I want to show that any Brownian Motion $(B(t))$ on a Probability space $(\Omega,\mathscr{A},\mathbb{P})$ is the a.s. result of a Levy's Construction, i.e. interpolation using a sequence of independent normally distributed random variables $\{Z_d\}_{d\in \mathcal{D}}$ with mean $0$ and variance $1$ where $\mathcal{D} = \{\text{dyadic numbers}\} = \{2^{-n}k: 1\leq k \leq 2^n,n\geq 1\}$. I think this is true but I would like to have this verified.
My attempted proof:
We use notation from the book Brownian Motion by Mörters and Peres. Let $\mathcal{D}_n = \{k2^{-n}: 1\leq k \leq 2^n\}$ and let $(\tilde{B}(t))$ be a Brownian Motion attained via Levy's construction using $\{\tilde{Z}_d\}_{d\in \mathcal{D}}$, where $\{\tilde{Z}_d\}_{d\in \mathcal{D}}$ are independent random variables with Gaussian distribution with mean $0$ and variance $1$. So that for $d\in \mathcal{D}_n$
$$\tilde B_n(d) = \sum_{i=0}^{n}\tilde F_i(d) = \sum_{i=0}^{\infty}\tilde F_i(d)$$ where for $n>1$ $$\tilde F_n(t) = \begin{cases}2^{-(n+1)/2}\tilde Z_t& \text{ for $t\in \mathcal{D}_n\setminus \mathcal{D}_{n-1}$} \\ 0& \text{ for $t\in \mathcal{D}_{n-1}$} \\ \text{linear interpolation between}\end{cases}$$ and $\tilde F_0(t) = t\tilde Z_1$.
Then $$\tilde B(d) = \frac{\tilde B(d-2^{-n})+\tilde B(d+2^{-n})}{2}+\frac{\tilde Z_d}{2^{(n+1)/2}}$$ and hence for every $n$ there is a $2^n\times 2^n$ invertible matrix $A_n$ such that
\begin{equation*} \begin{pmatrix} \tilde{Z}_{d_1} \\ \tilde{Z}_{d_2} \\ \vdots \\ \tilde{Z}_{d_{2^n}} \end{pmatrix} = A_n \begin{pmatrix}\tilde{B}_{d_1} \\ \tilde{B}_{d_2} \\ \vdots \\ \tilde{B}_{d_{2^n}} \end{pmatrix}. \end{equation*} where $d_k = k2^{-n}$ and where for any $m\leq n$ and $d_k\in \mathcal{D}_m$ the random variable $\tilde{Z}_{d_k}$ only depends on those $\tilde{B}_d$ where $d\in \mathcal{D}_m$. If we set \begin{equation*} \tilde{X} := \begin{pmatrix}\tilde{B}_{d_1} \\ \tilde{B}_{d_2} \\ \vdots \\ \tilde{B}_{d_{2^n}} \end{pmatrix} \quad \text{and } \quad X:= \begin{pmatrix}{B}_{d_1} \\ {B}_{d_2} \\ \vdots \\ {B}_{d_{2^n}} \end{pmatrix} \end{equation*} then we can define $A_nX = Z$. And since $\mathbb{P}_{X} = \tilde{\mathbb{P}}_{\tilde{X}}$ and $Z$ is a Gaussian random vector we conclude that the entries of $Z$ are independent with normal distributions with mean $0$ and variance $1$. Furthermore $X = A_n^{-1}Z$ and hence the Levy construction attained from $Z$ at step $n$ i.e. $$\sum_{i=0}^{n}F_i(d) = \sum_{i=0}^{\infty}F_i(d)$$ where we let $F_i$ be constructed as $\tilde F_i$ but changing $Z_t$ for $\tilde{Z}_t$, agrees with the Brownian Motion $(B(t))$ at all dyadic rationals of order less than $2^{-n}$: $\mathcal{D}_n$. Since $(B(t))$ are continuous a.s. the limit of the interpolation obtained via the Levy construction is precisely $(B(t))$.
Is there anything subtle I'm missing here?
The reason I wan't to have this result is that it is sometimes easier to prove a result about Brownian Motions as the limit of Levys Construction.