[proofplan] We derive the three properties of Gaussian white noise from the isometry $WN: L^2(\mathbb{R}_+) \to S$. Properties (1) and (2) are direct consequences of the isometry and the characterisation of independence for jointly Gaussian variables. Property (3) requires a convergence argument: the partial sums $\sum_{i=1}^n WN(A_i)$ form an $L^2$-bounded martingale, and we invoke the $L^2$ martingale convergence theorem for both $L^2$ and almost sure convergence, then identify the limit via continuity of the isometry. [/proofplan] [step:Verify $WN(A) \sim N(0, |A|)$ for Borel sets $A$ with finite Lebesgue measure] Let $A \subseteq \mathbb{R}_+$ be a Borel set with $|A| := \mathcal{L}^1(A) < \infty$. The indicator function $\mathbb{1}_A$ belongs to $L^2(\mathbb{R}_+)$ with $\|\mathbb{1}_A\|_{L^2}^2 = \int_{\mathbb{R}_+} \mathbb{1}_A^2 \, d\mathcal{L}^1 = \mathcal{L}^1(A) = |A|$. By definition, $WN(A) := WN(\mathbb{1}_A) = I(\mathbb{1}_A)$, where $I: L^2(\mathbb{R}_+) \to S$ is the Gaussian Hilbert space isometry from [Theorem 2067](/theorems/2067). That theorem gives $I(f) \sim N(0, \|f\|_{L^2}^2)$ for every $f \in L^2(\mathbb{R}_+)$, so: \begin{align*} WN(A) = I(\mathbb{1}_A) \sim N(0, \|\mathbb{1}_A\|_{L^2}^2) = N(0, |A|). \end{align*} [/step] [step:Verify independence of $WN(A)$ and $WN(B)$ for disjoint Borel sets] Let $A, B \subseteq \mathbb{R}_+$ be disjoint Borel sets with $|A|, |B| < \infty$. Since $A \cap B = \varnothing$, the indicators satisfy $\mathbb{1}_A(t) \cdot \mathbb{1}_B(t) = 0$ for all $t \in \mathbb{R}_+$, hence: \begin{align*} (\mathbb{1}_A, \mathbb{1}_B)_{L^2} = \int_{\mathbb{R}_+} \mathbb{1}_A(t) \mathbb{1}_B(t) \, d\mathcal{L}^1(t) = \int_{\mathbb{R}_+} 0 \, d\mathcal{L}^1(t) = 0. \end{align*} By the isometry property of [Theorem 2067](/theorems/2067), $\mathbb{E}[WN(A) \cdot WN(B)] = (\mathbb{1}_A, \mathbb{1}_B)_{L^2} = 0$. The random variables $WN(A)$ and $WN(B)$ are jointly Gaussian: any linear combination $\alpha WN(A) + \beta WN(B) = I(\alpha \mathbb{1}_A + \beta \mathbb{1}_B)$ is Gaussian by [Theorem 2067](/theorems/2067), since $\alpha \mathbb{1}_A + \beta \mathbb{1}_B \in L^2(\mathbb{R}_+)$. For jointly Gaussian centered random variables, zero covariance implies independence. [guided] Why does zero covariance imply independence here? In general, uncorrelated random variables need not be independent. However, for jointly Gaussian random variables, the joint distribution is completely determined by the mean vector and covariance matrix. If $WN(A)$ and $WN(B)$ have zero means and covariance $\mathbb{E}[WN(A) \cdot WN(B)] = 0$, then the covariance matrix of the vector $(WN(A), WN(B))$ is diagonal: \begin{align*} \operatorname{Cov}\bigl((WN(A), WN(B))\bigr) = \begin{pmatrix} |A| & 0 \\ 0 & |B| \end{pmatrix}. \end{align*} A multivariate Gaussian with diagonal covariance matrix factors as a product of its marginals, which