^2 \rightarrow t$ in $L^2$. I also know that by continuity, $\max |B_{j/n} - B_{(j-1)/n}| \rightarrow 0$ a.s. Because $L^2$ and a.s. convergence imply convergence in probability, then the product will also converge to 0 in probability. But why does the solution say that it converges to 0 a.s.?