Let \((X,\mu)\) and \((Y,\nu)\) be measure spaces and \(f\) a \(\mu\times\nu\)-measurable function on \(X\times Y\). The exact statements are on The statements.
Hypotheses of each theorem
Tonelli. If \(\mu\) and \(\nu\) are \(\sigma\)-finite and \(f \geq 0\), then the two iterated integrals and \(\int_{X\times Y} f\,d(\mu\times\nu)\) are equal, possibly all equal to \(+\infty\).
Fubini. If \(\mu\) and \(\nu\) are \(\sigma\)-finite and \(f \in L^1(\mu\times\nu)\), then the two iterated integrals are finite and equal to \(\int_{X\times Y} f\,d(\mu\times\nu)\).
If \(\mu\) and \(\nu\) are \(\sigma\)-finite and one iterated integral of \(\abs f\) is finite, then \(f\in L^1(\mu\times\nu)\), and the iterated integrals of \(f\) are equal.
By Tonelli’s theorem applied to \(\abs f\), \(\int_{X\times Y}\abs f\,d(\mu\times\nu)\) equals the finite iterated integral, so \(f\in L^1(\mu\times\nu)\), and Fubini’s theorem applies to \(f\).
Counterexamples without the hypotheses
(Without $\sigma$-finiteness)
Let \(\mu\) be Lebesgue measure and \(\nu\) counting measure on \([0,1]\), and let \(f\coloneqq\chi_\Delta\) for the diagonal \(\Delta\subseteq[0,1]^2\). Then \(\int\qty{\int f(x,y)\,d\nu(y)}d\mu(x) = \int_0^1 1\dx = 1\) and \(\int\qty{\int f(x,y)\,d\mu(x)}d\nu(y) = \sum_{y} 0 = 0\).
(Without nonnegativity or integrability)
On \([0,1]^2\) with Lebesgue measure, \(f(x,y) \coloneqq \frac{x^2-y^2}{(x^2+y^2)^2}\) has \[ \int_0^1\int_0^1 f(x,y)\dy\dx = \frac\pi4, \qquad \int_0^1\int_0^1 f(x,y)\dx\dy = -\frac\pi4, \] so \(f\notin L^1([0,1]^2)\).
Other applications
(Computing a single integral as a double integral)
Since \(\frac1x = \int_0^\infty e^{-xt}\dt\) for \(x>0\), applying Fubini’s theorem on \([0,A]\times[0,\infty)\) to \(e^{-xt}\sin x\) and letting \(A\to\infty\) gives \(\int_0^\infty \frac{\sin x}{x}\dx = \int_0^\infty\frac{\dt}{1+t^2} = \frac\pi2\).
(Layer cake formula)
If \(\mu\) is \(\sigma\)-finite and \(f\) is measurable, then \(\int \abs f\,d\mu = \int_0^\infty \mu(\theset{\abs f > t})\dt\).
Apply Tonelli’s theorem to \(\chi_{\theset{(x,t) \st 0<t<\abs{f(x)}}}\) on \(X\times(0,\infty)\).
For \(f,g\in L^1(\RR^n)\), Tonelli’s theorem gives \(\norm{f*g}_1\leq\norm f_1\norm g_1\), and Fubini’s theorem gives associativity of convolution; see Convolution.