A deterministic integral construction of a Gaussian process

For $m\ge 1$ and $t\ge 0$, put $R_m(t)=\frac{t}{m}{\sum }_{k=0}^{m-1}{B}_{kt\mathrm{/}m}\mathrm{.}$ This is a measurable random variable by [F2]. On $A$, [F3] makes $R_m(t)(\omega )$ converge to the displayed Riemann integral as $m\to \mathrm{\infty }$; for $t=0$, every sum and the integral are zero. Define $X_t$ by this limit on $A$ and by zero on $A^c$. It is measurable: the convergence set and limsup of the measurable sequence are measurable by [F2], and pasting that finite limit on the measurable set $A$ with zero on its complement preserves every Borel inverse image. Thus the statement defines a real stochastic process rather than an integral that might be undefined on exceptional paths.

For $s,t>0$, covariance bilinearity and [F1] give $C_m(s,t):=\mathrm{Cov}(R_m(s),R_m(t))=\frac{st}{m^2}{\sum }_{k,\mathrm{\ell }=0}^{m-1}\min (ks\mathrm{/}m,\mathrm{\ell }t\mathrm{/}m)\mathrm{.}$ This is the lower-corner product-grid Riemann sum for the continuous function $(u,v)\mapsto \min (u,v)$ on $[0,s]\times [0,t]$. Its mesh tends to zero, so [F4] gives $C_m(s,t)⟶C(s,t):={\int }_0^s{\int }_0^t\min (u,v)dvdu\mathrm{.}$ If $s=0$ or $t=0$, both $C_m(s,t)$ and the declared degenerate-rectangle integral are zero, so the same conclusion holds.

Fix $n\ge 1$, times $t_1,\dots ,t_n\ge 0$, coefficients $a_1,\dots ,a_n$, and set $Y_m={\sum }_i{a}_i{R}_m(t_i)$ and $Y={\sum }_i{a}_i{X}_{t_i}$. By [F1], $Y_m$ is centered normal. Step 1.2 and covariance bilinearity show that its variance $V_m={\sum }_{i,j=1}^n{a}_i{a}_j{C}_m(t_i,t_j)$ converges to $V={\sum }_{i,j=1}^n{a}_i{a}_{j}C(t_i,t_j)\mathrm{.}$ On $A$, step 1.1 gives $Y_m\to Y$, hence the convergence is almost sure. In particular, $V={\lim }_m{V}_m\ge 0$.

Now let $0<s\le t$. Every section of $\min (u,v)$ is continuous, so [F4] writes $C(s,t)={\int }_0^s({\int }_0^{u}vdv+{\int }_u^{t}udv)du={\int }_0^s(ut-\frac{u^2}{2})du\mathrm{.}$ The polynomial antiderivatives justified by [F9] give $C(s,t)=\frac{t{s}^2}{2}-\frac{s^3}{6}=\frac{s^2(3t-s)}{6}\mathrm{.}$ For $s=0$ both the double integral and polynomial are zero by their endpoint conventions.

For each real $z$, [F6] gives $E[e^{iz{Y}_m}]=e^{-z^2{V}_m\mathrm{/}2}$. The left side converges to $E[e^{izY}]$ by [F7], because $e^{iz{Y}_m}\to e^{izY}$ almost surely and every modulus is one; the right side converges to $e^{-z^{2}V\mathrm{/}2}$. By [F8], $Y\sim N(0,V)$. Since the finite list and coefficients were arbitrary, $X$ is a centered Gaussian process. This argument proves Gaussian closure from the actual almost-sure Riemann-sum limit; it does not assume that arbitrary pointwise limits of Gaussian variables remain Gaussian.

Apply step 3.1 to the singleton coefficients and to $(X_s+X_t)$. It gives $\mathrm{Var}(X_r)=C(r,r)$ and $\mathrm{Var}(X_s+X_t)=C(s,s)+2C(s,t)+C(t,t)$. Covariance bilinearity also gives $\mathrm{Var}(X_s+X_t)=\mathrm{Var}(X_s)+2\mathrm{Cov}(X_s,X_t)+\mathrm{Var}(X_t)$. Comparing and cancelling proves $\mathrm{Cov}(X_s,X_t)=C(s,t)$, including $s=0$ or $t=0$.

Steps 1.1--4.1 prove every claim. Repeated times, zero coefficients, and singular linear combinations are retained in steps 2.1 and 3.1; $t=0$ is handled without a nondegenerate rectangle, and the empty finite list has the unique empty-tuple law. Changing the chosen probability-one continuity event changes $X_t$ only on a null set for each $t$, so the asserted finite laws and covariance are unaffected. AC is used exactly through [F1], [F3], [F6], and [F8], including the countable-choice input inherited by the continuous-integrability interface in [F3]; the fixed uniform left-endpoint sums, pasting, and finite algebra require no further choices.