I have the following definitions:
Let $(A, \mathcal{A}, \mu)$ be a $\sigma$-finite measure space such that $(A, \mathcal{A})$ is a Polish space and $\mu$ is non-atomic. We set $\mathcal{A}_0 = \{B \in \mathcal{A} : \mu(B) < \infty\}$. A Gaussian measure on $(A, \mathcal{A})$ with intensity $\mu$ is a centered Gaussian family of the form $G = \{G(B) : B \in \mathcal{A}_0\}$ such that $E[G(B)G(C)] = \mu(B \cap C)\ \forall\ B, C \in \mathcal{A}_0$.
Let $B \in \mathcal{A}_0$ and let $B_n = \{B_{n,1}, \ldots, B_{n, k_n}\}, n \geq 1$ be a sequence of measurable partitions of $B$ such that $\max_{j = 1, \ldots, k_n} \mu(B_{n, j}) \rightarrow 0$ as $n \rightarrow \infty, k_n \rightarrow \infty$.
How can I show that $\sum\limits_{j = 1}^{k_n} G(B_{j, n})^2 \rightarrow \mu(B)$ in $L^2(\mathbb{P})$ as $n \rightarrow \infty$ ?
Many thanks for your help !
Since
$$\mu(B) = \sum_{j=1}^{k_n} \mu(B_{j,n})$$
we find
$$\begin{align*} I_n := \mathbb{E} \left| \mu(B)- \sum_{j=1}^{k_n} G(B_{j,n})^2 \right|^2 &= \mathbb{E} \left( \sum_{j=1}^{k_n} (\mu(B_{j,n})-G(B_{j,n})^2) \right)^2 \\ &= \sum_{i=1}^{k_n} \sum_{j=1}^{k_n} \mathbb{E} \big[ (G(B_{i,n})^2-\mu(B_{i,n})) (G(B_{j,n})^2-\mu(B_{j,n})) \big]. \end{align*}$$
Using that $G$ is a centered Gaussian process and $\mathbb{E}(G(A) G(B)) = 0$ for any disjoint $A,B \in \mathcal{A}_0$ it follows that the mixed terms in the above sum vanish. Thus,
$$I_n = \sum_{i=1}^{k_n} \mathbb{E} \underbrace{\big[ (G(B_{i,n})^2-\mu(B_{i,n}))^2 \big]}_{\text{var}(G(B_{i,n})^2)}.$$
As
$$\text{var}(X^2) = \mathbb{E}(X^4) - (\mathbb{E}(X^2))^2 = 2 \sigma^4$$
for any Gaussian random variable $X \sim N(0,\sigma^2)$, we get
$$I_n = 2 \sum_{j=1}^{k_n} \mu(B_{j,n})^2,$$
and so
$$I_n \leq 2 \max_{j} \mu(B_{j,n}) \underbrace{\sum_{j=1}^{k_n} \mu(B_{j,n})}_{=\mu(B)} \xrightarrow[]{n \to \infty} 0.$$