Integral defined as zero implies function is zero

[coolnessitself]

Homework Statement

\int\limits_{\theta}^{\infty} f(x) g(x) dx = 0
\theta > 0
f(x) = ke^{-k(x-\theta)}
Show g(x) is identically 0.

Homework Equations

The Attempt at a Solution

f is always >= 0 since it behaves exponentially in the region of interest.
From something like https://www.physicsforums.com/showthread.php?t=299145" I could say that IF f(x)g(x)\ge 0 in this region, then g(x)=0, but I don't know anything about g. I could say that assuming g is positive somewhere, then the integral wouldn't be zero, and that assuming g is negative somewhere, the integral wouldn't be zero, but what about the case where g(c_1) f(c_1) = k and g(c_2) f(c_2) = -k. Then g is positive somewhere and negative somewhere else such that the product cancels out. Wouldn't that allow the integral to be zero?

 

[LCKurtz]

What are your assumptions? g continuous? True for all k and \theta?

 

[HallsofIvy]

As it stands, this simply isn't true! Are you requiring that g be continuous?

 

[coolnessitself]

I guess I should explain a bit more. I'm trying to show that a family of pdfs is complete. I no longer have access to my paper (for the time being), so I'll reproduce what I can remember. The notation seems standard, so I'll just type out a definition:
Let the random variable Z of either the continuous type or the discrete type have a pdf or pmf that is one member of the family \left\{h(z;\theta) : \theta \in \Omega\right\}. If the condition E[u(Z)]=0, for every \theta \in\Omega requires that u(z) be zero except on a set of points that has probability zero for each h(z;\theta),\;\theta\in\Omega, then the family \left\{h(z;\theta) : \theta \in \Omega\right\} is called a complete family of probability density or mass functions.

Here, my pdf is the f I typed in the first post, and I need to show it's complete. As far as I can tell, I know nothing more about g(x). So, being the naive little statistician that I am, I would try and set the integral of g*f equal to zero and show that g is identically zero.

(ps the book sucks so there's a good chance the authors performed a little hand waving, although they stress this example so I would guess the error is on my part, not theirs)
(pps the book is hogg mckean and craig)
(ppps is there a way to use dollar signs $ instead of typing [ tex ] each time?)

 

[Dick]

Do you mean f(x)=k*exp(-k*|x-theta|)? Then as k->infinity f(x) behaves like a multiple of the delta function centered at theta. Is that what they are getting at?

 

[Mute]

The key to proving that is the fact that \theta is arbitrary. The point is not that the integral is zero for some \theta, it's that it's zero for every \theta.

You can prove it for you specific function by, for example, taking a derivative with respect to \theta. Note, however, I think that it will require that g(x) be continuous over the integration region. I'm not sure if there is a more general proof for your function.