Let $X_1,...,X_n$ be idd and uniformly distributed over $(0,\theta)$. Choosing between the null hypothesis which states that $\theta=\theta_0$ and the alternate hypothesis which states that $\theta=\theta_1$, with $\theta_1<\theta_0$, suppose we reject the null if and only if $x_{n:n}<c$ for some fixed, positive and known number $c$.
My question asks: for which values of $\alpha$ does there exist a MP level-$\alpha$ test?
By the Neyman-Pearson Lemma, I will reject the null for sufficiently large values of $L(x,\theta_0)/L(x,\theta_1)=(\frac{\theta_0}{\theta_1})^n*I(x_{n:n}<\theta_1)/I(x_{n:}<\theta_0)$. However, this expression can only take two values for a particular sample $x$: either the likelihood ratio is $1$, or the likelihood ratio is $0$.
My question is, does this mean that there are only two values of $\alpha$ for which this test can be MP level-$\alpha$? I would find this strange, since I'm pretty sure that for any $\alpha$ between $0$ and $1$, we can find a value $c$ to make the test level-$\alpha$.
The test should be a function of the minimal sufficient statistic, namely $X_{(n)} = \max\{X_1,...,X_n\}$, such that if $X_{(n)} > \theta_1$ then you reject $H_0$ with $\alpha = 0$ and where $X_{(n)} \le \theta_1$ then you should set a rejection region $c$. $$ \alpha = \mathbb{E}_{\theta_0}\Psi (x) = \mathbb{P}(\Psi (x) = 1) = \mathbb{P}( X_{(n)} \le c)=F_X(c)=\left(\frac{c}{\theta_0}\right)^n, $$ namely, for $X_{(n)} \le \theta_0$, reject $H_0$ if $$ X_{(n)} \le a^{1/n} \theta_0. $$