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The variance at zero is zero, so $X_0=0$ almost surely. For $s<t$, the increment is Gaussian with mean zero and varianceFor every $u\le s$,The increment and any finite vector of past values are jointly Gaussian. Zero cross-covariances imply their independence, by the factorization of the joint Gaussian characteristic function. The Monotone class theorem extends this independence to the sigma-field generated by all past values, ${\mathcal{F}}_s^X=\sigma (X_u:u\le s)$.

Together with the assumed continuity, these proveThis is the Gaussian-process characterization of Brownian motion. Completion of the natural filtration preserves the independence assertion.

For every finite collection of positive times, $(B_{t_1},\dots ,B_{t_m})$ is a linear transformation of the corresponding values of $W$, so $B$ is a centered Gaussian process. If $0<s\le t$, thenThus $B$ has the covariance of Brownian motion.

Its paths are continuous for $t>0$. As $t\downarrow 0$, put $u=1/t$. The given almost-sure limit givesThe Gaussian-process characterization of Brownian motion now proves the time inversion of Brownian motion:

Part (c), iterated over disjoint nested annuli, gives independent increments, and the law $X_s-X_t\stackrel{d}{=}{X}_{s-t}$ gives stationary increments. Every finite vector is jointly Gaussian by the definition of the Zero-boundary Gaussian free field, and a continuous version was assumed. Moreover $X_0=0$, because the field has zero boundary values.

Thus $X$ is a continuous centered Gaussian process with stationary increments and independent increments. Its variance is a continuous additive function on the nonnegative real numbers, so $\mathrm{Var}(X_t)={\sigma }^{2}t$ for some ${\sigma }^2\ge 0$. The Gaussian-process characterization of Brownian motion now givesfor standard Brownian motion $B$. This is the Circle-average process of the Gaussian free field is Brownian motion.

The process $Y_t=B_{T-t}-B_T$ is centered Gaussian and continuous. Its covariance isas follows by expanding Brownian covariances, or by reading its increments backwards. The Gaussian-process characterization of Brownian motion therefore shows that $(Y_t{)}_{0\le t\le T}$ has the same law as $(B_t{)}_{0\le t\le T}$.