White noise is not a signed measure for fixed $\omega$

I'm going over some materials on stochastic analysis, and stuck with a problem on Gaussian white noise:

Let $$(\mathbb{R}^d,\mathcal{B},m)$$ be the Borel measurable space on $$\mathbb{R}^d$$. A white noise is a mean zero Gaussian process $$\{W(A):A\in\mathcal{B},m(A)<\infty\}$$ with covariance $$\mathbb{E}[W(A)W(B)]=m(A\cap B)$$.

Fix positive numbers $$\{c_k\}_{k\in\mathbb{N}}$$ such that $$\sum c_k^2=1$$ but $$\sum c_k=\infty$$. Let $$A=\bigcup_k A_k$$ be a disjoint union of Borel sets with $$m(A_k)=c_k^2$$. Show that $$\sum|W(A_k)|=\infty$$ a.s.

Even though one can show that $$\sum W(A_k)=W(A)$$ a.s. by a standard result on random series, the problem above shows that $$\sum W(A_k)$$ doesn't converge absolutely and hence $$W(\cdot,\omega)$$ cannot be a signed measure.

Any thought will be appreciated.

• I am confused, is $W(A)$ a $N(0, m(A))$ rv? – Will M. Dec 13 '18 at 19:00
• @WillM. Yes. And $W(A_k)$'s are independent. – SimonChan Dec 13 '18 at 20:16
• I think this can be proved by Kolmogorov zero-one law and three series theorem. – SimonChan Dec 13 '18 at 21:16
• I agree with your last comment, and do not think there is any easier method. – Mike Earnest Dec 13 '18 at 23:52