Fubini's theorem

Fubini's theorem permits the order of integration of an absolutely integrable function on a product measure space to be exchanged.

The function $(x-y)/(x+y{)}^3$ on the unit square is not Lebesgue integrable. On $0<y<x/2$, its absolute value is at least $4/(27x^2)$, whose area integral diverges logarithmically. Its two signed iterated integrals nevertheless exist, ignoring null boundary sections, and equal $1/2$ and $-1/2$. This illustrates why absolute integrability is required for Fubini's theorem.

Define the displayed function away from the origin and set $f(0,0)=0$. Every section in either coordinate is integrable on $[-1,1]$ and is an odd function, so both orders of iterated integrals on $[-1,1{]}^2$ are zero. However, on the annulus $\epsilon <\sqrt{x^2+y^2}<1$, polar integration givesThus the function is not Lebesgue integrable jointly. Reflection interchanges its positive part and negative part, whose integrals are both infinite. Existence and even agreement of finite iterated integrals do not establish the absolute integrability hypothesis of Fubini's theorem.

Let $f$ be a Borel measurable function in L1 space on $\mathbb{R}$, and let $a,b$ be measurable real functions of $x$ with $b(x)>0$ and $W={\int }_{\mathbb{R}}|a(x)|/b(x)dx<\infty$. The joint integrand $G(x,y)=a(x)f(b(x)y)$ is measurable, and Tonelli's theorem followed by Lebesgue substitution on an interval givesFor each fixed $x$, substitute $z=b(x)y$ in the inner $y$ integral. The absolute Jacobian is $1/b(x)$, leaving $|a(x)|\Vert f{\Vert }_1/b(x)$. Integrating in $x$ proves the formula and absolute integrability. Fubini's theorem then permits both orders of signed integration, givingThe one-dimensional sections are integrable for almost every value of the other variable; the theorem does not assert integrability of every section.

For a planar Lebesgue measurable set $B$, its vertical section $B_x$ is a one-dimensional Lebesgue measurable set for $x$ almost everywhere. Write $B△C\subseteq Z$ with $C$ a planar Borel set and $Z$ a Borel Lebesgue null set. Every section $C_x$ and $Z_x$ is Borel; Tonelli's theorem gives $\int m(Z_x)dx=m_2(Z)=0$. Thus $m(Z_x)=0$ for $x$ almost everywhere, and $B_x△C_x\subseteq Z_x$ proves the assertion by completeness of Lebesgue measure. No finite-measure condition on $B$ is needed.

Exceptional sections need not be measurable: $\{0\}\times A$ is a planar Lebesgue null set for every $A\subseteq \mathbb{R}$, even when $A$ is a non-measurable set. Conversely, if both $A_1$ and $A_2$ are non-measurable, $A_1\times A_2$ cannot be measurable. Otherwise every $x\in A_1$ would have to lie in the null exceptional set, making $A_1$ measurable by completeness of Lebesgue measure.

For $0<a<b$, the nonnegative function $e^{-xy}{1}_{\{x>0,a\le y\le b\}}$ is measurable and has integral ${\int }_a^{b}dy/y=\log (b/a)$. Tonelli theorem first proves integrability without assuming it; Fubini theorem then permits the order of integration to be exchanged. Integrating in $y$ gives the stated Frullani integral. The cancellation between the exponentials is essential at $x=0$.

For $\alpha >1$, write $[\mathrm{sech}x-\mathrm{sech}(\alpha x)]/x={\int }_1^{\alpha }\mathrm{sech}(xy)\tanh (xy)dy$. The integrand is nonnegative, so Tonelli's theorem permits reversal. Substitution $u=xy$ makes the inner $x$ integral $1/y$ because ${\int }_0^{\infty }\mathrm{sech}u\tanh udu=1$. Integrating $dy/y$ proves the formula. The original difference is $O(x^2)$ at zero, making division by $x$ integrable. For $0<\alpha <1$, substitution and reversal of the two terms give the same $\ln \alpha$ result; at $\alpha =1$ both sides vanish.