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 [articular, says that integrability over the product space implies the iterated integral is well defined ad gives the same result.