Preimage of the product of two functions, Tonelli-Fubini's Theorem, Measurable functions.

Question

Let $F=fg$ where $f,g $ are functions. What is the relation between the preimage of $F$ and the preimage of $f$ and $g$ ?

Just in case, if you are able to understand the following, the main problem that i'm trying to solve is,

Let $f:\mathbb{R}^n\to \mathbb{R}$ be a measurable function, define $F:\mathbb{R}^{2n}\to \mathbb{R}$ by $F(x,y)=\chi_{Q_i}(x)f(x)-\chi_{2Q}(y)f(x+y)$ where $x,y\in\mathbb{R}^n$, $\chi_{Q_1},\chi_{Q_2}$ is the characteristic function of the open cubes $Q_1,Q_2\subset\mathbb{R}^n$ and $f:\mathbb{R}^n\to \mathbb{R}$ is measurable, then $F$ is measurable.

and i'm also trying to solve this for be able to use Tonelli-Fubini's theorem for the following integral, $$\int\limits_{Q_1}\int\limits_{Q_2}\lvert f(x)-f(x+y)\rvert^p dydx$$ where $f\in L^p(\mathbb{R^n})$

Answer 1

Hint: $fg=\frac{(f+g)^2 - f^2 - g^2}{2}$ This helps for seeing $F$ is measurable.

Don't expect to find a relationship between preimage of $F$ and preimages of $f, g$! Since $F$ does not have uniue representation ! for example :

$$F=fg= e^xf \times e^{-x}g$$

Answer 2

Indeed, I just realize my first question is really naive, and the second one is really bad formulated, but the thrid one, that integral where I'm trying to apply Fubini-Tonelli's theorem is the one that I need to answer, I need to justify why that function $(x,y)\to f(x+y)-f(x)$ is measurable in $R^{2n}$, which is the big integration space. I remark my knowledged in product spaces is kind of null. If anyone can explain me, I would appreciate it.