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:
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.
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.