Why the convergence of $\sum_{k=0}^{n-1}B_{t_{i}}(B_{t_{i+1}}-B_{t_i})$ depend on the partition pointwise but not in $L^2$?

Question

Let $(B_t)$ a Brownian motion. Let $0=t_0<t_1<...<t_n=T$ a partition on $[0,T]$. I know that $$\lim_{n\to \infty }\sum_{k=0}^{n-1}B_{t_{i}}(B_{t_{i+1}}-B_{t_i})$$ doesn't depend on the partition in $L^2$, but it does depend on the partition pointwise. I don't understand how to prove that pointwise it depend on the partition. How can I find two sequences of partition $0=t_0^{(n)}<t_1^{(n)}<...<t_{m_n}^{(n)}=T$ and $0=s_0^{(n)}<s_1^{(n)}<...<s_{p_n}^{(n)}=T$ such that $$\sum_{k=0}^{m_n-1}B_{t_i^{(n)}}(B_{t_{i+1}^{(n)}}-B_{t_i^{(n)}})\quad \text{and}\quad \sum_{k=0}^{p_n-1}B_{s_i^{(n)}}(B_{s_{i+1}^{(n)}}-B_{s_i^{(n)}}),$$ converges a.s., but the limit are different ?

Answer 1

That's not possible ! If both of your sum converges a.s., then they will have the same limit. Indeed, they have the same limit in $L^2$. Now, a basic result of measure theory says that if $f_n(x)\to g(x)$ pointwise and $f_n(x)\to h(x)$ in $L^2$, then $g(x)=h(x)$ a.e.