Here I cite [Remark 4.1 p 169, Karatzas, Ioannis, and Steven E. Shreve. "Brownian motion." Brownian Motion and Stochastic Calculus. Springer, New York, NY, 1998. 47-127.]:
"Our first representation theorem involves the notion of the extension of a probability space. Let $X$ be an $\mathcal{F}_t$ adapted process on some $(\Omega,\mathcal{F},P)$. We may need a $d$-dimensional Brownian motion independent of $X$, but because $(\Omega,\mathcal{F},P)$ may not be rich enough to support this Brownian motion we need to extend the probability space to construct this."
Then he proceed with the construction of the extension: "Let $(\hat{\Omega},\hat{\mathcal{F}},\hat{P})$ be another probability space on which we consider a d-dimensional $\hat{\mathcal{F}}_t-$Brownian Motion $\hat{B}$. Set $\tilde{\Omega}=\Omega \times \hat{\Omega}$, $\tilde{\mathcal{F}}=\mathcal{F}\otimes \hat{\mathcal{F}}$, $\tilde{P}=P \times \hat{P}$, define a new filtration by $\tilde{\mathcal{F}}_t=\mathcal{F}_t \otimes \hat{\mathcal{F}}_t$ then augment it and make it right continuous.
We may also extend $X$ and $B$ by defining for $(\omega,\hat{\omega}) \in \tilde{\Omega}$ $$\tilde{X}_t(\omega,\hat{\omega})=X_t(\omega)$$ $$\tilde{B}_t(\omega,\hat{\omega})=B_t(\hat{\omega})"$$
My question is why do we need this extension? What does he mean with "$(\Omega,\mathcal{F},P)$ may not be rich enough to support this Brownian motion" ?
Thanks in advance.
Your probability space could be something like $\Omega = \{0\}$, in which case there aren't enough elements to define any non-trivial random variables, let alone something like a Brownian motion.