Let \(f\colon\RR^n\times\RR^m\to\CC\) be integrable with respect to Lebesgue measure on \(\RR^{n+m}\). For \(x\in\RR^n\) let \(f_x\colon\RR^m\to\CC\) be the slice \(f_x(y)\coloneqq f(x,y)\). Then:
for almost every \(x\in\RR^n\), the slice \(f_x\) is integrable on \(\RR^m\);
the function \(F(x)\coloneqq\int_{\RR^m} f(x,y)\,dy\), defined for almost every \(x\), is integrable on \(\RR^n\);
\[ \int_{\RR^{n+m}} f = \int_{\RR^n}\qty{\int_{\RR^m} f(x,y)\,dy}dx , \] and the same holds with the roles of \(x\) and \(y\) exchanged.
Moreover, if \(E\subseteq\RR^{n}\times\RR^m\) is Lebesgue measurable, then for almost every \(x\in\RR^n\) the slice \(E_x\coloneqq\theset{y\in\RR^m \suchthat (x, y) \in E}\) is Lebesgue measurable.
The hypothesis is \(f\) integrable on \(\RR^{n+m}\); \(f\) may take values of both signs. Tonelli's theorem instead assumes \(f\geq0\) measurable, with no integrability hypothesis.