I'd be astounded if this isn't a duplicate, but I've yet to find anything equivalent. Whenever I've seen the natural filtration of a Brownian Motion stated explicitly, it's been in the form of $\mathcal{F}_t=\sigma\{W_s:s\leq t\}$ for any $t\geq0$. In other words, the natural filtration of a Brownian Motion up to some point $t$ is the sigma-field generated by all of the Brownian Motions from every time not after $t$.
My question is, why is that last bit necessary? Why do we not just have $\mathcal{F}_t=\sigma\{W_t\}$ for any $t\geq0$? By definition $W_t$ is $\sigma\{W_t\}$-measurable and $W_t$ can clearly take all of the values that any $W_s$ (with $s\leq t$) can, so I can't see what is gained by adding the fields generated by the "earlier" random variables. Even in cases where we want to talk about some point $b\leq t$, I can't see why we would use $\mathcal{F}_b=\sigma\{W_a:a\leq b\}$ when we could use $\mathcal{F}_b=\sigma\{W_b\}$. Indeed, I can't see why these two fields would not be identical. Is there an obvious element that I'm missing?
I think I've got it, or at least something along the right lines.
The proposed filtration is not a filtration at all. If it was, then we would have $\mathcal{F}_s\subseteq\mathcal{F}_t$ for all $s\leq t$. However, this is clearly not the case. Consider $\{W_0=0\}\subseteq\mathcal{F}_0$ where $\mathcal{F}_0=\sigma\{W_0\}$. Clearly this set is in $\mathcal{F}_1$ if we take $\mathcal{F}_1=\sigma\{W_s:s\leq 1\}$ (and therefore we have $\mathcal{F}_0\subseteq\mathcal{F}_1$), but this is not the case if we take $\mathcal{F}_1=\sigma\{W_1\}$ (and therefore we have $\mathcal{F}_0\not\subseteq\mathcal{F}_1$), meaning that the proposed filtration is not a filtration.