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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.

Suppose an unknown has prior distribution $\Pi$ and its conditional observation law $P_u$ has jointly measurable Radon-Nikodym derivative $L(u,m)$ relative to a fixed observation law $P_0$. If $L>0$ for $\Pi (du)P_0(dm)$-almost every pair, thendefines the posterior distribution for almost every observation. Indeed, $\int L(u,m)P_0(dm)=1$ for almost every $u$, so Tonelli theorem gives $\int Z(m)P_0(dm)=1$. Hence $Z$ is finite almost everywhere, and positivity of $L$ and Fubini's theorem give $Z>0$ almost everywhere. The data law is $Z{P}_0$, and integration of the proposed posterior against this data law recovers the joint law. This proves the conditional distribution property and also shows that the data law is an equivalent probability measure to $P_0$.

For $p,q>0$, express the product of the two Gamma integrals as the nonnegative double integralUse $r=x+y$ and $u=x/(x+y)$, with inverse $x=ru$, $y=r(1-u)$ and Jacobian determinant of absolute value $r$. The new domain of integration is $r>0$, $0<u<1$, and the integrand with this factor becomes $r^{p+q-1}{e}^{-r}{u}^{p-1}(1-u{)}^{q-1}$. Tonelli's theorem and separation of the two integrals therefore give $\Gamma (p+q)B(p,q)$. The positive real parameter hypothesis ensures that all the original integrals are finite.

For complex parameters with positive real parts, the same proof first gives the undivided identity $\Gamma (p+q)B(p,q)=\Gamma (p)\Gamma (q)$. Absolute integrability follows by taking real parts in the exponents, so Fubini's theorem applies. The nonvanishing of the Gamma function then permits division by $\Gamma (p+q)$.

Let $a\in BV(\Omega )$, $b\in BV(\Omega )\cap L^{\infty }(\Omega )$ and ${\delta }_{t}c=c\circ {\Phi }_t-c$, where ${\Phi }_t$ is the local flow of $\phi e_j$ with nonnegative $\phi \in C_c^{\infty }(\Omega )$. The formula uses common oriented BV traces on a hypersurface; the trace difference of $a$ is zero almost everywhere on $J_b\setminus J_a$. It also holds for both negative-time increments with denominator positive $t$.

The BV slicing theorem reduces the calculation to one dimension. With a right-continuous representative of $a$ and $\mu =Da$, Fubini's theorem yieldsThe inner factor tends to $\phi (s)[b](s)$ and is bounded in absolute value by $2\Vert b{\Vert }_{\infty }\Vert \phi {\Vert }_{\infty }$. Dominated convergence against $|\mu |$ leaves only its measure atoms at the countable slice jumps of $b$. That same bound times $|D{a}_y|$ permits integration over the transverse coordinates. No essential bound on $a$ is used, even if its slice values grow without bound. The formula is useful for opposite-flow fidelity identity for quadratic data with unbounded data and a bounded comparison function.

For finite complex measures on the additive circle group, convolution is the pushforward measure of their product under addition modulo $2\pi$. EquivalentlyBounded tests are integrable against the product variation measure, so Fubini's theorem applies. Testing the factorization $e^{-in(x+y)}=e^{-inx}{e}^{-iny}$ proves the displayed Fourier coefficient of a measure on the circle identity. The convolution of two probability measures is again a probability measure. On these measures convolution preserves simultaneous convergence in the weak-star topology: multiply the converging coefficients and use their common unit variation bound and dense trigonometric tests.

For Lebesgue integrable functions $f,g$ on $\mathbb{R}$, use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. ThenThus the integral combining all translated products becomes an ordinary multiplication at each frequency. Different Fourier transform normalizations can place a constant in this identity; fixing the convention prevents that ambiguity. Its use in filtering is illustrated by ^[ref].

For the indicator function $q=1_{[0,1]}$,with $\mathrm{sinc}(0)=1$. Therefore the triangular function $q\ast q$ has Fourier transform $e^{-2\pi i\xi }{\mathrm{sinc}}^2(\xi )$. The extra factor of the sinc function improves the large-frequency bound from a constant times $1/|\xi |$ to one times $1/|\xi {|}^2$, matching the smoother output in the overlap picture. Fourier decay of repeated interval convolutions extends this calculation to further repetitions.

For a proof outline, Fubini's theorem permits exchanging the two integrals because the integral of $|f(y)g(x-y)|$ over $(x,y)$ is $\Vert f{\Vert }_1\Vert g{\Vert }_1<\infty$. Set $u=x-y$ in the inner integral:The factorization of the complex exponential is the mechanism behind the simplification. The discrete convolution version similarly turns coefficient sums into products of generating functions.

