Condition for Equality of Two Integrals

Question

If I have an arbitrary scalar function f(x,y,z) that is non-zero and positive within a volume $V$ of $\Bbb R^3$, and I have a constraint that requires

$$\int_{V} f(x,y,z)\, dV =\int_{V} f(x,y,z) \cdot g(x,y,z)\, dV$$

where g(x,y,z) is a unique given function and f(x,y,z)>0 inside $V$ and =0 outside $V$ .

It would appear to me that this requires g(x,y,z)=1 given that f(x,y,z) is arbitrary.

What would be the best way to strictly prove this mathematically?

Thanks in advance.

Answer 1

Since $g$ is fixed it can come before the integral(as any number would).

Answer 2

Is there not a uniqueness theorem for integration? This would reduce the problem to identifying the g function as a multiplicative identity by inspection which is 1.