The Construction of Brownian Motion

Uniform convergence. Write $U_j(t)=\sum_{k}\xi_{j,k}S_{j,k}(t)$ for the level-$j$ term. Because the tents $S_{j,k}$ at a fixed level have disjoint interiors and peak at height $2^{-j/2-1}$,

The Gaussian tail obeys $\P(\abs\xi\ge x)\le e^{-x^2/2}$ for $x\ge 1$, since $\P(\xi\ge x)\le\frac{1}{\sqrt{2\pi}}\int_x^\infty\frac ux e^{-u^2/2}\,du=\frac{e^{-x^2/2}}{x\sqrt{2\pi}}$ and doubling stays below $e^{-x^2/2}$ for $x\ge 1$. With $x=2\sqrt j$ (so $x\ge 1$ for $j\ge 1$), the union bound over the $2^j$ variables at level $j$ gives

a convergent series in $j$. By the Borel-Cantelli lemma, with probability one there is a finite random threshold $J(\omega)$ with $\max_k\abs{\xi_{j,k}}\le 2\sqrt j$ for all $j\ge J(\omega)$, so by Equation (3) the bounds $\sup_t\abs{U_j(t)}\le\sqrt j\,2^{-j/2}=:M_j$ hold for $j\ge J(\omega)$. Fix such an $\omega$ and split the sum as $W=(\xi_{-1}t+\sum_{j<J}U_j)+\sum_{j\ge J}U_j$. The first bracket is a finite sum of continuous functions, hence continuous. For the tail $\sup_t\abs{U_j(t)}\le M_j$ with $\sum_{j\ge J}M_j<\infty$, so by the Weierstrass M-test the tail partial sums converge uniformly to a continuous function. A uniformly convergent series of continuous functions has a continuous sum, so the path $W(\omega,\cdot)$ is continuous, and this holds for every $\omega$ in the almost-sure event.

Covariance. For fixed $s$, the coefficients of $\mathbf 1_{[0,s]}$ against the Haar basis are $\ip{\mathbf 1_{[0,s]}}{h_{j,k}}=\int_0^s h_{j,k}=S_{j,k}(s)$, and $\ip{\mathbf 1_{[0,s]}}{1}=s$, so by Bessel's inequality $\sum_n S_n(s)^2\le\norm{\mathbf 1_{[0,s]}}^2=s<\infty$, where $S_n$ runs over all basis functions. Hence the series Equation (2) converges in $L^2(\Omega)$ at each $t$ to a limit that agrees almost surely with the uniform limit $W_t$, since a subsequence of the partial sums converges almost surely to both limits, forcing them to coincide. By continuity of the $L^2(\Omega)$ inner product and the orthonormality $\E[\xi_n\xi_m]=\delta_{nm}$ of the coefficients,

the middle equality being the Parseval identity for the orthonormal Haar basis and the last the integral $\int_0^1\mathbf 1_{[0,s]}\mathbf 1_{[0,t]}=\min(s,t)$. The mean is zero because every coefficient has mean zero.

Gaussianity. Each partial sum of Equation (2) is a finite linear combination of the independent Gaussians $\xi_n$, hence Gaussian, and a finite linear combination of finitely many $W_{t_i}$ is the $L^2$ limit of the corresponding combinations of partial sums, each Gaussian. A mean-square limit of Gaussians is Gaussian, since the characteristic functions $e^{-\sigma_n^2 t^2/2}$ converge to $e^{-\sigma^2t^2/2}$ because $L^2$ convergence forces the variances $\sigma_n^2=\E[X_n^2]$ to converge, and the Levy continuity theorem identifies the limit. So every finite vector $(W_{t_1},\dots,W_{t_n})$ is Gaussian, and $W$ is a Gaussian process with the mean and covariance computed above.

At $t=0$, $\Var(W_0)=\min(0,0)=0$, so $W_0=0$. For $s<t$ the increment is Gaussian with mean zero and

using $K(s,t)=\min(s,t)=s$. Take any disjoint intervals $(s_1,t_1],\dots,(s_m,t_m]$. The increment vector $(W_{t_1}-W_{s_1},\dots,W_{t_m}-W_{s_m})$ is a linear image of the Gaussian vector $(W_{s_1},W_{t_1},\dots,W_{s_m},W_{t_m})$, hence jointly Gaussian. For any non-overlapping pair $(s_i,t_i]$, $(s_l,t_l]$,

so the covariance matrix of the increment vector is diagonal. For a jointly Gaussian vector a diagonal covariance matrix is equivalent to mutual independence of the components by the Gaussian equivalence of uncorrelated and independent, so the increments over disjoint intervals are mutually independent.