Let \((X,\mathcal M,\mu)\) and \((Y,\mathcal N,\nu)\) be \(\sigma\)-finite measure spaces with product measure \(\mu\times\nu\) on \(\mathcal M\otimes\mathcal N\). Tonelli’s and Fubini’s theorems both equate \(\int_{X\times Y} f\,d(\mu\times\nu)\) with the two iterated integrals; Tonelli’s theorem assumes \(f\geq0\) and Fubini’s theorem assumes \(f\in L^1(\mu\times\nu)\). Which one applies? compares the hypotheses.
Tonelli: nonnegative measurable functions
For \(f\geq0\) measurable on \(X\times Y\), the iterated integrals and the integral over \(X\times Y\) are equal as elements of \([0,\infty]\). For a double sequence \(a_{mn}\geq0\) (counting measure on \(\NN\times\NN\)), this gives \(\sum_m\sum_n a_{mn} = \sum_n\sum_m a_{mn}\).
Tonelli's theorem for nonnegative measurable functions (Tag T-6PRW3)
Let \(f\colon\RR^n\times\RR^m\to[0,\infty]\) be Lebesgue measurable on \(\RR^{n+m}\). For \(x\in\RR^n\) let \(f_x\colon\RR^m\to[0,\infty]\) be the slice \(f_x(y)\coloneqq f(x,y)\). Then:
for almost every \(x\in\RR^n\), the slice \(f_x\) is Lebesgue measurable on \(\RR^m\);
the function \(F(x)\coloneqq\int_{\RR^m} f(x,y)\,dy\in[0,\infty]\), defined for almost every \(x\), is Lebesgue measurable on \(\RR^n\);
\[ \int_{\RR^{n+m}} f = \int_{\RR^n} F(x)\,dx = \int_{\RR^n}\qty{\int_{\RR^m} f(x,y)\,dy}dx \in[0,\infty], \] and the same holds with the roles of \(x\) and \(y\) exchanged [@Fol13].
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.
Tonelli's theorem (Tag FT-4JRQX)
Let \(f\colon\RR^{n}\times \RR^k\to[0,\infty]\) be Lebesgue measurable on \(\RR^{n+k}\). For \(y\in\RR^k\), let \(f^y\colon\RR^n\to[0,\infty]\) be the slice \(f^y(x) \coloneqq f(x, y)\). Then:
For almost every \(y\in \RR^k\), the slice \(f^y\) is measurable on \(\RR^n\).
The function \(F(y) \coloneqq \int_{\RR^n} f^y(x) \, dx\), defined for almost every \(y\), is measurable on \(\RR^k\).
In \([0,\infty]\), \[ \int_{\RR^{n+k}} f(x,y) \, d(x,y) = \int_{\RR^k} \qty{\int_{\RR^n} f(x,y) \,dx}\, dy . \]
The same statements hold with the roles of \(x\) and \(y\) exchanged, so the two iterated integrals are equal.
The hypotheses are that \(f\) is nonnegative and measurable; \(f\) need not be integrable, and all three integrals in (3) may be \(+\infty\). Fubini’s theorem instead assumes \(f\in L^1(\RR^{n+k})\) and allows \(f\) to take values of both signs.
Fubini: integrable functions
For \(f\in L^1(\mu\times\nu)\), the sections \(f(x,\wait)\) are in \(L^1(\nu)\) for \(\mu\)-almost every \(x\), the function \(x\mapsto\int_Y f(x,y)\,d\nu(y)\) is in \(L^1(\mu)\), and the iterated integrals are finite and equal to \(\int_{X\times Y} f\,d(\mu\times\nu)\); the same holds with the roles of \(X\) and \(Y\) exchanged.
Fubini's theorem for integrable functions (Tag T-4GPEF)
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.
Fubini--Tonelli theorem (Tag T-X7XZX)
Let \((X,\mcm,\mu)\) and \((Y,\mcn,\nu)\) be \(\sigma\)-finite measure spaces, and let \(f\colon X\times Y\to\CC\) be \(\mcm\otimes\mcn\)-measurable. Suppose that one of the iterated integrals of \(\abs{f}\) is finite: \[ \int_X\qty{\int_Y\abs{f(x,y)}\,d\nu(y)}d\mu(x)<\infty\quad\text{or}\quad\int_Y\qty{\int_X\abs{f(x,y)}\,d\mu(x)}d\nu(y)<\infty. \] Then \(f\in L^1(X\times Y,\mu\times\nu)\), and \[ \int_{X\times Y}f\,d(\mu\times\nu)=\int_X\qty{\int_Y f(x,y)\,d\nu(y)}d\mu(x)=\int_Y\qty{\int_X f(x,y)\,d\mu(x)}d\nu(y), \] where the inner integrals exist for almost every value of the outer variable.
