[proofplan] We construct Brownian motion on $[0,1]$ by defining it on dyadic rationals via midpoint interpolation with independent Gaussian perturbations, then extend to all of $[0,1]$ by proving Hölder [continuity](/page/Continuity) via Kolmogorov's criterion. The Gaussian structure and independent-increment property pass from dyadics to the continuum by Lévy's [continuity theorem](/theorems/1145). Finally, we extend to $\mathbb{R}_+$ by concatenation and to $\mathbb{R}^d$ by taking independent copies. [/proofplan]

[step:Define the process on dyadic rationals in $[0,1]$ by midpoint interpolation] Let $\mathcal{D}_n = \{k 2^{-n} : 0 \leq k \leq 2^n\}$ and $\mathcal{D} = \bigcup_{n \geq 0} \mathcal{D}_n$. Let $(Z_d)_{d \in \mathcal{D} \setminus \{0\}}$ be an independent family of $\mathcal{N}(0,1)$ random variables on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$. [Set](/page/Set) $B_0 = 0$ and $B_1 = Z_1$. For each $n \geq 1$ and each $d \in \mathcal{D}_n \setminus \mathcal{D}_{n-1}$, let $d_- = d - 2^{-n}$ and $d_+ = d + 2^{-n}$ be the neighbours of $d$ in $\mathcal{D}_{n-1}$, and define \begin{align*} B_d = \frac{B_{d_-} + B_{d_+}}{2} + \frac{Z_d}{2^{(n+1)/2}}. \end{align*} Set $N_d = (B_{d_+} - B_{d_-})/2$ and $N_d' = Z_d / 2^{(n+1)/2}$. By induction, $N_d$ is $\sigma(Z_e : e \in \mathcal{D}_{n-1} \setminus \{0\})$-measurable and $N_d'$ depends only on the fresh variable $Z_d$, so $N_d$ and $N_d'$ are independent. Both are centred Gaussian with variance $2^{-(n+1)}$: for $N_d$, $\operatorname{Var}(N_d) = \operatorname{Var}(B_{d_+} - B_{d_-})/4 = 2^{-n+1}/4 = 2^{-(n+1)}$; for $N_d'$, $\operatorname{Var}(N_d') = 1/2^{n+1}$. Since $B_d - B_{d_-} = N_d + N_d'$ and $B_{d_+} - B_d = N_d - N_d'$, these half-increments are jointly Gaussian with \begin{align*} \operatorname{Cov}(N_d + N_d', \, N_d - N_d') = \operatorname{Var}(N_d) - \operatorname{Var}(N_d') = 0. \end{align*} Uncorrelated jointly Gaussian random variables are independent, so the two half-increments are independent $\mathcal{N}(0, 2^{-n})$. By induction on $n$, all increments $(B_d - B_{d-2^{-n}})_{d \in \mathcal{D}_n, d > 0}$ are independent $\mathcal{N}(0, 2^{-n})$, and for any $s, t \in \mathcal{D}$ with $s < t$, the increment $B_t - B_s \sim \mathcal{N}(0, t-s)$.[/step]

[guided]The idea is to build Brownian motion "from the top down": first define $B_0$ and $B_1$, then fill in the midpoint $B_{1/2}$, then the quarter-points $B_{1/4}$ and $B_{3/4}$, and so on. At each stage, the new value at a dyadic midpoint $d$ is the average of its two already-defined neighbours plus an independent Gaussian perturbation whose variance is chosen to produce the correct increment [distribution](/page/Distribution). Why the specific variance $2^{-(n+1)}$ for the perturbation? We need the two half-increments $B_d - B_{d_-}$ and $B_{d_+} - B_d$ to be independent $\mathcal{N}(0, 2^{-n})$ (matching the interval length $d - d_- = d_+ - d = 2^{-n}$). Writing $B_d - B_{d_-} = N_d + N_d'$ and $B_{d_+} - B_d = N_d - N_d'$ where $N_d = (B_{d_+} - B_{d_-})/2$ and $N_d' = Z_d / 2^{(n+1)/2}$, independence of $N_d$ and $N_d'$ follows because $N_d$ is built from previously defined variables while $N_d'$ uses only the fresh Gaussian $Z_d$. Both have the same variance $2^{-(n+1)}$, so \begin{align*} \operatorname{Var}(B_d - B_{d_-}) = \operatorname{Var}(N_d) + \operatorname{Var}(N_d') = 2^{-(n+1)} + 2^{-(n+1)} = 2^{-n}, \end{align*} and the covariance \begin{align*} \operatorname{Cov}(B_d - B_{d_-}, \, B_{d_+} - B_d) = \operatorname{Cov}(N_d + N_d', \, N_d - N_d') = \operatorname{Var}(N_d) - \operatorname{Var}(N_d') = 0 \end{align*} vanishes. Since the pair is jointly Gaussian (both are linear combinations of independent Gaussians), zero correlation implies independence. This is precisely the balanced-variance condition that makes the midpoint construction work. The inductive conclusion: at level $n$, the increments $(B_{k 2^{-n}} - B_{(k-1)2^{-n}})_{k=1}^{2^n}$ are independent $\mathcal{N}(0, 2^{-n})$ random variables. Additivity of independent Gaussians then gives $B_t - B_s \sim \mathcal{N}(0, t-s)$ for all $s, t \in \mathcal{D}$.[/guided]

