Challenging-conventional-wisdom question based on an answer to my previous question.
If $X \in L^1 (\Omega, \mathscr{F}, \mathbb{P})$ has pdf $f_X$, $Y \in L^1 (\Omega, \mathscr{F}, \mathbb{P})$ has pdf $f_Y$, and they have joint pdf $f_{XY}(x,y)$, we can say that:
Since $f_{XY}(x,y)$ is non-negative when $x$ is non-negative, and $-x\;f_{XY}(x,y)$ is non-negative when $x$ is negative, we can apply Tonelli's/Fubini's Theorem for non negative functions by separating the integral into two.
$$\begin{align}\int_\Bbb R\int_\Bbb R x\;f_{XY}(x,y)\operatorname d x\operatorname d y & =\int_\Bbb R\left(\int_{\Bbb R^+} x f_{XY}(x,y) \operatorname d x-\int_{\Bbb R^-} (-x)f_{XY}(x,y)\operatorname d x\right)\operatorname d y \\[1ex] ~ & = \int_{\Bbb R^+} x\int_{\Bbb R}f_{XY}(x,y)\operatorname d y\operatorname d x-\int_{\Bbb R^-}(-x)\int_\Bbb R f_{XY}(x,y)\operatorname d y\operatorname d x \\[1ex] ~ & = \int_\Bbb R x\int_\Bbb R f_{XY}(x,y)\operatorname d y\operatorname d x\end{align}$$
Soooooo why can't we just decompose every integrand like extending from nonnegative functions to general functions like in Lebesgue integrals i.e. the following?
$$\int \int g(x,y) dx dy = \int \int g^{+}(x,y) dx dy + \int \int g^{-}(x,y) dx dy$$
What exactly is the problem otherwise? Counterexample pls? I'm guessing in three dimensions '$g^{+}(x,y)$' or '$g^{-}(x,y)$' don't necessarily exist. Not quite sure how I would define those. I guess the integrand must change sign w/rt at most 1 variable in order for Tonelli's/Fubini's to apply?