For functions $f,g$ on $[0,\infty )$ with $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ Lebesgue integrable functions, their unilateral convolution, defined almost everywhere if necessary, satisfiesTaking a causal extension by zero of both functions gives the existing whole-line convolution definition. Absolute convergence and Fubini's theorem justify the product rule: put $t=\tau +u$ in the double integral, so $e^{-st}=e^{-s\tau }{e}^{-su}$. For piecewise continuous functions of exponential order $a$ and $b$, the hypotheses hold for $\mathrm{Re}s>\max (a,b)$. The Laplace transform article works out the concrete example $h(t)=e^{-2t}$ and the Heaviside step function input, whose output is $(1-e^{-2t})/2$. ^[ref].

For this sign convention, cross-correlation compares a function with translated copies of another, conjugating the first factor. If $f,g\in L^1(\mathbb{R})$, Fubini's theorem gives Fourier transform $\widehat{f\star g}(k)=\overline{\hat{f}(k)}\hat{g}(k)$. Indeed, the substitution $s=x+y$ factors the double integral into $\int \overline{f(y)}{e}^{iky}dy$ and $\int g(s)e^{-iks}ds$. The zero-lag value for $f=g\in L^2$ is the squared L2 norm. Fourier inversion applied to this correlation proves the Parseval identity, first for Schwartz functions and then by density for L2 space.

Write $e^u\varphi (u)=\int \psi (t)e^{-iut}dt$. The double integral $\iint \psi (t)\psi (t^{\prime })(1+it)(1+i{t}^{\prime })/(2+i(t+t^{\prime }))dtd{t}^{\prime }$ equals the displayed constant. Express the denominator as a Laplace integral and each differentiated Fourier factor as $-{\varphi }^{\prime }(u)$, then use Fubini's theorem. For a real cutoff the resulting square is positive even though the Fourier transform factors have no complex conjugation. The Cauchy-Schwarz inequality and $\varphi (0)=1$, $\varphi (1/3)=0$ give $c_{\varphi }\ge 3$.

Suppose $h$ is continuously differentiable on $[0,L]$. For $0<x\le L$, the improper integral $g(x)={\int }_0^{x}h(t)(x-t{)}^{-1/2}dt$ exists, but substituting directly into the ordinary Leibniz integral rule produces an undefined endpoint term. First use integration by parts to obtainThe new endpoint term at $t=x$ vanishes, and the differentiated kernel is locally integrable. Differentiating gives the displayed formula. One rigorous justification rewrites the second term as ${\int }_0^x[{\int }_0^s{h}^{\prime }(t)(s-t{)}^{-1/2}dt]ds$ by Fubini's theorem; the inner integral is continuous for $s>0$, so the fundamental theorem of calculus applies. Moreover $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$: if $h(0)\ne 0$, the right difference quotient at zero diverges rather than defining a finite derivative.

For independent integrable variables, $\mathbb{E}{\prod }_i{X}_i={\prod }_i\mathbb{E}{X}_i$. Tonelli theorem first gives $\mathbb{E}{\prod }_i|X_i|={\prod }_i\mathbb{E}|X_i|<\infty$; then Fubini's theorem factors the signed joint integral.

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.

If the destination transition kernel has probability density function $m_t(x,y)$, the formal adjoint operator of $L_{t}f=a{f}^{\prime }+\alpha (K_{t}f-f)$ is $(L_t^{\ast }p)(x)=-{\partial }_x(a(t,x)p(x))-\alpha (x)p(x)+\int \alpha (y)m_t(y,x)p(y)dy$. Multiply by a smooth compactly supported test function, use integration by parts on transport and Fubini's theorem on the jump gain to prove the formula when the integrals are finite. The general formulation is $\frac{d}{dt}\int fd{\nu }_t=\int L_{t}fd{\nu }_t$ for probability measures ${\nu }_t$. A deterministic initial value leaves positive no-jump atomic mass at finite times, so a full smooth density cannot be assumed for that initialization.

Let $f$ be compactly supported and continuous, let $g,\hat{g}\in L^1(\mathbb{R})$, and use $\hat{g}(t)=\int g(x)e^{itx}dx$. If $g(0)\ne 0$, thenThis follows by Fubini's theorem, a change of variables, Fourier inversion theorem, and dominated convergence theorem. It gives a direct proof of the uniqueness theorem for characteristic functions because the right side depends only on ${\phi }_Z$.

For Schwartz functions $f,g$ and the Fourier transform convention $\hat{f}(\xi )=\int f(x)e^{-ix\xi }dx$, the displayed formula exchanges a pointwise product for a convolution. Substitute Fourier inversion for $g$ into the Fourier transform of $fg$. Fubini's theorem, justified by $\Vert f{\Vert }_1\Vert \hat{g}{\Vert }_1<\infty$, givesUnlike the convolution theorem $\widehat{f\ast g}=\hat{f}\hat{g}$, this version cannot be asserted for arbitrary pairs in $L^1$ without additional hypotheses, since their pointwise product need not be an integrable function.

