Brownian motion properties In my notes, the quadratic variation is defined as 
$$Q[f]=\lim_{N\to \infty} \sum_{i=0}^{N-1} |f_{t_{i+1}}-f_{t_{i}}|^2$$
Take time period $[0,T]$ with $N$ partitions, so $dt=T/N$; $t_i=idt$.
Now, for Brownian Motion $W_t$,
$$Q[W_t]=\lim_{N\to \infty} \sum_{i=0}^{N-1} |W_{t_{i+1}}-W_{t_{i}}|^2=\lim_{N\to \infty} \sum_{i=0}^{N-1} |\pm \sqrt{dt}|^2=\lim_{N\to \infty} Ndt=T$$
My firstquestion is why $W_{t_{i+1}}-W_{t_{i}}=\pm \sqrt{dt}$ ?
I konw that according o the definition of Brownian motion, 
$$dW_t=W_{t+dt}-W_t \text{ ~ } N(0,dt)$$
But this just imply that $Var[W_{t_{i+1}}-W_{t_{i}}]=dt$, so why the variance is dropped in the above? Or is this due to another reason?
And then it says, 
for the variation, we have:
$$Q[W_t]=\lim_{N\to \infty} \sum_{i=0}^{N-1} |W_{t_{i+1}}-W_{t_{i}}|=\lim_{N\to \infty} \sum_{i=0}^{N-1} |\pm \sqrt{dt}|=\lim_{N\to \infty} N \sqrt{dt} \to \infty$$
"The Quadratic Variation property implies ${dWt}^2=dt$."
My second question is where this implication come from?
Thanks for help!
 A: First of all some advice useful to keep in mind when you are dealing with math-related stuff: you have to be cautious about the real meaning of the notation.
In your case the first thing to notice is that $Q[W_t]=T$ doesn't make sense, since on the left of the equality you have a random variable and on the right a constant. The problem here is that you are defining $Q[W_t]$ as a limit but you are not saying in what sense you have to interpret this notation.
To be more concrete let's focus on the proof of your statement.
Let's fix some notation. Let $W:[0,T]\times\Omega\rightarrow\mathbb{R}$ be the standard Brownian motion, let $\pi:=\{0<\frac{T}{N}...<\frac{(N-1)T}{N}< T\}$ be your partition of $[0,T]$. We set $t_i=i\frac{T}{N}$, for $i=0,..,N$. We define the quadratic variation of $W$ with respect to the partition $\pi$ as $$S_{\pi}(W):=\sum_{i=1}^{N} (W_{t_{i}}-W_{t_{i-1}})^2.$$ What we want to  show is that $$\lim_{N\rightarrow\infty}S_{\pi}=T\ \ \ \  in \ \ \ L^2(\Omega),$$
which means that we have to show $$\lim_{N\rightarrow\infty}E[(S_\pi-T)^2].$$
Defining $Y_i=(W_{t_{i}}-W_{t_{i-1}})^2-(t_i-t_{i-1})$ we can write $S_\pi-T=\sum_{i=1}^{N}Y_i$. We observe that the random variables $Y_1,...,Y_N$ are independent since each $Y_i$ is a function of $W_{t_{i}}-W_{t_{i-1}}$ and by the definition of Brownian motion its increments are independent. Moreover
$$E[Y_i]=E[(W_{t_{i}}-W_{t_{i-1}})^2]-(t_i-t_{i-1})=Var(W_{t_{i}}-W_{t_{i-1}})-(t_i-t_{i-1})=0.$$
Using these two facts we obtain
$$E[(S_\pi-T)^2]=\sum_{i=1}^{N}E[Y_i^2].$$
We know that the distribution of $W_{t_{i}}-W_{t_{i-1}}$ is $N(0,t_{i}-t_{i-1})$ thus the distribution of $$Z:=\frac{W_{t_{i}}-W_{t_{i-1}}}{\sqrt{t_{i}-t_{i-1}}}$$ is $N(0,1)$. (Z doesn't depend on $i$ because we are interested only in its distribution and for every $i$ it is a standard normal). Using this fact we write
$$\sum_{i=1}^{N}E[Y_i^2]=\sum_{i=1}^N E[\ [(W_{t_{i}}-W_{t_{i-1}})^2-(t_{i}-t_{i-1})]^2\ ]=\\ \sum_{i=1}^N (t_{i}-t_{i-1})^2\  E[\ [\left( \frac{W_{t_{i}}-W_{t_{i-1}}}{\sqrt{t_i-t_{i-1}}}\right)^2-1]^2\ ]=\\ 
 \sum_{i=1}^N (t_{i}-t_{i-1})^2\  E[(Z^2-1)^2]=C\sum_{i=1}^N (t_{i}-t_{i-1})^2= \\ C \frac{T}{N}\sum_{i=1}^N t_{i}-t_{i-1}=C\frac{T^2}{N}$$ and the result follows for $N\rightarrow\infty$. $C:=E[(Z^2-1)^2]$ is a constant.
Hope this answer helps you clarify your totally justified doubts.
