Let $\Lambda \subset \mathbb{Z}^{d}$ be a finite set and $\varphi = (\varphi_{x})_{x\in \Lambda} \in \mathbb{R}^{|\Lambda|}$. Let $F^{\Lambda}=F^{\Lambda}(\varphi)$ be a real-valued *global function*, because it depends on every entry of $\varphi$. The renormalization group setup goes like this: first, for a fixed $N$, we define $Z_{0}: \overbrace{\mathbb{R}^{|\Lambda|}\times\cdots\times\mathbb{R}^{|\Lambda|}}^{\mbox{$N$ times}} \to \mathbb{R}$ by $Z_{0}(\xi_{1},...,\xi_{N}) := F^{\Lambda}(\xi_{1}+\cdots+\xi_{N})$. Now, suppose $\mu_{C}$ is a Gaussian measure on $\mathbb{R}^{|\Lambda|}$ associated to the (positive-definite) matrix $C$, which has a decomposition $\sum_{j=1}^{N}C_{j}$, where each $C_{j}$ is again positive-definite and let $\mu_{j}$ be the Gaussian measure associated to $C_{j}$. Then, because $\xi_{j} \sim N(C_{j}) \Rightarrow \sum_{j=1}^{N}\xi_{j} \sim N(C)$, we have:
\begin{eqnarray}
\int d\mu_{C}(\varphi)F^{\Lambda}(\varphi) = \int d\mu_{N}(\xi_{N})\int d\mu_{N-1}(\xi_{N-1})\cdots \int d\mu_{1}(\xi_{1})Z_{0}(\xi_{1},...,\xi_{N}) \tag{1}\label{1}
\end{eqnarray}
Now, the left hand side of (\ref{1}) is well-defined provided $F^{\Lambda}$ is measurable and $\mu_{C}$-integrable. In this case, $Z_{0}$ is again measurable because it is the composition of $F^{\Lambda}$ with the continuous function $\mathbb{R}^{\Lambda}\times\cdots\times \mathbb{R}^{\Lambda} \to \mathbb{R}^{\Lambda}$ given by $(\xi_{1},...,\xi_{N}) \mapsto \xi_{1}+\cdots+\xi_{N}$.

**My question is:** Is the right hand side of (\ref{1}) well-defined? Is $Z_{0}$ automatically $\mu_{j}$-integrable (for each $j$) given that $F^{\Lambda}$ is measurable and $\mu_{C}$-integrable? Do I need more assumptions?