The r.h.s. of (12) provides and writing this in more concise notation Since f$_{x}$ is continuous, then ˜ ψ ( x ) = ψ ( F ( x )) f ( x ) is continuous on R $^{d$_{x}$}$. For an arbitrary ˜ ψ ∈ D$_{0}$ ( R d$_{x}$ ) consider Do the transformation, then Define a continuous function ψ ( F$_{x}$$_{1}$, ..., F$_{x}$$_{d}$$_{x}$ ) = ˜ ψ ( x 1 ,...,x$_{d}$$_{x}$ ) f$_{x}$ ( x 1 ,...,x$_{d}$$_{x}$ ) on (0 , 1) d$_{x}$ , then this equals (13) . ■ Suppose now that F$_{x}$ is absolutely continuous with continuous density function, f$_{y}$$_{|}$$_{x}$ ; then the support of the density function is an open set in R $^{d$_{x}$}$, S$_{y}$$_{|}$$_{x}$. The Lemma applies by considering ˜ ψ ( x ) = f$_{x}$ ( x ) ψ ( F ( x )) ∈ D$_{0}$ ( S$_{y}$$_{|}$$_{x}$ ) 22 ( F$_{y}$$_{|}$$_{x}$, ˜ ψ ) = ∫ ∂$_{x}$F$_{x,y}$ ( x, y ) f$_{x}$ ( x ) ˜ ψ ( x ) dx$_{1}$...dx$_{d}$$_{x}$. ( F$_{y}$$_{|}$$_{x}$, ˜ ψ ) = ∫ ∂$_{z}$F$_{x,y}$ ( F − 1 x ( z ) , y ) ˜ ψ ( F − 1 x ( z )) f$_{x}$ ( F − 1 x ( z )) dz. = ∫ ∂ $^{d$_{x}$}$F$_{x,y}$ ( x, y ) f$_{x}$ ( x ) ψ ( F ( x )) f$_{x}$ ( x ) dx = ∫ ∂ $^{d$_{x}$}$F$_{x,y}$ ( x, y ) f$_{x}$ ( x ) ˜ ψ ( x ) dx. ( − 1) d$_{x}$ ∫ ... ∫ F$_{x,y}$ ( F − 1 x 1 ( z$_{1}$ ) , ..., F − 1 x$_{d}$$_{x}$ ( z$_{d}$$_{x}$ ) , y ) ∂ d$_{x}$ z ψ ( z$_{1}$, ..., z$_{d}$$_{x}$ )) dz$_{1}$...dz$_{d}$$_{x}$ = ∫ ... ∫ ∂ d$_{x}$ z F$_{x,y}$ ( F − 1 x 1 ( z$_{1}$ ) , ..., F − 1 x$_{d}$$_{x}$ ( z$_{d}$$_{x}$ ) , y ) ψ ( z$_{1}$, ...z$_{d}$$_{x}$ ) dz$_{1}$...dz$_{d}$$_{x}$ = ... ∂ d$_{x}$ x F$_{x,y}$ ( x, y ) f$_{x}$ ( x ) ψ ( F$_{x}$$_{1}$ ( x$_{1}$ ) , ..., F$_{x}$$_{d}$$_{x}$ ( x$_{d}$$_{x}$ ) ) f$_{x}$ ( x ) dx$_{1}$...dx$_{d}$$_{x}$,