Concluding $g(t)=0$ almost everywhere based on an integral Let $g(x,y)$ be a measurable function. Suppose
$$\int_0^\infty \int_0^\infty g(x,y)h(x,y)y^n dy dx = 0$$
where $h(x,y) > 0$ for all  $x,y \in [0,\infty)\times[0,\infty)$ and the above holds for all $ n > 1$. 
Is it possible to conclude that $g(x,y) = 0$ everywhere on the area integrated?
This seems like a continuous version of the problem
$$\sum_{i=1}^n g(i)z^i$$
in which case you can conclude that $g(i)=0$ for $i=1,2,\dots,n$ using the fundamental theorem of algebra.
Does this hold for the above example as well? And if so, how would I go about showing this?

Consider $g(x,y)=\frac 1 {h(x,y)}(\chi_{[0,1]}(x)-\chi_{[1,2]}(x) )e^{-y}$.
We have :
$$\int_0^\infty \int_0^\infty g(x,y)h(x,y)y^n dy dx = \int_0^1  \int_0^\infty e^{-y}y^n dy dx -\int_1^2  \int_0^\infty e^{-y}y^n dy dx =0.$$
But $g \not \equiv 0$.
