I am trying to show that the order statistic $X_{(n)}$ for a set of RV $\{X_i\}_{1}^{n}$ where $X_i\overset{iid}\sim \text{Unif}(0,\theta)$ is complete when $\theta \in (0,\infty)$ but not when $\theta \in [1,\infty)$.
By completeness $E[g(X_{(n)})]=0$ iff $g(X_{(n)})=0$ a.e.
$X_{(n)}\sim n\theta^{-n}X_{(n)}^{n-1}$ then if
$$E[g(X_{(n)})]=\int_{0}^{\theta}g(X_{(n)})n\theta^{-n}X_{(n)}^{n-1}dX_{(n)}=0$$
this implies
$$g(\theta)=0, \forall \theta \in(0,\infty)$$
Since for any $X_{(n)}$ there exists a $\theta=X_{(n)}$ one can conclude $g(X_{(n)})=0$
when the parameter space is restricted to $\theta \in [1,\infty)$ then by the statement above one can conclude
$$g(\theta)=0, \forall \theta \in [1,\infty)$$
not guaranteeing that $g(X_{(n)})=0$ for $X_{(n)}\in(0,1)$
I am having trouble using this to justify non completeness.
To demonstrate that $X_{(n)}$ is not complete, you need to come up with a nonzero function $g$ such that $$E_\theta[g(X_{(n)})]=0\quad\text{for all $\theta\ge 1$}.\tag1$$ You have already shown that any $g$ that satisfies (1) must have $g(\theta)=0$ for all $\theta\ge1$, so your job is to think of a $g$ that's nonzero but still satisfies $$ \int_0^1 g(x)n\theta^{-n}x^{n-1}dx=0\quad\text{for all $\theta\ge1$}.\tag2 $$ (Note that you should be using a consistent variable of integration throughout (2), and not alternating between $x$ and $X_{(n)}$.) You can factor out the $n\theta^{-n}$ from (2), which simplifies your requirement to $$\int_0^1g(x)x^{n-1}dx=0.\tag3$$ Many choices of $g$ exist for (3). Hint: You could take $g$ to be piecewise constant, for instance. Maybe the quickest choice (making the integration in (3) as painless as possible) would be $$g(x):=\begin{cases}1-cx&0<x<1\\0&x\ge 1\end{cases}$$ for a constant $c$ that you have to determine.