I understand that the Quadratic Variation of Brownian Motion $B_t$ is $[B_t,B_t]=t$ and I know that the equality is under the meaning of $\mathcal{L}^2$ convergence. Yet I saw in some book saying that $$[B_t,B_t]=t, \text{ a.s.}$$ and I'm just wondering if that is true or not? For one thing I know that $\mathcal{L}^2$ convergence always ensure a sub-sequence which converges almost surely. But that is just a sub-sequence, not the whole. Right?

In general the variation sums

$$\sum_{t_{j-1},t_j \in \Pi_n} |B(t_j)-B(t_{j-1})|^2$$

do not converge almost surely as $|\Pi_n| \to 0$, but there are sufficient coniditions to ensure the pointwise convergence, e.g.

1. The sequence $(\Pi_n)_n$ satisfies $\sum_n |\Pi_n| < \infty$. This follows rather directly from Borel-Cantelli's Lemma and Markov's inequality. Actually, one can show that $|\Pi_n| = O(1/\log n)$ is sufficient.
2. The sequence is nested, i.e. $\Pi_n \subseteq \Pi_{n+1}$.

(cf. René L. Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes.)

• Sorry, I don't get you. What I'm really wondering is if $[B,B]_t$ converges almost surely. I understand it converges in $L^2$ and hence a sub-sequence of the partition exists to ensure almost surely convergence. Yet what I'm confused is it is just a sub-sequence and so there is still some chance for $[B,B]_t$ to deconverge for some partition. Is that right? – user90846 Sep 14 '13 at 11:39
• @user90846 Yes, that's correct. I rewrote my answer, hopefully it's clearer now. – saz Sep 14 '13 at 12:00