In the page 8 of the book "The Malliavin Calculus and Related Topics" from Nualart one reads:
to simplify our ideas, take $\mu$ to be the lebesgue measure on $\Bbb{R}_+$.
In this case then we can see the white noise as the derivative of the Brownian motion $(B_t)_{t\geq 0}$ and define for intervals $[a,b)$:
$$W\big([a,b)\big)= B_b - B_a $$
So for intervals we have the sigma additivity ($[a_i,b_i)\cap [a_j, b_j) = \emptyset$ for $i \neq j$) $$\cup_i[a_i,b_i) = [a,b) \Rightarrow W(\cup_i[a_i,b_i)) = \sum_i B_{b_i} - W_{a_i} = B_b - B_a =W\big([a,b)\big)$$
I imagine it is not the case for arbitrary $\cup_i A_i = A$
That is, one should have $\cup_i A_i = A$, $A_i \cap A_j = \emptyset$ for $i \neq j$ and yet
$$W(A) \neq \sum_i W(A_i) $$
Do you have any idea of why is this the case?