# Sigma additvity of Gaussian White Noise

Let $(E,\mathcal{E},\mu)$ be a $\sigma$-finite measure space and $(\Omega,\mathcal{F},\mathbb{P})$ a probability space. A Gaussian White Noise is a linear isometry $G \colon L^2(E,\mathcal{E},\mu) \rightarrow H$, where $H \subseteq L^2(\Omega,\mathcal{F},\mathbb{P})$ is a centered Gaussian space.

Let $A \in \mathcal{E}$ s.t. $\mu(A) < \infty$ and assume that $A$ can be written as the disjoint union of measurable subsets $(A_j)_{j \geq 1}$ of $E$. Then I could verify that $\mathbb{1}_{A} = \sum_{j=1}^{\infty} \mathbb{1}_{A_j}$ in $L^2(E,\mathcal{E},\mu)$ and hence by the isometry property $$G(A) = \sum_{j=1}^{\infty} G(A_j)$$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$.

Now I don't understand the following sentence:

Since the random variables $G(A_j)$ are independent, an easy application of the convergence theorem for discrete martingales also shows that the series converges a.s.

I can explain why the random variables $G(A_j)$ are independent, but how can I establish almost sure convergence?

• Apply the convergence theorem for discrete martingales. Aug 9, 2017 at 11:46
• @LoveTooNap29 The sequence $(M_n)_{n \geq 1}$ defined by $M_n := \sum_{j=1}^{n} G(A_j)$ is a martingale. If we can show that $\sup_{n}[(M_n)^{-}] < \infty$, then $M_n$ converges a.s to $\sum_{j=1}^{\infty} G(A_j)$. What I don't understand is: How can we show that the limit $\sum_{j=1}^{\infty} G(A_j)$ is equal to $G(A)$ and why is $\sup_{n}[(M_n)^{-}] < \infty$? Aug 9, 2017 at 12:10

1. The martingale $(M_n)$ is $L^2$ bounded: $E[M_n^2]=\sum_{k=1}^n\mu(A_k)\le\mu(A)<\infty$. This permits application of the martingale convergence theorem.
2. $E[(G(A)-M_n)^2]=\sum_{k=n+1}^\infty\mu(A_k)$.