This is not a measure in the usual sense, but in a probabilistic sense.
Specifically, let $\mathcal B_f(\mathbb{R})$ denote the family of Borel sets having finite Lebesgue measure.
An independently scattered Gaussian measure (with Lebesgue control measure) is a random set function $$\{B(A,\omega),A\in \mathcal B_f(\mathbb{R}), \omega\in \Omega\}$$ (you can understand this a $\mathcal B_f(\mathbb{R})$-indexed stochastic process) such that
for any $A\in \mathcal B_f(\mathbb{R})$, $B(A)$ is a centered Gaussian random variable with variance $\lambda(A)$;
for any disjoint $A_1,A_2,\dots,A_n \in \mathcal B_f(\mathbb{R})$, the random variables $B(A_1),B(A_2),\dots,B(A_n)$ are independent;
for any disjoint $A_1,A_2,\ldots \in \mathcal B_f(\mathbb{R})$ such that $\bigcup_{n=1}^\infty A_n \in \mathcal B_f(\mathbb{R})$,
$$
B\biggl(\bigcup_{n=1}^\infty A_n\biggr) = \sum_{n=1}^\infty B(A_n)\quad \text{ a.s.}
$$
Using it, you can define various objects like
However, here are some important remarks.
In the Gaussian case, the independence is equivalent to orthogonality, and the almost sure convergence of a series with independent terms is equivalent (thanks to the Kolmogorov three series theorem) to the mean-square convergence. So any reference on measures with orthogonal increments will do.
Despite the $\sigma$-additivity is in almost-sure sense, the exceptional set depends on the particular sequence of sets. And there is no version of $B$ which is a signed measure for almost all $\omega$, so it is not possible to define the integral w.r.t. $B$ as that w.r.t. a signed measure; it can just be defined in the mean-square sense (you should know this already).