I believe I understood the concepts of weak and strong solution of a SDE. But I don't understand why should I care about strong solution. Let me put it more precisely.
Suppose a have a Brownian motion $B$ defined on a filtered proabibility space $(\Omega, \mathcal{F}, (\mathcal{F_t})_{t>0},\mathbb{P})$. Suppose I have the following SDE
$$dX_t=\mu(X_t)dt+\sigma(X_t)dB_t,$$
where $\mu(\cdot)$ and $\sigma(\cdot)$ are integrable functions.
Now suppose I have a weak, but not strong solution to this SDE. Then, if I understood correctly, having a weak solution means that if I enrich my probability space by replacing the original filtration $(\mathcal{F_t})_{t>0}$ by some richer admissible filtration for the Brownian motion $B_t$, say $(\mathcal{\hat{F}_t})_{t}$, then I would get a strong solution on the filtered probability space $(\Omega, \mathcal{F}, (\mathcal{\hat{F}_t})_{t>0},\mathbb{P})$. Importantly, the new filtration is still admissible, and therefore, it doesn't "rely on future information".
Therefore, whenever I have a weak I solution, but not strong, all I should do is assume a filtration that is "rich enough"? I guess I must have got something wrong.