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Fubini's theorem states that an absolutely integrable measurable function on a product of two $\sigma$-finite measure spaces has integrable sections almost everywhere, the corresponding iterated integrals exist, and both orders equal its product integral. In particular, for $f\in L^1({\mathbb{R}}^2)$,To establish integrability of the nonnegative function here, first use Tonelli's theorem: for a nonnegative measurable function the product integral equals either iterated integral with infinity allowed. This avoids assuming the conclusion needed for Fubini's theorem.

If $\alpha \ge -1$, ${\int }_0^{\infty }(1+x+y{)}^{\alpha }dx=\infty$ for every fixed $y$. If $\alpha <-1$, it isThe outer integral is finite exactly when $\alpha +1<-1$, that is, $\alpha <-2$. In that caseFor $-2\le \alpha <-1$ the outer integral diverges, including logarithmic divergence at $\alpha =-2$. Nonnegativity means finiteness of this integral is exactly absolute integrability; after proving it, Fubini's theorem allows either order.

Fubini's theorem for the product of two sigma-finite measure spaces states that an integrable measurable function on the product has integrable sections almost everywhere, the section integrals define Lebesgue integrable functions, and both iterated integrals equal the product integral. The integrability hypothesis is absolute integrability. For nonnegative measurable functions, Tonelli's theorem gives equality of the possibly infinite product and iterated integrals without first assuming finiteness.

The displayed function is not Lebesgue integrable. Its value at the origin can be assigned arbitrarily, since a point is a null set. On $0<x<1$ and $0<y<x/2$,Consequently Tonelli's theorem gives the divergent lower boundThe signed iterated integrals illustrate why the hypothesis matters. For $x>0$, $f(x,y)={\partial }_y[y/(x+y{)}^2]$, so integration first in $y$ and then in $x$ gives $1/2$. Integrating first in $x$ gives $-1/2$, using $f(x,y)=-{\partial }_x[x/(x+y{)}^2]$. Exceptional boundary sections have measure zero. Thus finite iterated values do not establish product integrability.

On a product of sigma-finite measure spaces, Tonelli's theorem equates product and iterated integrals for a nonnegative measurable function, allowing infinite values. Fubini's theorem says that an absolutely integrable measurable function has integrable sections almost everywhere, their section integrals are integrable, and both iterated integrals equal the product integral. Absolute integrability may be checked first by applying Tonelli to the absolute value.

On a product of sigma-finite measure spaces, Tonelli's theorem equates product and iterated integrals for a nonnegative measurable function, allowing infinite values. Fubini's theorem says that an absolutely integrable measurable function has integrable sections almost everywhere, their section integrals are integrable, and both iterated integrals equal the product integral. Absolute integrability may be checked first by applying Tonelli to the absolute value.

The function is continuous and nonnegative, bounded above by $x$ on $(0,1)$ and by $x{e}^{-x}$ on $(1,\infty )$, so it is integrable. For $x>0$ the geometric series givesThe series is absolutely summable in integral becauseEach integral follows by substituting $t=nx$ and integrating $t{e}^{-t}$ by parts. Tonelli's theorem applied to the absolute terms, followed by Fubini's theorem, therefore justifies termwise integration and gives

The function is nonnegative and continuous on the positive quadrant. In polar coordinates its product integral is, by the nonnegative change of variables formula and Tonelli's theorem,The angular factor is at most one, while the supplied radial integral is $1/2$, so the product integral is at most $\pi /4$ and is finite. This establishes integrability before any signed interchange is used.

Let $A$ denote the angular integral. The polar evaluation is $A/2$. In Cartesian coordinates Fubini's theorem instead factors the integral asusing $u=x^2$, $v=y^2$. All factors converge: their powers exceed $-1$ at zero and the exponential controls infinity. Equating the evaluations gives

Both terms of each one-dimensional section are integrable by a positive scaling substitution, since $f\in L^1(0,\infty )$. For fixed $y>0$,For fixed $x>0$,The strict inequalities hold because $a>1$ and the measurable integrand is strictly positive on each nonempty interval. If $g$ were integrable on the stated product domain, Fubini's theorem would make both iterated integrals finite and equal. But the first order of integration would give a strictly negative value, and the other a strictly positive value. This contradiction proves $g$ is not integrable on that product.

On a product of sigma-finite measure spaces, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, allowing $+\infty$.

Fubini's theorem says that if a measurable function is absolutely integrable on the product, almost every section is integrable, the integrals of sections are integrable, and both iterated integrals equal the product integral. In particular, to establish absolute integrability one can first apply Tonelli's theorem to the absolute value and show its iterated integral is finite.

