Most powerful test of simple vs. simple in $\mathrm{Unif}[0, \theta]$ 
Say $X \sim \mathrm{Unif}[0, \theta]$. Denote the observations as $x_i$ $(i=1, \cdots, n)$.
Show that any test $\phi$ that satisfies the following two conditions is most powerful test of level $\alpha$ for $H_0 : \theta = \theta_0$ vs. $H_1 : \theta = \theta_1$ $(\theta_1 > \theta_0)$.

*

*$\phi^{*}(x_1, \cdots, x_n) = 1 \ \ (\mathrm{if} \max\{x_1, \cdots, x_n\}) > \theta_0$


*$\mathbb{E}_{\theta_0} \phi^{*}(X) = \alpha$


Try
By Neyman-Pearson lemma, any test that satisfies

*

*$\phi^{*}(x_1, \cdots, x_n) = \begin{cases} 1 & (\mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n) > k) \\ 
\gamma & (\mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n) = k) \\
0 & (\mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n) < k) \end{cases} $
(for some $\gamma \in [0,1]$)

*

*$\mathbb{E}_{\theta_0} \phi^{*}(X) = \alpha$
is an MP test for $H_0$ vs. $H_1$.
Since
$$
\begin{aligned}
\mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n)  &= (\theta_0 / \theta_1)^n I(\max x_i \le \theta_1)/I(\max x_i \le \theta_0) \\
&= \begin{cases} (\theta_0/\theta_1)^n & (\max x_i > \theta_0) \\ \infty & (\max x_i \le \theta_0) \end{cases}
\end{aligned}
$$
we have
$$
\phi^{*}(x_1, \cdots, x_n) = \begin{cases}1 & (\max x_i > \theta_0) \\ \alpha & (\max x_i \le \theta_0) \end{cases}
$$
is an MP test for $H_0$ vs. $H_1$.
But this does not mean ANY test that satisfies


*

*$\phi^{*}(x_1, \cdots, x_n) = 1 \ \ (\mathrm{if} \max\{x_1, \cdots, x_n\}) > \theta_0$


*$\mathbb{E}_{\theta_0} \phi^{*}(X) = \alpha$

is an MP test, but rather just provides an example of MP test.
Is there anyone to help me out?
 A: The logic is:

*

*Any test which satisfies the two conditions has the same significance level $\alpha$ and power $1-\left(\frac{\theta_0}{\theta_1}\right)^n +\alpha\left(\frac{\theta_0}{\theta_1}\right)^n$ as the particular test you found; the


*Given that your test is most powerful of all tests with significance level $\alpha$, all the other tests satisfying the two conditions are also most powerful with significance level $\alpha$
If you want to check, consider the significance level and power of these two deterministic tests

*

*$\phi^{*}(x_1, \cdots, x_n) = \begin{cases} 1 &  \text{ when }\max x_i > \theta_0 \sqrt[n]{1-\alpha} \\
0 & \text{ when }\max x_i \le \theta_0 \sqrt[n]{1-\alpha} \end{cases}$

*$\phi^{*}(x_1, \cdots, x_n) = \begin{cases} 1 &  \text{ when }\max x_i > \theta_0  \\
0 &  \text{ when }\min x_i \gt \theta_0 \left(1- \sqrt[n]{1-\alpha}\right)   \text{ and } \max x_i \le  \theta_0 \\
1 & \text{ when }\min x_i \le \theta_0 \left(1- \sqrt[n]{1-\alpha}\right)  \end{cases}$
