= Solution
Put $r(x)=f_{\theta_1}(x)/f_{\theta_0}(x)$. The proposed first density can be written
$$
g_{0,c}(x)=(1-\epsilon)
\max\!\left\{f_{\theta_0}(x),\frac{f_{\theta_1}(x)}c\right\}.
$$
Its integral is continuous and strictly decreasing in $c$, tends to infinity as $c\downarrow0$, and tends to $1-\epsilon$ as $c\to\infty$. Hence a unique $c>0$ makes its integral one. Since $g_{0,c}\geq(1-\epsilon)f_{\theta_0}$,
$$
g_{0,c}=(1-\epsilon)f_{\theta_0}+\epsilon h_0
$$
for the density $h_0=[g_{0,c}-(1-\epsilon)f_{\theta_0}]/\epsilon$.
Similarly,
$$
g_{1,d}(x)=(1-\epsilon)
\max\{f_{\theta_1}(x),d f_{\theta_0}(x)\}.
$$
Its integral is continuous and strictly increasing from $1-\epsilon$ to infinity as $d$ ranges from zero to infinity. The unique normalizing $d>0$ gives
$$
g_{1,d}=(1-\epsilon)f_{\theta_1}+\epsilon h_1
$$
for a density $h_1$. Thus $G_0\in\mathcal P_\epsilon(F_{\theta_0})$ and $G_1\in\mathcal P_\epsilon(F_{\theta_1})$.
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