For $\alpha >0$, put $a=\alpha /2$. The regular distribution of $(1+x^2{)}^{-a}$ is a tempered distribution, since the function is bounded. Its Fourier transform is the integrable functionwith arbitrary finite value at zero when the displayed integral diverges there. To prove the identity, the gamma function gives $(1+x^2{)}^{-a}=\Gamma (a{)}^{-1}{\int }_0^{\infty }{t}^{a-1}{e}^{-t}{e}^{-t{x}^2}dt$. Pair this with a Schwartz function, use the Fourier transform of a Gaussian, and interchange integrals by Fubini's theorem. The absolute bound for the transformed pairing follows from Tonelli's theorem:The positive kernel and dominated convergence give positivity and continuity away from zero. It has a continuous representative at zero exactly when $\alpha >1$, in which case $h_{\alpha }(0)=\sqrt{\pi }\Gamma ((\alpha -1)/2)/\Gamma (\alpha /2)$. For $\alpha \le 1$, Fatou's lemma gives $h_{\alpha }(\xi )\to \infty$ as $\xi \to 0$, so no change on a set of zero Lebesgue measure can give continuity.

With $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, the Fourier transform of a tempered distribution is defined byContinuity of the Fourier transform on the Schwartz space makes this well defined. For regular integrable functions, Fubini's theorem shows that it agrees with the ordinary transform. The inverse convention has factor $(2\pi {)}^{-n}$ and the positive exponential.

Green’s theorem converts circulation around a positively oriented planar boundary into the area integral of scalar curl.

Let $D$ be a bounded region in the Euclidean plane whose boundary is a finite union of disjoint, piecewise smooth simple closed curves. Suppose the vector field $F=(P,Q)$ is continuously differentiable on an open set containing the whole closed region. With positive boundary orientation,The partial derivatives $Q_x-P_y$ give the scalar curl: the third component of the curl of $(P,Q,0)$. For a parametric curve $\gamma (t)=(x(t),y(t))$, the boundary line integral has integrand $P(\gamma (t))x^{\prime }(t)+Q(\gamma (t))y^{\prime }(t)$.^[ref]

Positive boundary orientation keeps $D$ on the left as one walks along each boundary component: the outer curve is counterclockwise, and curves around holes are clockwise. Every component belongs in the line integral. The vector field must be regular throughout $D$, but need not extend into a hole.^[ref]

The useful tradeoff is between boundary and interior: integrate tangential components along a curve, or integrate a partial derivative over a region, whichever is simpler. The boundary formula for planar area uses this freedom in the opposite direction, recovering area from a line integral. As a planar case of Stokes theorem, Green theorem also explains why a local curl can determine a complete boundary measurement while leaving room for the global obstruction in Green theorem with holes.

A short proof starts with a rectangle $D=[a,b]\times [c,d]$. Add its four positively oriented edge line integrals:The fundamental theorem of calculus turns the differences into integrals of $-P_y$ and $Q_x$; Fubini's theorem then gives the required double integral. On subdividing a region, adjacent edge line integrals cancel because their directions oppose. For curved boundaries, the proof uses pieces bounded by graphs and double integrals; cancellation works the same way. This is the geometric reason only the region's complete boundary survives.^[ref]

For a strongly continuous unitary representation, integration against $f\in L^1(G)$ defines an operator satisfying $\Vert \pi (f)\Vert \le \Vert f{\Vert }_1$. Fubini's theorem gives $\pi (f\ast g)=\pi (f)\pi (g)$, and the group-algebra involution gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. On an Abelian group, the operator-norm closure is a commutative C-star algebra. Its algebra characters produce continuous unitary group characters, allowing the Commutative Gelfand--Naimark theorem and Riesz-Markov-Kakutani representation theorem to turn a positive vector functional into a finite spectral measure.

An iterated integral performs successive one-variable integrals. In the displayed expression, $y$ is held fixed during the inner integration over $x$, and the result is then integrated over $y$. The inner bounds may depend on that fixed outer variable. By contrast, an inner bound depending on an already integrated variable does not specify such a nested operation without further interpretation. For example, the triangle $0\le y\le 1$, $2y\le x\le 2$ gives ${\int }_0^1{\int }_{2y}^{2}f(x,y)dxdy$, or, in the other order, ${\int }_0^2{\int }_0^{x/2}f(x,y)dydx$. Fubini's theorem and Tonelli's theorem supply conditions under which these iterated expressions equal the same multiple integral.