# Why is $g(x)=0$ a.s. if $p(x | \theta)$ is a density and $\int_0^\infty g(x) p(x | \theta) dx = E[g(x)] = 0$ for all $\theta$?

I am trying to prove that a statistic $$T$$ is complete, where $$T \sim \text{Gamma}(n, e^\theta)$$. I am stuck at deducing that $$g(T)$$ must be zero almost surely, just by looking at the expectation: $$E[g(T)] = \int_0^\infty g(t) t^{n-1} e^{-te^\theta} \frac{e^{-n \theta}}{\Gamma(n)} dt= 0$$ The above shoulld imply that $$g(t)=0$$ almost surely.

Why is $$g(t)=0$$ if $$\int_0^\infty g(t) t^{n-1} e^{-te^\theta} dt = 0$$ for all $$\theta$$? I know that $$t^{n-1} e^{-te^\theta}$$ is a positive function, but $$g(t)$$ does not have to be? Is it because the integral must be zero for the whole gamma family?

• It might help to recognize your last integral as the Laplace transform of the function $t\to g(t)t^{n-1}$ evaluated at $e^\theta\gt 0.$ – whuber Nov 28 '18 at 18:53
• Thank you for the hint, I was not familiar with Laplace transformations, and I managed to find an answer here, but, I think I'm out of my league.. math.stackexchange.com/questions/47507/… – alekdimi Nov 28 '18 at 19:52