= Solution
Let $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ be <measure spaces> with <sigma-finite measures>, and use their <product measure>.
\b[<Fubini's theorem>:] if $u$ is a measurable absolutely integrable <function> on $X\times Y$, then its sections are integrable for <almost every> fixed value of the other coordinate, the resulting section-integral <functions> are integrable, and
$$
\boxed{\int_{X\times Y}u\,d(\mu\times\nu)
=\int_X\left(\int_Yu(x,y)\,d\nu(y)\right)d\mu(x)
=\int_Y\left(\int_Xu(x,y)\,d\mu(x)\right)d\nu(y).}
$$
All three <integrals> are finite. Values of an inner <integral> on an exceptional <null set> can be assigned arbitrarily.
\b[<Tonelli's theorem>:] if $u:X\times Y\to[0,\infty]$ is measurable and nonnegative, the same equality holds for its nonnegative product and iterated <integrals>, allowing the common value $+\infty$. No advance <absolute integrability> assumption is needed. On a completed <Lebesgue measure> product, the section and measurability assertions are interpreted <almost everywhere>.
Thus <Tonelli's theorem> can first establish that $\int|u|<\infty$; after this, <Fubini's theorem> permits signed integration in either order.
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