is the definition of independence. To verify joint Gaussianity rigorously: any linear combination $\alpha WN(A) + \beta WN(B) = I(\alpha \mathbb{1}_A + \beta \mathbb{1}_B)$, and since $\alpha \mathbb{1}_A + \beta \mathbb{1}_B \in L^2(\mathbb{R}_+)$, [Theorem 2067](/theorems/2067) gives that this is $N(0, \|\alpha \mathbb{1}_A + \beta \mathbb{1}_B\|_{L^2}^2)$, confirming that every linear combination is Gaussian. This is precisely the definition of joint Gaussianity. [/guided] [/step] [step:Establish $L^2$ and pointwise convergence of $\mathbb{1}_A = \sum_{i=1}^\infty \mathbb{1}_{A_i}$] Let $A = \bigsqcup_{i=1}^\infty A_i$ be a countable partition of $A$ into disjoint Borel sets with $|A| < \infty$ and $|A_i| < \infty$ for all $i$. Define the partial sums $f_n := \sum_{i=1}^n \mathbb{1}_{A_i}$. Since the $A_i$ are pairwise disjoint, $f_n(t) = \mathbb{1}_{\bigcup_{i=1}^n A_i}(t) \nearrow \mathbb{1}_A(t)$ pointwise for every $t \in \mathbb{R}_+$. We verify $L^2$ convergence. Since $A = \bigsqcup_{i=1}^\infty A_i$ and the sets are disjoint: \begin{align*} \|\mathbb{1}_A - f_n\|_{L^2}^2 = \left\|\sum_{i=n+1}^\infty \mathbb{1}_{A_i}\right\|_{L^2}^2 = \left\|\mathbb{1}_{A \setminus \bigcup_{i=1}^n A_i}\right\|_{L^2}^2 = \mathcal{L}^1\!\left(A \setminus \bigcup_{i=1}^n A_i\right) = \sum_{i=n+1}^\infty |A_i|. \end{align*} Since $\sum_{i=1}^\infty |A_i| = |A| < \infty$, the tail $\sum_{i=n+1}^\infty |A_i| \to 0$ as $n \to \infty$. Therefore $f_n \to \mathbb{1}_A$ in $L^2(\mathbb{R}_+)$. [/step] [step:Show the partial sums $M_n = \sum_{i=1}^n WN(A_i)$ form an $L^2$-bounded martingale] Define $M_n := \sum_{i=1}^n WN(A_i)$ and let $\mathcal{G}_n := \sigma(WN(A_1), \ldots, WN(A_n))$ be the natural filtration. We verify that $(M_n, \mathcal{G}_n)_{n \geq 1}$ is a martingale. **Adaptedness.** Each $M_n$ is a finite sum of $\mathcal{G}_n$-measurable random variables, hence $\mathcal{G}_n$-measurable. **Integrability.** Since $WN(A_i) \sim N(0, |A_i|)$ by property (1) and the $WN(A_i)$ are independent by property (2), we have $\mathbb{E}[M_n^2] = \sum_{i=1}^n \mathbb{E}[WN(A_i)^2] = \sum_{i=1}^n |A_i| \leq |A| < \infty$, so $M_n \in L^2(\mathbb{P}) \subset L^1(\mathbb{P})$. **Martingale property.** For the conditional expectation: \begin{align*} \mathbb{E}[M_{n+1} \mid \mathcal{G}_n] = \mathbb{E}[M_n + WN(A_{n+1}) \mid \mathcal{G}_n] = M_n + \mathbb{E}[WN(A_{n+1}) \mid \mathcal{G}_n]. \end{align*} Since $WN(A_{n+1})$ is independent of $(WN(A_1), \ldots, WN(A_n))$ (by property (2), as $A_{n+1}$ is disjoint from $A_1, \ldots, A_n$), and $\mathcal{G}_n = \sigma(WN(A_1), \ldots, WN(A_n))$, independence gives $\mathbb{E}[WN(A_{n+1}) \mid \mathcal{G}_n] = \mathbb{E}[WN(A_{n+1})] = 0$. Hence $\mathbb{E}[M_{n+1} \mid \mathcal{G}_n] = M_n$. **$L^2$ bound.