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In Stochastic Calculus for Finance II: Continuous-time Models by Steve Shreve,

Theorem 3.4.3. Let $W$ be a Brownian motion. Then $[W, W](T) = T$ for all $T > 0$ almost surely.

where $[W, W](T)$ is the quadratic variation of $W$ up to time T $$ [W, W](T): = \lim_{||\Pi|| \to 0} \sum_{j=0}^{n-1} [W(t_{j+1}) - W(t_j)]^2 $$ where $$\Pi=[t_0, t_1, \dots, t_n] \text{ and } 0 \leq t_0 < t_1 < \cdots < t_n = T$$ and $$||\Pi||= \max_{j=0,\dots, n-1} (t_{j+1} -t_j).$$

One fellow student also says a stronger conclusion is also true, i.e.

$[W, W](T)$ converges to $T$ in $L^2$ or in $L^p, p>1$, for all $T > 0$.

By "stronger", I mean I heard it implies the original theorem for convergence a.e..

I wonder if there are some texts or online tutorials for proving this stronger conclusion, and/or if you could kindly provide the proof here?

Thanks!

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An $L^2$ convergence argument is given here: math.stackexchange.com/questions/59547/… –  Byron Schmuland Nov 7 '11 at 21:31
    
@ByronSchmuland: Thanks! Does the link talk about convergence in $L^2$ or a.e.? All I see is latter. –  Ethan Nov 7 '11 at 21:44
    
Sorry I wasn't clearer. In the problem description, before he gets to points 1) and 2), the OP gives an argument that concludes "The quadratic variation thus converges in mean-square to t". This is the $L^2$ convergence. –  Byron Schmuland Nov 7 '11 at 21:48
    
@ByronSchmuland: Thanks! In this case, does convergence in $L^2$ imply convergence in a.e., or the other way around, or neither implies the other? –  Ethan Nov 7 '11 at 22:26
    
Neither implies the other. –  Byron Schmuland Nov 7 '11 at 22:49

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