[step:Verify Hölder continuity on $\mathcal{D}$ via Kolmogorov's criterion and extend to $[0,1]$] For $s, t \in \mathcal{D}$ with $s < t$, the increment $B_t - B_s \sim \mathcal{N}(0, t-s)$, so for any $p > 0$, \begin{align*} \mathbb{E}[|B_t - B_s|^p] = |t - s|^{p/2} \cdot \mathbb{E}[|N|^p], \end{align*} where $N \sim \mathcal{N}(0,1)$. Taking $p = 4$ gives $\mathbb{E}[|B_t - B_s|^4] = 3|t - s|^2$, since $\mathbb{E}[N^4] = 3$ (computed via the moment generating [function](/page/Function) of the standard Gaussian). By [Kolmogorov's Continuity Criterion](/theorems/1170) applied with exponent $\alpha = 4$ and the bound $3|t-s|^{1+1}$ (so $\beta = 1$, giving $\beta/\alpha = 1/4$), the process admits a modification with $\gamma$-Hölder continuous paths for every $\gamma < 1/4$. To reach all $\gamma < 1/2$, take any $p > 2$: the bound $\mathbb{E}[|B_t - B_s|^p] = C_p |t-s|^{p/2}$ yields $\beta/\alpha = (p/2-1)/p = 1/2 - 1/p$, which approaches $1/2$ as $p \to \infty$. In particular, $d \mapsto B_d$ is uniformly continuous on $\mathcal{D} \cap [0,1]$ almost surely and extends uniquely to a continuous function on $[0,1]$: \begin{align*} B_t = \lim_{\substack{d \to t \\ d \in \mathcal{D}}} B_d, \qquad t \in [0,1]. \end{align*} [/step]

[step:Verify independent Gaussian increments on $[0,1]$ via characteristic functions] Fix $0 = t_0 < t_1 < \cdots < t_k \leq 1$. For each $j$, choose dyadic [sequences](/page/Sequence) $t_j^{(m)} \in \mathcal{D}$ with $t_j^{(m)} \to t_j$ as $m \to \infty$. By path continuity, $(B_{t_1^{(m)}}, \ldots, B_{t_k^{(m)}}) \to (B_{t_1}, \ldots, B_{t_k})$ almost surely. The joint characteristic function of the increments at dyadic times factors by independence: \begin{align*} \mathbb{E}\!\left[\exp\!\left(i \sum_{j=1}^k u_j (B_{t_j^{(m)}} - B_{t_{j-1}^{(m)}})\right)\right] = \prod_{j=1}^k \exp\!\left(-\frac{(t_j^{(m)} - t_{j-1}^{(m)}) u_j^2}{2}\right). \end{align*} Almost sure convergence implies convergence in distribution, so the left-hand side converges to $\mathbb{E}[\exp(i \sum_j u_j (B_{t_j} - B_{t_{j-1}}))]$. The right-hand side converges to $\prod_{j=1}^k \exp(-(t_j - t_{j-1}) u_j^2 / 2)$. By the [Uniqueness of Characteristic Functions](/theorems/530), the increments $(B_{t_j} - B_{t_{j-1}})_{j=1}^k$ are independent with $B_{t_j} - B_{t_{j-1}} \sim \mathcal{N}(0, t_j - t_{j-1})$. [/step]

[step:Extend from $[0,1]$ to $\mathbb{R}_+$ and from $\mathbb{R}$ to $\mathbb{R}^d$] For the extension to $\mathbb{R}_+$: construct independent copies $(B^{(n)})_{n \geq 0}$, each a standard Brownian motion on $[0,1]$ built by the above procedure, and define \begin{align*} B_t = \sum_{j=0}^{\lfloor t \rfloor - 1} B_1^{(j)} + B_{t - \lfloor t \rfloor}^{(\lfloor t \rfloor)}, \qquad t \geq 0. \end{align*} Path continuity holds on each interval $[n, n+1]$ by construction, and at the junction $t = n$ we have $B_n = \sum_{j=0}^{n-1} B_1^{(j)}$ from both the left and right formulas, so the paths are globally continuous. The independent-increment and Gaussian-increment properties follow from the independence of the copies and the additivity of Gaussians. For $\mathbb{R}^d$: take $d$ independent one-dimensional Brownian motions $B^{(1)}, \ldots, B^{(d)}$ and set $B_t = (B_t^{(1)}, \ldots, B_t^{(d)})$. The increments $B_t - B_s$ are $\mathcal{N}(0, (t-s) I_d)$, and independence of increments follows from the independence of the components. [/step]