Question on complete sufficient statistic

Question

In my textbook, I have this example which says

$X\sim U(0,\theta)$ we show that the family of PDFs of X is complete. We need to show that $E(g(x))=\int_{0}^{\theta}\frac{1}{\theta}g(x)dx=0\ \ \ \ \ \forall \ \ \ \theta>0$

iff $g(x)=0 $ for al $x$. They applied lebnitz integral rule(differentiation under integral sign) and got $g(\theta)=0$

But we had to prove $g(x)=0$ isn't it? How does $g(\theta)=0 $ makes it a complete sufficient statistic?

Answer 1

Using either $θ$ or $x$ as a dummy variable of the function $g$ does not make a difference. That is, using the notation $g(x)$ or the notation $g(θ)$ does not change the definition of $g$. Hence proving that $$g(x)=0\qquad\qquad ∀x$$ is exactly the same as proving that $$g(θ)=0\qquad\qquad ∀θ$$