Fubini's Theorem: Let $\mu,\nu$ be two $\sigma$-finite measures and $f$ be a measurable function. If $f \ge 0$ or $f \in L^1(\mu\times \nu)$, then $$ \int f ~d \mu \times \nu = \iint f ~d\mu ~d\nu = \iint f ~d\nu ~d\mu.$$
When $f^+ \in L^1(\mu\times \nu)$, but $f$ may be not integrable: using Fubini's Theorem for $f^+,f^-$, we have that $$ \int f^+ ~d \mu \times \nu = \iint f^+ ~d\mu ~d\nu = \iint f^+ ~d\nu ~d\mu,$$ $$ \int f^- ~d \mu \times \nu = \iint f^- ~d\mu ~d\nu = \iint f^- ~d\nu ~d\mu.$$ Note that $\int f^+~ d\mu$ is finite $\nu$-a.e. Thus we obtain that \begin{align*} \int f ~d \mu \times \nu & = \int f^+ ~d \mu \times \nu - \int f^- ~d \mu \times \nu \\ & = \iint f^+ ~d\mu ~d\nu - \iint f^- ~d\mu ~d\nu \\ & = \int \left(\int f^+ ~d\mu - \int f^- ~d\mu \right)~d\nu \\ & = \iint f ~d\mu ~d\nu. \end{align*} Therefore, the equality in Fubini's Theorem is valid for $f$, where we only have that $f^+ \in L^1(\mu\times \nu)$.
Is there any wrong steps? I'm not sure whether it is true.