The Setup:
Let $X_1, \dots, X_n$ be IID with pdf $f_{\theta}$, where $\theta \in {0, 1}$ is unknown.
Given $f_{0}(x) \equiv 1_{0 < x < 1}$ and $f_{1}(x) \equiv \frac{1}{\sqrt{2x} }1_{0 < x < 1}$, find the MLE.
The Given Solution
$\hat{\theta}_n = 1_A$, where
$$A = \prod_{i=1} ^{n} \frac{1}{\sqrt{2x_i}} \geq 1 $$
My Question: So I do understand how to get to the indicator form of the MLE. What I do not understand is why the indicator region of $A$ uses a "greater than or equal" sign rather than a "strictly greater than" sign.
In other words, why is the solution not
$\hat{\theta}_n = 1_A$, where
$$A = \prod_{i=1} ^{n} \frac{1}{\sqrt{2x_i}} > 1 $$
Theoretically, it does appear that $L(\theta)$ = 1 can occur for both $\theta = 0$ or $\theta = 1$. So I can see that the given solution does make sense, but does it matter than we are not giving the boundary to $\theta = 0$? Is there even a difference between the $\geq$ and $>$ cases?
Thank you in advance.