Fubini--Tonelli theorem (Tag FT-H6AWV)
Let \(f\colon\RR^n\times\RR^k\to\CC\) be Lebesgue measurable on \(\RR^{n+k}\), and suppose that one of the iterated integrals of \(\abs{f}\) is finite: \[ \int_{\RR^k}\qty{\int_{\RR^n}\abs{f(x,y)}\,dx}dy<\infty \quad\text{or}\quad \int_{\RR^n}\qty{\int_{\RR^k}\abs{f(x,y)}\,dy}dx<\infty. \] Then \(f\in L^1(\RR^{n+k})\), and \[ \int_{\RR^{n+k}} f(x,y)\,d(x,y)=\int_{\RR^k}\qty{\int_{\RR^n} f(x,y)\,dx}dy=\int_{\RR^n}\qty{\int_{\RR^k} f(x,y)\,dy}dx, \] where the inner integrals exist for almost every value of the outer variable.
By Tonelli's theorem applied to the nonnegative measurable function \(\abs{f}\), \(\int_{\RR^{n+k}}\abs{f}\) equals each iterated integral of \(\abs{f}\), so it is finite and \(f\in L^1(\RR^{n+k})\). Fubini’s theorem for \(f\in L^1(\RR^{n+k})\) then gives the equality of \(\int_{\RR^{n+k}}f\) with both iterated integrals of \(f\).
Hypotheses of Tonelli's and Fubini's theorems (Tag FT-VHK2H)
Let \(f\) be a Lebesgue measurable function on \(\RR^n\times\RR^k\).
Tonelli. If \(f\) takes values in \([0,\infty]\), then \(\int_{\RR^{n+k}}f\) equals both iterated integrals of \(f\), as elements of \([0,\infty]\) (FT-4JRQX).
Fubini. If \(f\in L^1(\RR^{n+k})\), then \(\int_{\RR^{n+k}}f\) equals both iterated integrals of \(f\), whose inner integrals exist for almost every value of the outer variable (T-4GPEF).
Other interchanges
For nonnegative terms, sums and integrals commute by the monotone convergence theorem, or by Tonelli’s theorem with counting measure. For differentiation under the integral sign: if \(f(\wait,t)\in L^1(\mu)\) for each \(t\in(a,b)\), \(\partial_t f(x,t)\) exists for all \(x\in X\) and \(t\in(a,b)\), and \(\abs{\partial_t f(x,t)}\leq g(x)\) for some \(g\in L^1(\mu)\), then \(F(t)\coloneqq\int_X f(x,t)\,d\mu(x)\) is differentiable on \((a,b)\) with \(F'(t)=\int_X\partial_t f(x,t)\,d\mu(x)\), by the mean value theorem and the dominated convergence theorem.
Differentiation under the integral sign (Tag PR-V4MOK)
Let \((X,\mcm,\mu)\) be a measure space, let \(I\subseteq\RR\) be an open interval, and let \(f\colon X\times I\to\CC\) be such that \(x\mapsto f(x,t)\) is integrable for each \(t\in I\). Put \(F(t)\coloneqq\int_X f(x,t)\dmu(x)\). Suppose \(\frac{\partial f}{\partial t}(x,t)\) exists for all \(x\in X\) and \(t\in I\), and there is \(g\in L^1(\mu)\) with \(\abs{\frac{\partial f}{\partial t}(x,t)}\leq g(x)\) for all \(x\in X\) and \(t\in I\). Then \(F\) is differentiable on \(I\) and \[ F'(t) = \int_X \frac{\partial f}{\partial t}(x, t) \dmu(x) \] [@Fol13].
Series of nonnegative measurable functions can be integrated term by term (Tag PR-JW3QE)
Let \((X,\mcm,\mu)\) be a measure space and let \(f_n\in\) $L^+$ for \(n\geq1\). Then \[ \int_X \sum_{n=1}^\infty f_n\dmu = \sum_{n=1}^\infty\int_X f_n\dmu \] in \([0,\infty]\) [@Fol13].
Integration by parts, special case (Tag E-WXIRH)
\[ F(x):=\int_{0}^{x} f(y) d y \quad \text { and } \quad G(x):=\int_{0}^{x} g(y) d y \\ \implies \int_{0}^{1} F(x) g(x) d x=F(1) G(1)-\int_{0}^{1} f(x) G(x) d x .\]
Solution
Let \(f, g \in L^1[0,1]\) and let \(T = \theset{(x,y) : 0 \le y \le x \le 1}\).
<1>1. \(\iint_{T} |f(y)||g(x)|\,dy\,dx \le \|f\|_1\|g\|_1 < \infty\).
By Tonelli’s theorem, \(\iint_T |f(y)||g(x)|\,dy\,dx \le \int_0^1\int_0^1 |f(y)||g(x)|\,dy\,dx = \|f\|_1\|g\|_1\).
<1>2. \(\int_0^1 F(x)g(x)\,dx = \int_0^1 f(y)\big(G(1) - G(y)\big)\,dy\).
By the definition of \(F\), \(\int_0^1 F(x)g(x)\,dx = \int_0^1\int_0^x f(y)g(x)\,dy\,dx\). By step <1>1, Fubini’s theorem allows integrating over \(T\) in the other order, with \(y \in [0,1]\) and \(x \in [y,1]\), giving \(\int_0^1 f(y)\int_y^1 g(x)\,dx\,dy\). Finally \(\int_y^1 g = G(1) - G(y)\).
<1>3. Q.E.D.
By step <1>2 and \(F(1) = \int_0^1 f\), \(\int_0^1 Fg = G(1)\int_0^1 f - \int_0^1 fG = F(1)G(1) - \int_0^1 fG\).