I am trying to prove that a shifted expolential distribution is a complete set.
i.e.
$$X_1,...,X_n \sim_{iid} f_X(x;\theta)=e^{-(x-\theta)}, 0<\theta<x$$ and if $E[u(X)]=0$ then $$u(X) = 0$$.
The procedure I am taking is to first look at the expectation
$$\int_{\theta}^{\infty} u(x)e^{-(x-\theta)}dx=0$$
and take the derivative on both sides with respect to $\theta$.
I see that the result will be
$$\int_{\theta}^{\infty}u(x)e^{-(x-\theta)}dx=-u(\theta)$$
but I do not know how this leads to $u(X)=0$.
The notes that I am reading mentions that $Pr[u(X)=0]=1$ instead of $u(X)=0$ so I am thinking that I am missing something.
I would appreciate your input.
The equation $E_{\theta}\left[u(X)\right]=\int_\theta^\infty u(x)e^{-(x-\theta)}\,dx=0$ (here $\theta$ need not be positive) is the same as $$\int_\theta^\infty u(x)e^{-x}\,dx=0\quad\forall\,\theta\in\mathbb R\tag{1}$$
Differentiating $(1)$ with respect to $\theta$ one ends up with $$e^{-\theta}u(\theta)=0\quad\forall\,\theta$$
As $e^{-\theta}>0$ for any real $\theta$, the function $u(\cdot)$ must be identically zero.
In terms of random variables, this is equivalent as saying $u(X)=0$ with probability $1$.
This proves $u(X)$ is a complete statistic and the fact that the shifted exponential family is complete.