Suppose $P_0 $ and $P_1$ are singular distributions. Show that $N=[0,1]^2$.

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Suppose we want to test $H_0 :\mathbf{X} \sim P_0$ against $H_1 : \mathbf{X} \sim P_1$, where $P_0,P_1$ are two different laws of $\mathbf{X}$.

Let $N$ be the set of points $(\alpha,\beta)$ for which there exists a test $\phi$ such that $E_{P_0}(\phi)=\alpha $ and $E_{P_{1}}(\phi)=\beta$.

Now suppose $P_0 $ and $P_1$ are singular distributions.

Show that $N=[0,1]^2$.

Now, all I could say that as $P_0$ and $P_1$ are singular, the supports of $P_0$ and $P_1$ are disjoint.

This means $\phi(\mathbf{X})=1(f_{0}(\mathbf{x})=0)$ has size $0$ and power $1$.

But what to do next? That is how can I show that $N$ is actually containing all points within the unit square?