[proofplan] We verify each of the four Brownian motion properties for $B_t = WN([0,t])$ by reducing them to the corresponding properties of Gaussian white noise established in [Theorem 2068](/theorems/2068). Property (1) is immediate. Property (3) follows from the distribution of white noise on intervals. Property (2) requires showing that increments over disjoint intervals, which are white noise values on disjoint sets, are independent of the past. Property (4) uses joint Gaussianity of the image of $L^2$ functions under the isometry, together with an explicit covariance computation via inner products of indicator functions. [/proofplan]

[step:Verify $B_0 = 0$ a.s.] By definition, $B_0 = WN([0,0]) = WN(\mathbb{1}_{[0,0]})$. Since $[0,0] = \{0\}$ has Lebesgue measure $\mathcal{L}^1(\{0\}) = 0$, we have $\mathbb{1}_{\{0\}} = 0$ in $L^2(\mathbb{R}_+)$ (the function is nonzero only on a set of measure zero). Since the isometry $I: L^2(\mathbb{R}_+) \to S$ is linear and $I(0) = 0$: \begin{align*} B_0 = WN(\mathbb{1}_{\{0\}}) = I(\mathbb{1}_{\{0\}}) = I(0) = 0 \quad \text{a.s.} \end{align*} [/step]

[step:Verify the increment distribution $B_t - B_s \sim N(0, t-s)$] For $t \geq s \geq 0$, we write the increment as a white noise evaluation on a single interval. The key identity is: \begin{align*} B_t - B_s = WN([0,t]) - WN([0,s]) = WN(\mathbb{1}_{[0,t]} - \mathbb{1}_{[0,s]}) = WN(\mathbb{1}_{(s,t]}), \end{align*} where the second equality uses linearity of the isometry $I$, and the third uses $\mathbb{1}_{[0,t]} - \mathbb{1}_{[0,s]} = \mathbb{1}_{(s,t]}$ (valid $\mathcal{L}^1$-a.e., since the two functions differ only at the single point $\{s\}$, which has measure zero). By property (1) of [Theorem 2068](/theorems/2068): \begin{align*} B_t - B_s = WN((s,t]) \sim N(0, |(s,t]|) = N(0, t - s). \end{align*} [/step]

[step:Verify independence of $B_t - B_s$ from $\sigma(B_r : r \leq s)$]We must show that $B_t - B_s = WN(\mathbb{1}_{(s,t]})$ is independent of $\sigma(B_r : r \leq s)$. The $\sigma$-algebra $\sigma(B_r : r \leq s)$ is generated by the random variables $\{B_r : 0 \leq r \leq s\}$. Each $B_r = WN(\mathbb{1}_{[0,r]}) = I(\mathbb{1}_{[0,r]})$. For any finite collection $0 \leq r_1 < r_2 < \cdots < r_m \leq s$, we can write each $B_{r_k}$ as a sum of increments over disjoint intervals: \begin{align*} B_{r_k} = \sum_{j=1}^k WN((r_{j-1}, r_j]), \end{align*} where $r_0 := 0$. Each $WN((r_{j-1}, r_j])$ corresponds to the indicator of a subinterval of $[0, s]$, so all these intervals are disjoint from $(s, t]$. By property (2) of [Theorem 2068](/theorems/2068), $WN(\mathbb{1}_{(s,t]})$ is independent of each $WN(\mathbb{1}_{(r_{j-1}, r_j]})$. Since $(B_t - B_s, WN((r_0, r_1]), \ldots, WN((r_{m-1}, r_m]))$ is a vector of jointly Gaussian random variables (each is in the Gaussian subspace $S$) with pairwise zero covariance, the vector is mutually independent. As each $B_{r_k}$ is a measurable function of $(WN((r_0, r_1]), \ldots, WN((r_{k-1}, r_k]))$, we conclude that $B_t - B_s$ is independent of $(B_{r_1}, \ldots, B_{r_m})$ for every finite subcollection. Since this holds for all finite subcollections with $r_k \leq s$, it follows that $B_t - B_s$ is independent of $\sigma(B_r : r \leq s)$.[/step]

