 Solution

Put $r(x)=f_{{\theta }_1}(x)/f_{{\theta }_0}(x)$. The proposed first density can be writtenIts integral is continuous and strictly decreasing in $c$, tends to infinity as $c\downarrow 0$, and tends to $1-ϵ$ as $c\to \infty$. Hence a unique $c>0$ makes its integral one. Since $g_{0,c}\ge (1-ϵ)f_{{\theta }_0}$,for the density $h_0=[g_{0,c}-(1-ϵ)f_{{\theta }_0}]/ϵ$.

Similarly,Its integral is continuous and strictly increasing from $1-ϵ$ to infinity as $d$ ranges from zero to infinity. The unique normalizing $d>0$ givesfor a density $h_1$. Thus $G_0\in {\mathcal{P}}_{ϵ}(F_{{\theta }_0})$ and $G_1\in {\mathcal{P}}_{ϵ}(F_{{\theta }_1})$.