Quadratic variation of Brownian motion and almost-sure convergence 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 $\displaystyle\mathop{\text{max}}_{0\leq j<n} (t_{j+1} - t_j) \to 0$.
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 $\Pi$?

 A: I would like to comment on your 2nd question. The independence of $\langle M,M\rangle_T$ ($[W,W](T)$) of the partition you chose.
Let's see the fundamental steps from the beginning :


*

*Let $M$ be a martingale

*consider a sequence  $\Pi_n = \{t^n_0 ,\ldots, t^n_{p_n}\}$ of nested partitions of $[0,T]$ ($\Pi_{n} \subset \Pi_{n+1}$) such that $\big\vert \Pi_n \big\vert \underset{n \to \infty}{\longrightarrow} 0$  

*define $$X^n_t : = \sum_{i = 1}^{p_n} M_{t^n_{i-1}} (M_{t^n_{i-1}\wedge t} - M_{t^n_{i-1}\wedge t})$$
Note that $X^n_t$ is a martingale $(\mathbb{E} [X^n_t \vert \mathcal{F}_s] = X^n_s)$.
Note also that $M^n_T - X^n_T  = \sum_{i=1}^{p_n} (M_{t^n_i} - M_{t^n_{i-1}})^2$ (It may take a while)

*Prove that $\lim_{n,m \to \infty}{(X^n_T - X^m_T)^2} = 0$ (It's a bit hard)

*Use Doob's $L^2$ inequality and 4 (above)
$$\lim_{n,m \to \infty}\mathbb{E}[{\sup_{t \leq T}(X^n_t - X^m_t)^2}] \leq 4 \mathbb{E}[{\sup_{t \leq T}(X^n_T - X^m_T)^2}] \underset{n,m \to \infty}{\longrightarrow} 0$$

*There is a subsequence $X^{n_k}$ sucht that 
$$\mathbb{E}[{\sup_{t \leq T}(X^{n_k}_T - X^{n_{k+1}}_T)^2}] < 2^{-k} $$


*Consider the events $A_k = \{\omega: \sup_{t \leq T} \vert{X^{n_k}_t - X^{n_{k+1}}_t}\vert > \frac{1}{k^2}\}$
note (by Chebyshev's inequality) that $\mathbb{P} (A_k) \leq k^4 2^{-k}$
Note that $\sum_k k^4 2^{-k} < \infty$ 

*Now use Borel cantelli lemma to obtain that $\mathbb{P}(A_k i.o) = 0$(the event $\{A_k i.o.\}$ is the event $\{\omega : \;\{k:\omega \in A_k\}$ is an infinite set$\}$.

*Now there is a full measure set $\Omega^*$ ($\mathbb{P} (\Omega^*) = 1$) such that $\omega \in \Omega^* \Rightarrow \exists \;K_0(\omega) $, $k > K_0(\omega) \Rightarrow  \sup_{t \leq T} \vert{X^{n_k}_t - X^{n_{k+1}}_t}\vert < \frac{1}{k^2}$ therefore $ X^{n_k}_t$ converges uniformly (to a continuous function say $Y_t(\omega)$ on the interval $[0,T]$)

*note that $Y_t$ is the $L^2$ limit of $X^{n}$ this implies that $Y_t$ is also a martingale

*Note $$M^n_{t^n_j} - X^n_{t^n_j}  = \sum_{i=1}^{j} (M_{t^n_i} - M_{t^n_{i-1}})^2$$
Therefore conclude that $M^n_t - Y_t = A_t$ is a continuous increasing process (almost surely) and since $M^n_{T} - X^n_{T} = Q_n$ (in your notation) we obtain that $\mathbb{E}[(Q_n - A_T)^2] = \mathbb{E}[(X^n_T - Y_T)^2] \to 0$

*Now to the main point, what if you had chosen a different sequence of partitions? Then you might get a different limit increasing process say $A'_t$. But it suffices to observe that $A_t - A'_t = M_t - A'_t  - (M_t - A_t)$ is a martingale of bounded variation and therefore it is indistinguishable from 0 therefore $\{A'_t = A_t$ for all $t\}$ almost surely
And you conclude that the limit is independent of the partitions you chose.
For reference I used Le Gall's book (Mouvement Brownien, Martingales Et Calcul Stochastique - is in French). 
A: If the diameter of the $n$-th partition converges (to zero) fast enough, namely if it is of order less than $1/\log(n)$, then the quadratic variation converges almost surely. If not, then not in general (for un-nested partitions). More precisely, Wrobel (1982) shows that that there exists a sequence of partitions with diameters of order less than $1/\log(n)$ raised to the power $\alpha$, for every positive $\alpha$ less than $1$, such that the Brownian square variation diverges almost surely. 
For this and related results see also Stoyanov (2013, Sec. 24.8).
Stoyanov, J.M. (2013, 3rd edition): Counterexamples in probability. New York: Dover Publ.
Wrobel, A. (1982): On the almost sure convergence of the square variation of the Brownian motion, Probability and Mathematical Statistics, 3, 97-101. PDF from http://www.math.uni.wroc.pl/~pms/publications.php?nr=3.1
A: 1) Since your quantity converges in $L^2$ it converges in Probability, so a subsequence will converge a.s. and you need not impose any further conditions.
2) The limit does not depend on the partition, because if you choose any other partition with mesh going to 0, you would still have convergence in Probability to a certain limit (your proof works for any such partition) and thus you still have a subsequence converging a.s.
Thus, the two limits must be the same.
A: I dont know whether it is correct:
{Q: [0, T] is spaced equally, as usual notation:$\Delta B_i=B_{t_i}-B_{t_{i-1}}$, $t_i-t_{i-1}=T/n$.
 Show $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $}
$Proof$: 
for all $\varepsilon$,
$\mathbb{P}\left(\{\omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-t|>\varepsilon\}\right)$
$\leq  {\frac{E(\sum_{i=1}^{n}\Delta B_i^2-t)^4}{\varepsilon^4}}$(by Markov inequality)
$=12T^4(1/n^2+4/n^4)=O(1/n^2)$ .
then $\sum \mathbb{P}\left(\{ \omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-T|>\varepsilon\}\right)$ is a convergent series, satisfying the definition of c.c. (complete convergence),  
So $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $, that also implies $a.s.$ (or by Borel-Cantelli lemma).
Is there any mistake?  welcome someone to point it out. 
