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?



Gaussian Property of the Renormalization Group
The following (between the horizontal lines) is a statement of what one may call the abstract change of variable theorem. It is taken from some exercise I gave my students a while ago:




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Let $(X,\mathcal{M},\mu)$ be a measure space and let $(Y,\mathcal{N})$ be a measurable space.
Let $f:X\rightarrow Y$ be an $(\mathcal{M},\mathcal{N})$-measurable map.
For $B\in\mathcal{N}$ define
$$
f\_{\ast}\mu(B)=\mu(f^{-1}(B))\ .
$$


a) Show that $f\_{\ast}\mu$ is well defined and gives a measure on $(Y,\mathcal{N})$.


b) Let $\phi:Y\rightarrow\mathbb{R}$ be a nonnegative simple function.
Show that
$$
\int\_X \phi\circ f\ {\rm d}\mu= \int\_Y \phi\ {\rm d}(f\_{\ast}\mu)
$$


c) Using the Monotone Convergence Theorem, prove that the last equality holds without the assumption of $\phi$ being a simple function.


d) Show that a non necessarily nonnegative measurable function $\phi$ from $Y$ to $\mathbb{R}$
is $f\_{\ast}\mu$-integrable iff $\phi\circ f$ is $\mu$-integrable. In this case, show that the equality in b) still holds.




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Here $f$ is the map $(\xi\_1,\ldots,\xi\_N)\mapsto \xi\_1+\cdots+\xi\_N$. The measure $\mu$ is the product measure
$$
\otimes\_{j=1}^{N}d\mu\_j(\xi\_j) 
$$
on $X=\mathbb{R}^{|\Lambda|}\times\cdots\times\mathbb{R}^{|\Lambda|}$, $N$ times.
Finally the push-forward measure is $\mu\_C$. So you see that there is no need for extra hypotheses on $F^{\Lambda}$. Morover, being able to integrate successively over the $\xi\_j$ is a consequence of Fubini's Theorem which, in particular, says that integrability over the product space implies the iterated integral is well defined and gives the same result as the integral over the product space.


