Say that $W(t)$ is a Brownian motion. The quadratic variation $[W,W](t)$ is defined in terms of a partition $\Pi = \{0 = t_0 < t_1 < \cdots < t_n = t\}$ by
$ \begin{split} [W,W](t) &= \lim_{|\Pi|\to 0} \sum_{j=0}^{n-1} \Big( W(t_{j+1}) - W(t_j) \Big)^2\\ &= \lim_{|\Pi|\to 0} Q_n \end{split} $
Here $|\Pi|\to 0$ means that $\displaystyle\mathop{\text{max}}_{0\leq j
One can argue that $[W,W](t) = t$ by noting that the expected value and variance of the $j$th summand are $t_{j+1} - t_j$ and $2(t_{j+1} - t_j)^2$, respectively, so that $E[Q_n] = t$. One argues that $Var[Q_n]$ is proportional to the maximum partition length, and so vanishes in the limit. The quadratic variation thus converges in mean-square to $t$: $ \lim_{|\Pi|\to 0} E[(Q_n - t)^2] = 0. $
1) What additional arguments, if any, are needed to locate a subsequence of $\{Q_n\}$ so that the convergence is almost-sure?
2) How do you verify that the definition of the quadratic variation is independent of the choice of partition $\Pi$?