Brownian time inversion

For any finite positive times $t_1,\dots ,t_n$ and coefficients $u_j$, the linear combination ${\sum }_{j=1}^n{u}_j{Y}_{t_j}={\sum }_{j=1}^n{u}_j{t}_j{B}_{1\mathrm{/}{t}_j}$ is normal by [F1]. Appending any occurrences of $t=0$ only appends deterministic zero coordinates. Hence $Y$ is a centered Gaussian process, including repeated-time and singular finite vectors.

For $0<s\le t$, $\mathrm{Cov}(Y_s,Y_t)=st\mathrm{Cov}(B_{1\mathrm{/}s},B_{1\mathrm{/}t})=st\min (1\mathrm{/}s,1\mathrm{/}t)=s\mathrm{.}$ If $s=0$, both sides of the required identity are zero. Thus $\mathrm{Cov}(Y_s,Y_t)=\min (s,t)$ for all $s,t\ge 0$. By [F2], $Y$ and $B$ have the same finite-dimensional laws.

Let $I=\mathbb{Q}\cap (0,\mathrm{\infty })$ and define ${\mathrm{\Phi }}_B,{\mathrm{\Phi }}_Y:\mathrm{\Omega }\to {\mathbb{R}}^I$ by their coordinates. Each map is measurable for the cylinder sigma-algebra: the sets whose inverse images are measurable form a sigma-algebra containing every coordinate inverse image, hence all finite-coordinate cylinders and their generated sigma-algebra. Their pushforward laws ${\lambda }_B,{\lambda }_Y$ are therefore probabilities by [F3]. Step 2.1 makes them equal on every finite-coordinate cylinder. The class of cylinder-measurable sets on which they agree is a lambda-system by normalization, nested finite differences, and continuity from below; [F3]--[F4] therefore give ${\lambda }_B={\lambda }_Y$.

In ${\mathbb{R}}^I$ put $H={\bigcap }_{m\ge 1}{\bigcup }_{N\ge 1}{\bigcap }_{q\in I,0<q<1\mathrm{/}N}\{x:\mathrm{\mid }x(q)\mathrm{\mid }<1\mathrm{/}m\}\mathrm{.}$ This is cylinder-measurable because all three index sets are countable by [F5]. The event $A_0=A\cap \{B_0=0\}$ has probability one by [F1] and [F6]. On $A_0$, continuity of $B$ at zero gives ${\mathrm{\Phi }}_B\in H$, so ${\lambda }_B(H)=1$. Equality from step 3.1 gives $P({\mathrm{\Phi }}_Y\in H)={\lambda }_Y(H)=1$.

On $A$, $t\mapsto Y_t=t{B}_{1\mathrm{/}t}$ is continuous for every $t>0$. Fix $\omega \in A$ with ${\mathrm{\Phi }}_Y(\omega )\in H$ and $\epsilon >0$. Choose $m$ with $1\mathrm{/}m<\epsilon$ by [F5], and then $N$ from the definition of $H$. For every $t\in (0,1\mathrm{/}N)$, fix an enumeration of the rationals and, for each integer $k\ge 1$, let $q_k$ be its least-indexed member of $I\cap (0,1\mathrm{/}N)$ within $1\mathrm{/}k$ of $t$; density in [F5] makes this canonical sequence converge to $t$. Continuity gives $Y_{q_k}(\omega )\to Y_t(\omega )$. Since $\mathrm{\mid }{Y}_{q_k}(\omega )\mathrm{\mid }<1\mathrm{/}m$ for all $k\ge 1$, we get $\mathrm{\mid }{Y}_t(\omega )\mathrm{\mid }\le 1\mathrm{/}m<\epsilon$. Therefore $Y_t(\omega )\to 0=Y_0(\omega )$ as $t\downarrow 0$. By [F6], the set of such $\omega$ has probability one, so $Y$ has one common continuity event on $[0,\mathrm{\infty })$.

Steps 1.1--2.1 give the centered Gaussian covariance characterization, and step 5.1 gives path continuity; [F1] therefore makes $Y$ standard Brownian motion. The formula at $t=0$ is separately defined because $1\mathrm{/}t$ is unavailable there. Empty finite lists are vacuous, singleton and repeated-time laws were included in step 1.1, and AC is used only through [F1]--[F2], not in the fixed countable rational argument.