The Problem
Let $X_1, \dots, X_n \sim U(\theta, 5)$ where $0 < \theta < 5$ with pdf
$$ f(x;\theta) = \dfrac{1}{5-\theta} \quad \theta < x < 5 $$
Find the generalized likelihood ratio test for testing $H_0 : \theta = 4$ vs. $H_1 : \theta > 4$. Determine the critical region if $\alpha = 0.05$.
My attempt
For my own understanding, another way to write the pdf is $\frac{1}{5-\theta}\cdot I(x\in(\theta, 5))$. Since we're dealing with the GLRT, we need to first find the test statistic of the following form:
$$ \lambda(x) = \dfrac{\sup_{H_0} L(\theta | x)}{\sup_{H_0 \cup H_1} L(\theta | x)} $$
To find the supremums, we find the MLE of $f$ for both the restricted $H_0$ and unrestricted $H_0 \cup H_1$. The likelihood is
$$ L(\theta | x) = \dfrac{1}{(5-\theta)^n}\cdot I(x_i\in(\theta, 5)) $$
Restricted: We are effectively given the cut off for $\theta$ so our MLE is just $\frac{1}{(5-4)^n} = 1$ for $x_i\in (4,5)$. Maybe I am misunderstanding something here, but I don't think we need to worry about the $x_i$'s.
Unrestricted: This is where I get confused... We can take the log-likelihood and get the MLE for $\theta$ but it is unnecessary since we can directly see that the likelihood will be maximized when $\hat{\theta} = 5$ (I think approaches it is more accurate).
My issue
I know that the numerator of my test statistic $\lambda(x)$ is 1. I am unsure of what the denominator should be.
Your restricted likelihood is generally correct, but we need to keep track of the indicators--that's why you rewrote the density that way to begin with.
The likelihood function given the sample $\boldsymbol x = (x_1, \ldots, x_n)$ is $$\mathcal L(\theta \mid \boldsymbol x) =\frac{1}{(5-\theta)^n} \prod_{i=1}^n \mathbb 1(\theta < x_i < 5) = (5-\theta)^{-n} \mathbb 1(\theta < x_{(1)} \le x_{(n)} < 5),$$ where we have used the order statistics $x_{(1)} = \min_i x_i$ and $x_{(n)} = \max_i x_i$ for convenience. This also implies that $x_{(1)}$ is a sufficient statistic.
Then the restricted likelihood under $H_0$ is
$$\mathcal L(\theta \mid \boldsymbol x, H_0) = \mathcal L(4 \mid \boldsymbol x) = \mathbb 1(4 < x_{(1)} \le x_{(n)} < 5).$$ It is, strictly speaking, not identically $1$, because we must also have a sample whose observations must lie in the interval $(4, 5)$.
The unrestricted likelihood under the union of hypotheses $H_0 \cup H_1$ is
$$\begin{align} \mathcal L(\theta \mid \boldsymbol x, H_0 \cup H_1) &= \mathcal L(\theta \mid \boldsymbol x) \mathbb 1 (4 \le \theta < 5) \\ &= (5-\theta)^{-n} \mathbb 1(4 \le \theta < x_{(1)} \le x_{(n)} < 5). \end{align}$$ Now we maximize this likelihood by noting that $4 \le \theta < x_{(1)}$ is the relevant restriction. Since $(5-\theta)^{-n}$ is a monotonically increasing function of $\theta$ on the interval $(-\infty, 5)$, it now becomes apparent that $$\sup_{H_0 \cup H_1} \mathcal L(\theta \mid \boldsymbol x) = \begin{cases} (5-x_{(1)})^{-n}, & 4 < x_{(1)} < 5 \\ 0, & \text{otherwise} \end{cases}$$ and this in fact makes intuitive sense: since $x_{(1)}$ represents in a fashion an upper bound on what value $\theta$ is permitted to take, we expect that the likelihood under $H_0 \cup H_1$ would have a supremum that is a function of $x_{(1)}$. Hence
$$\lambda(\boldsymbol x) = \lambda(x_{(1)}) = (5 - x_{(1)})^n, \quad 4 < x_{(1)} < 5.$$ The test statistic is not defined if $x_{(1)} \le 4$ because we cannot observe such a sample under the condition $\theta \ge 4$.
I leave it to you to compute the critical region for such a test.