Why do we define the Brownian motion asociated to a Filtration? In every book that I've seen, the authors define the Brownian motion with respect to a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ where $\mathcal{F}$ is a filtration. Also, I have seen that the natural filtration induces a brownian motion and that every filtration can be extended to one with usual conditions.
So I want to know what's the role of the filtration when defining the Brownian motion. Why do we need/use it?
 A: Filtrations keep track of information and the information available to us is not always the information generated by a particular Brownian motion.
For example, let's suppose that $W_t$ is a Brownian motion and $\mathcal{F}_t^W = \sigma(W_s : s \leq t)$ is its natural filtration. This encodes the information we get by observing the Brownian motion. What happens to the process $W$ if we are told the value of $W_1$? This changes the information available to us at time $t \leq 1$ to be $\mathcal{G}_t = \mathcal{F}_t \vee \sigma(W_1)$ (i.e. the $\sigma$ algebra generated by $\mathcal{F}_t$ and $W_1$).
The process $W_t$ has a semi-martingale decomposition with respect to the filtration $\mathcal{G}_t$ given for $t<1$ by (see Section 2.2.1)
$$
W_t = \underbrace{W_t - \int_0^t \frac{W_1 - W_s}{1-s}ds}_{B_t} + \int_0^t\frac{W_1-W_s}{1-s}
$$
where $B_t$ is $\mathcal{G}_t$ Brownian motion for $t \leq 1$. This is one way (in my mind the "right way") of constructing the Brownian Bridge.
The point here is that $W_t$ and $B_t$ are both Brownian motions, but in different filtrations, where different pieces of information are available to us. Clearly, though, $W_t$ is not $\mathcal{G}_t$ Brownian motion (and it shouldn't be! We know its value at time $1$ in $\mathcal{G}_t$!) and so the property of "being Brownian Motion" is always with reference to some filtration.
You might now be wondering why we don't just throw away $\mathcal{G}_t$ when studying the $B$ above and work with its natural filtration. A good reason is that if we did so, we would throw away an important piece of information, namely that $\sigma(W_1) \subset \mathcal{G}_0$, which implies by Blumenthal's 0-1 law (really it's just the Markov property plus the fact that $B_0 =0$) that $B$ is independent of $W_1$. This means among other things that in the SDE describing $W$ in $\mathcal{G}_t$, we can treat $W_1$ as a constant. The fact that $B$ is independent of $W_1$ is a key observation if you want to rigorously show that it makes sense to say that the law of the Brownian motion $W$ conditioned on $W_1 = 0$ is given by the SDE
$$
dW_t = \frac{-W_t}{1-t}dt + dB_t \\
W_0 = 0
$$
which you will often see as a definition of the Brownian Bridge.