[guided]The argument has two layers. First, we decompose $B_{r_k}$ into sums of white noise increments over disjoint subintervals of $[0, s]$. These are all independent of $WN((s,t])$ because the intervals are disjoint from $(s,t]$. The subtle point is passing from pairwise independence to mutual independence of the Gaussian vector. Why does this work? The vector $(B_t - B_s, WN((r_0, r_1]), \ldots, WN((r_{m-1}, r_m]))$ consists of elements of the Gaussian subspace $S = I(L^2(\mathbb{R}_+))$, so any linear combination is Gaussian. The covariance matrix is block-diagonal: $\mathbb{E}[WN(\mathbb{1}_{(s,t]}) \cdot WN(\mathbb{1}_{(r_{j-1}, r_j]})] = (\mathbb{1}_{(s,t]}, \mathbb{1}_{(r_{j-1}, r_j]})_{L^2} = 0$ since $(s,t] \cap (r_{j-1}, r_j] = \varnothing$. For a jointly Gaussian centered vector, a block-diagonal covariance matrix means the corresponding blocks are independent. Finally, we use a standard measure-theoretic fact: if $X$ is independent of $(Y_1, \ldots, Y_m)$ for every finite subcollection of a generating family, then $X$ is independent of the $\sigma$-algebra generated by that family. This is because the collection of events $F$ such that $\mathbb{P}(X \in A, F) = \mathbb{P}(X \in A)\mathbb{P}(F)$ for all Borel $A$ forms a Dynkin system, and it contains the $\pi$-system of finite-dimensional cylinder sets, so by the $\pi$-$\lambda$ theorem it contains $\sigma(B_r : r \leq s)$.[/guided]

[step:Verify joint Gaussianity and the covariance formula $\mathbb{E}[B_s B_t] = s \wedge t$] **Joint Gaussianity.** For any $0 \leq t_1 \leq \cdots \leq t_n$ and $c_1, \ldots, c_n \in \mathbb{R}$, the linear combination $\sum_{k=1}^n c_k B_{t_k} = \sum_{k=1}^n c_k I(\mathbb{1}_{[0,t_k]}) = I\!\left(\sum_{k=1}^n c_k \mathbb{1}_{[0,t_k]}\right)$, using linearity of $I$. Since $\sum_{k=1}^n c_k \mathbb{1}_{[0,t_k]} \in L^2(\mathbb{R}_+)$, [Theorem 2067](/theorems/2067) gives that $I\!\left(\sum c_k \mathbb{1}_{[0,t_k]}\right) \sim N(0, \|\sum c_k \mathbb{1}_{[0,t_k]}\|_{L^2}^2)$. Hence every linear combination of $(B_{t_1}, \ldots, B_{t_n})$ is a centered Gaussian, so the vector is jointly Gaussian. **Covariance.** For $s, t \geq 0$, the isometry property of [Theorem 2067](/theorems/2067) gives: \begin{align*} \mathbb{E}[B_s B_t] = \mathbb{E}[I(\mathbb{1}_{[0,s]}) \cdot I(\mathbb{1}_{[0,t]})] = (\mathbb{1}_{[0,s]}, \mathbb{1}_{[0,t]})_{L^2(\mathbb{R}_+)}. \end{align*} Computing the inner product: \begin{align*} (\mathbb{1}_{[0,s]}, \mathbb{1}_{[0,t]})_{L^2(\mathbb{R}_+)} = \int_{\mathbb{R}_+} \mathbb{1}_{[0,s]}(u) \, \mathbb{1}_{[0,t]}(u) \, d\mathcal{L}^1(u) = \int_{\mathbb{R}_+} \mathbb{1}_{[0,s] \cap [0,t]}(u) \, d\mathcal{L}^1(u) = \mathcal{L}^1([0, s \wedge t]) = s \wedge t. \end{align*} Therefore $\mathbb{E}[B_s B_t] = s \wedge t$. This completes the verification that $(B_t)_{t \geq 0}$ satisfies properties (1)-(4) of Brownian motion, establishing that it is a Brownian motion in all respects except possibly the continuity of sample paths. [/step]