** The sequence $\mathbb{E}[M_n^2] = \sum_{i=1}^n |A_i| \leq |A| < \infty$ is bounded uniformly in $n$. [guided] The key structural observation is that orthogonality of the indicators $\mathbb{1}_{A_i}$ in $L^2(\mathbb{R}_+)$ translates, via the isometry $I$, into independence of the Gaussian variables $WN(A_i)$. This independence is what makes the partial sums a martingale with orthogonal increments. Why is the $L^2$ bound important? The $L^2$ martingale convergence theorem requires that $\sup_n \mathbb{E}[M_n^2] < \infty$. Here this bound is $|A|$, the total Lebesgue measure of $A$, which is finite by hypothesis. A subtlety: we need $\mathbb{E}[WN(A_{n+1}) \mid \mathcal{G}_n] = \mathbb{E}[WN(A_{n+1})]$. This uses the fact that $WN(A_{n+1})$ is independent of the entire $\sigma$-algebra $\mathcal{G}_n$, not just of each individual $WN(A_i)$. This follows because $WN(A_{n+1})$ is independent of the vector $(WN(A_1), \ldots, WN(A_n))$ — mutual independence of a jointly Gaussian family with pairwise zero covariance gives full independence. [/guided] [/step] [step:Apply the $L^2$ martingale convergence theorem and identify the limit as $WN(A)$] Since $(M_n)$ is an $L^2$-bounded martingale, the Martingale Convergence Theorem (for $L^2$-bounded martingales) guarantees that $M_n$ converges both in $L^2(\mathbb{P})$ and $\mathbb{P}$-almost surely to some limit $M_\infty \in L^2(\mathbb{P})$. It remains to identify $M_\infty = WN(A)$. By linearity of the isometry $I$: \begin{align*} M_n = \sum_{i=1}^n WN(A_i) = \sum_{i=1}^n I(\mathbb{1}_{A_i}) = I\!\left(\sum_{i=1}^n \mathbb{1}_{A_i}\right) = I(f_n), \end{align*} where $f_n = \sum_{i=1}^n \mathbb{1}_{A_i}$. We showed in the previous step that $f_n \to \mathbb{1}_A$ in $L^2(\mathbb{R}_+)$. Since $I$ is an isometry (hence continuous): \begin{align*} \|M_n - WN(A)\|_{L^2(\mathbb{P})} = \|I(f_n) - I(\mathbb{1}_A)\|_{L^2(\mathbb{P})} = \|f_n - \mathbb{1}_A\|_{L^2(\mathbb{R}_+)} \to 0. \end{align*} Therefore $M_n \to WN(A)$ in $L^2(\mathbb{P})$. Since $L^2$ limits are unique (up to a.s. equality), $M_\infty = WN(A)$ almost surely. Combined with the a.s. convergence $M_n \to M_\infty$, we conclude: \begin{align*} WN(A) = \sum_{i=1}^\infty WN(A_i) \end{align*} with convergence in both $L^2(\mathbb{P})$ and almost surely. [guided] Why do we get both modes of convergence? The $L^2$ martingale convergence theorem gives a.s. convergence $M_n \to M_\infty$ and $L^2$ convergence $M_n \to M_\infty$. Separately, the isometry argument gives $L^2$ convergence $M_n \to WN(A)$. Since the $L^2$ limit is unique (if $Y_n \to Y$ and $Y_n \to Z$ both in $L^2$, then $\|Y - Z\|_{L^2} = 0$, so $Y = Z$ a.s.), we must have $M_\infty = WN(A)$ a.s. Therefore the a.s. limit of $M_n$ is also $WN(A)$. Note that a.s. convergence does not follow from $L^2$ convergence alone — in general, $L^2$ convergence only implies convergence in probability. The a.s. convergence comes from the martingale structure, which provides the additional regularity needed. [/guided] [/step]