The raw natural Brownian filtration need not be right-continuous

Define $A:={\bigcup }_{m\ge 1}{\bigcap }_{s\in [0,1\mathrm{/}m]}\{{\pi }_s=0\}={\bigcup }_{m\ge 1}{\bigcap }_{q\in \mathbb{Q}\cap [0,1\mathrm{/}m]}\{{\pi }_q=0\}$, the second description using continuity of every path; thus $A$ is measurable and nonempty, since the zero path belongs to it, while the path $s\mapsto s$ does not.

For every $t>0$ one has $A\in {\mathcal{F}}_t^0$: choose $m$ with $1\mathrm{/}m<t$; then ${\bigcap }_{q\in \mathbb{Q}\cap [0,1\mathrm{/}m]}\{{\pi }_q=0\}\in {\mathcal{F}}_{1\mathrm{/}m}^0\subseteq {\mathcal{F}}_t^0$, and the union over $m$ lies in ${\mathcal{F}}_t^0$. Hence $A\in {\mathcal{F}}_{0+}^0={\bigcap }_{t>0}{\mathcal{F}}_t^0$ The Brownian germ sigma-algebra at zero.

$A\mathrm{\notin }{\mathcal{F}}_0^0=\sigma ({\pi }_0)$: the zero path is in $A$ and the path $s\mapsto s$ is not, while both have ${\pi }_0=0$, so membership in $A$ is not determined by ${\pi }_0$.

$\mathsf{W}(A)=0$: $A$ is contained in ${\bigcup }_{m\ge 1}\{{\pi }_{1\mathrm{/}(2m)}=0\}$, and each of those events has probability zero because the law of ${\pi }_{1\mathrm{/}(2m)}$ has the strictly positive density $p_{1\mathrm{/}(2m)}(0,\cdot )$ and is therefore atomless The Brownian kernels form a semigroup Standard normal and normal laws; countable subadditivity gives the claim Basic identities for a probability measure.

Combining steps 2.1, 2.2 and 2.3, $A\in {\mathcal{F}}_{0+}^0\setminus {\mathcal{F}}_0^0$, so the raw natural filtration is not right-continuous at time zero; its usual augmentation contains $A$ as a null event and is right-continuous by construction Natural and usual augmented Brownian filtrations. The witness is nonempty and of probability zero, so it is invisible to any probability computation alone. AC is used only through the ambient Brownian and Wiener interfaces.