Completeness of a statistic in a truncated distribution

Question

Suppose a random sample $x_1,\dots, x_n$ (i.i.d.) from a random variable $X$ defined over $(\Omega,\mathcal{F},P)$ whose probability density function is $f(x_1,\dots, x_n;\theta)$ and $T(x_1,\dots, x_n)$ is a complete statistic for this family. Let $f_A(x_1,\dots, x_n;\theta)$ be the probability density function associated with $P$ when we truncate it over $A$, a subset of the sample space $\Omega$, given by $$f_A(x_1,\dots, x_n;\theta) = \frac{f(x_1,\dots, x_n;\theta)I_{A}(x_1,\dots, x_n)}{P_\theta (A)}$$ such that $I_{A}(x_1,\dots, x_n)$ is the indicator function that is 1 when every point belongs to $A$ and 0 if at least one isn't and $P_\theta(A)$ is the probability of set $A$ that acts as a normalizing constant. Show that $T(x_1,\dots, x_n)$ is also complete for the truncated family.

I was studying mathematical statistics and found this problem. The problem set also had a exercise about showing that if $T$ is sufficient for a distribution then it is sufficient for the truncated one that is easily solved by using Fisher-Neyman factorization. I tried to show it through the definition of completeness, $\mathbb{E}_\theta(g(t))0 \xrightarrow{} P(g(t)=0)=1$, but this takes me nowhere as we know nothing about the pdf of $T$.

Answer 1

Consulting the Theory of Point Estimation (Lehmann and Casella, 1999) from which this exercise is taken

the second question contains the term in addition, which means that $T$ is sufficient.

If $T$ is incomplete and sufficient for the family of truncated distributions, there exists a non-zero function $g$ such that the expectation of $g(T)$ is zero for all $f_A(\cdot;\theta)$'s. This implies $$\mathbb E_\theta[g(T(X_1,\ldots,X_n)\mathbb I_A(X_1,\ldots,X_n)]=0\quad\forall\theta$$ for the untruncated distributions.

Hint 1: Apply the law of the total expectation to the above to derive a function of $T$ with expectation zero

Hint 2: Show that $g$ is not constant.