Exercise: Let $X_1,\ldots,X_n$ be a sample from the distribution with density $$f(x|\theta) = \dfrac{2x}{\theta^2}\mathbb{1}_{(0,\theta)}(x)$$ w.r.t. the Lebesgue measure.
Show that the statistic $T(X) = \max(X_1,\ldots,X_n)$ is complete.
What I've tried: I know that a statistic $T$ is said to be complete for $\theta\in\Omega$ if for any Borel function $f$, $\operatorname{E}_\theta f(T) = 0$ for all $P_\theta \in \Omega$, implies that $f(T)=0$, $P_\theta$-a.s. for all $\theta$. So what I want to do is take $\operatorname{E}_\theta f(T)$ and show that this is only equal to zero when $f(T) = 0$. If I'm correct, that means that I need to show that $$\displaystyle\int f(T(X))\dfrac{2x}{\theta^2}\mathbb{1}_{(0,\theta)}(x)dx = 0$$ implies $f(T(X)) = 0$. I'm not sure how to proceed from here though.
Question: How do I solve this exercise?
Thanks in advance!
$\def\d{\mathrm{d}}$As is already derived in this answer, the density function of $T$ is$$ f_T(t; θ) = \frac{2n}{θ^{2n}} t^{2n - 1} I_{(0, θ)}(t), $$ thus for any Borel function $g$,$$ E_θ(g(T)) = \int_0^θ g(t) \cdot \frac{2n}{θ^{2n}} t^{2n - 1} \,\d t = \frac{2n}{θ^{2n}} \int_0^θ g(t) t^{2n - 1} \,\d t. \tag{1} $$
If Borel function $g$ satisfies that$$ E_θ(g(T)) = 0, \quad \forall θ \in {\mit Θ} = (0, +\infty) $$ then by (1) there is$$ \int_0^θ g(t) t^{2n - 1} \,\d t = 0, \quad \forall θ > 0 $$ which implies for any interval $(a, b) \subseteq (0, +\infty)$,$$ \int_a^b g(t) t^{2n - 1}\,\d t = 0. $$ Therefore, $g(t) t^{2n - 1}$ equals $0$ a.e. on $(0, +\infty)$, so $g(t)$ equals $0$ a.e. on $(0, +\infty)$. Thus for any $θ > 0$,$$ P_θ(g(T) = 0) = \int_0^θ I_{\{t' \mid g(t') = 0\}}(t) \cdot \frac{2n}{θ^{2n}} t^{2n - 1} \,\d t = \int_0^θ \frac{2n}{θ^{2n}} t^{2n - 1} \,\d t = 1. $$ Hence $T$ is complete.