 Solution

In the Lévy construction of Brownian motion, start with $B_0=0$ and $B_1=Z_0$, where $Z_0$ has the standard normal distribution. Suppose values are defined at the endpoints $u=k{2}^{-n}$ and $v=(k+1)2^{-n}$. Setusing fresh independent standard normal random variables $Z_{n,k}$, also independent of everything previously used. Join the values by straight segments to obtain a continuous piecewise linear process $B^{(n)}$ on the dyadic grid.

To verify the Brownian increments, write $D=B_v-B_u$ and $\ell =v-u$. The two new increments are $D/2+\sqrt{\ell }{Z}_{n,k}/2$ and $D/2-\sqrt{\ell }{Z}_{n,k}/2$. If $D\sim N(0,\ell )$, their variances are $\ell /2$ and their covariance is zero. They are Jointly Gaussian, so they are independent. Induction proves the required Brownian increments on every dyadic grid.

For convergence, successive interpolants differ by tents of disjoint interiors, soThe Gaussian tail bound and the union bound giveThese probabilities are summable. The First Borel-Cantelli lemma and ${\sum }_{n}n{2}^{-n/2}<\infty$ therefore prove uniform convergence of the interpolants almost surely. Their limit has continuous sample paths. Approximating any finite set of times by dyadic times passes the Jointly Gaussian increment laws to the limit, giving a standard Brownian motion on $[0,1]$.

Equivalently, with ${\psi }_{n,k}(t)={\int }_0^t{H}_{n,k}(s)ds$, these tents give the Lévy-Ciesielski expansion