Verify that certain statistic is complete (namely, $X_{(1)}$ for $f_\theta(x)=e^{-(x-3\theta)}I_{(3\theta,\infty)}(x)$) Given the density function of a certain population
$$f_\theta(x)=e^{-(x-3\theta)}I_{(3\theta,\infty)}(x)$$
I am asked to find a complete statistic for a random simple sample $(X_1,X_2,\ldots,X_n)$ of it.
From the nature of the given density function, I would like to prove that $X_{(1)}$ is complete.
To do so, the density of $X_{(1)}$ is given by
$$g(x)=ne^{-n(x-3\theta)}$$
And if $\operatorname E[f(S)]=0$, where $S=X_{(1)}$, then
$$\operatorname E[f(S)]=\int_{3\theta}^\infty f(x)ne^{-n(x-3\theta)}\, dx=0$$
How to proceed from this point, and prove that $f=0$ almost everywhere? Thanks in advance.
 A: Firstly, the CDF of shifted exponential distribution is 
$$
F_{X_1}(x)=I_{(3\theta,\infty)}(x)\int_{3\theta}^x e^{-(t-3\theta)}\, dt = - e^{-(t-3\theta)} \big|_{3\theta}^x\cdot I_{(3\theta,\infty)}(x)=(1-e^{-(x-3\theta)})I_{(3\theta,\infty)}(x).
$$
It cannot be $1+e^{-(x-3\theta)}$ as assumed in your calculations since this value is greater than $1$, and $F_{X_1}(x)$ is a probability.
Next, the CDF of $X_{(1)}$ is 
$$
F_{X_{(1)}}(x)=1-(1-F_{X_1}(x))^n = (1-e^{-n(x-3\theta)})I_{(3\theta,\infty)}(x).
$$
Finally, the PDF of $X_{(1)}$ is 
$$
g(x)=ne^{-n(x-3\theta)}I_{(3\theta,\infty)}(x)
$$
To prove completeness we need to suggest that for any $\theta\in\mathbb R$
$$
\int_{3\theta}^\infty f(x)ne^{-nx}e^{3n\theta}\, dx =0
$$
and derive that $f(x)=0$ a.e. 
Note that multiplier $e^{3n\theta}$ cannot be zero. Therefore
$$
\int_{3\theta}^\infty f(x)ne^{-nx}e^{3n\theta}\, dx \equiv 0 \iff 
\int_{3\theta}^\infty f(x)ne^{-nx}\, dx \equiv 0
$$
Since the last integral equals zero for any $\theta$, we can subtract two integrals for different $\theta$ and get: for any $a<b$
$$\tag{1}\label{1}
\int_{a}^b f(x)ne^{-nx}\, dx = 0
$$
If you can apply Radon-Nikodym theorem here, do it and complete. Other case we need to follow the rest way. 
Prove that for any Borel set $B$
$$\tag{2}\label{2}
\int\limits_B f(x)ne^{-nx}\, dx = 0.
$$
To do it, let $$\mathcal A=\left\{B\in \mathfrak B(\mathbb R): \int\limits_B f(x)ne^{-nx}\, dx = 0\right\}.$$
We collect into $\mathcal A$ all Borel sets with the property: integrals over these sets are zero. 
Prove that $\mathfrak B(\mathbb R)\subseteq\mathcal A$. Indeed, any interval $(a,b)$ belongs to $\mathcal A$ due to (\ref{1}). Next, you can check that $\mathcal A$ is a sigma-algebra. Therefore it contains Borel sigma-algebra since Borel sigma-algebra is the smallest sigma-algebra that contains any interval on $\mathbb R$. 
So, the proved (\ref{2}).
Finally, consider Borel sets $B_1=\{x: f(x)>0\}$ and $B_2=\{x: f(x)<0\}$. From (\ref{2}) we have 
$$
\int\limits_{B_1} f(x)ne^{-nx}\, dx = 0 \text{ and } \int\limits_{B_2} f(x)ne^{-nx}\, dx = 0.
$$
If Lebesgue measure $\lambda(B_i)$ is not equal to zero than the first integral is strictly positive, and the second is strictly negative. So, both measures are zero and $f(x)=0$ almost everywhere. 