For a continuously differentiable diffeomorphism $\varphi :J\to I$ between intervals, the integration by substitution statesIt holds as an identity element of extended integrals for nonnegative measurable $h$. For signed or complex-valued measurable $h$, integrability on one side is equivalent to integrability after substitution, and the identity element then holds. The absolute derivative accounts for a decreasing substitution; on oriented finite intervals the equivalent formula uses ${\varphi }^{\prime }$ and oriented endpoints.

Put $C={\int }_0^{\infty }(1+x^4{)}^{-1}dx=\pi /(2\sqrt{2})$. The map $x=y^{-1/4}$ is a decreasing diffeomorphism of the positive half-line onto itself andApplying the nonnegative version before assuming finiteness proves both existence and value:Tonelli's theorem permits either iterated integral of a nonnegative measurable function on a product of sigma-finite measure spaces and identifies both with its extended product integral. Fubini's theorem states that for an absolutely integrable measurable function on such a product, almost every section is integrable, the section-integral functions are integrable, and the two iterated integrals equal the product integral.

The given two-variable function is continuous and positive on the positive quadrant. For fixed $y>0$, substituting $t=x{y}^{1/4}$ givesTonelli now givesThis proves absolute integrability, so Fubini is applicable.

For $x\ne 1$, integrate the displayed partial fraction identity element in $y$ up to $R$:At $x=1$ the section integral is ${\int }_0^{\infty }(1+y{)}^{-2}dy=1$, agreeing with the continuous limit of this expression. Therefore Fubini identifies the remaining integral and givesThe apparent singularity at one is removable, with limiting integrand value $1/4$. Positivity and the finite product integral already establish convergence at both ends.

For sigma-finite measures, Tonelli's theorem states that the double integral of a nonnegative measurable function equals either iterated integral, allowing $+\infty$. Fubini's theorem states that if a measurable function is absolutely integrable, its sections are integrable almost everywhere, both iterated integrals exist, and they equal the double integral.

Assign any value at $(0,0)$, which does not affect Lebesgue integrability. Near that corner set $s=x+y$ and $t=(x-y)/(x+y)$. Then $x=s(1+t)/2$, $y=s(1-t)/2$, and the absolute Jacobian determinant is $s/2$. On $0<s<1$, the full range is $-1<t<1$, soThe remaining part of the unit square has $x+y\ge 1$ and is harmless. The improper power integral is finite precisely when $\alpha <3$. Under the given restriction,For those values, the change $(x,y)\mapsto (y,x)$ preserves the square and changes the sign of $F$. Absolute integrability justifies this change of variables, givingFor $\alpha \ge 3$ the positive and negative parts each diverge, so symmetry does not define a Lebesgue integral.

Put $u=x-y$ and $v=x+y$. The square becomes $(-1,1{)}^2$, with $x=(u+v)/2$, $y=(v-u)/2$ and absolute Jacobian determinant $1/2$. On its support the denominator $v+2$ exceeds one and $|u|<1$, so the nonnegative integrand is bounded by one on a set of finite area; it is therefore Lebesgue integrable. The change of variables formula and Fubini's theorem give

Use the corrected piecewise-deterministic Markov process from part (b). For a smooth test function $f$, its time-dependent infinitesimal generator isIndeed, over an interval of length $h$, deterministic motion contributes $ha(t,x)f^{\prime }(x)$, while a jump has probability $h\alpha (x)+o(h)$ and changes the conditional mean of $f$ by $K_{t}f(x)-f(x)$. Terms involving both drift and a jump contribute only to the remainder. Thus $P_{t,t+h}f=f+h{L}_{t}f+o(h)$.

The Dynkin formula, or this infinitesimal expansion integrated against the current probability distribution, gives the weak evolution identityTake $f\in C_c^{\infty }(\mathbb{R})$, so integration by parts has no boundary contribution. Assuming the displayed jump integrals are finite, Fubini's theorem givesConsequently the forward equation of a piecewise-deterministic Markov process isHere $A_t=L_t^{\ast }$ is the formal adjoint operator acting on probability density functions. Its terms are transport, loss from $x$, and incoming mass from every $y$. In particular, the incoming rate is $\alpha (y)$, evaluated at the departure state. The operator denoted $A$ in the source is generally time-dependent. Because $\int m_t(y,x)dx=1$, the gain and loss integrals cancel; with vanishing transport flux at infinity, $\frac{d}{dt}\int p_t=0$.

The deterministic initial state in part (b) does not actually yield an everywhere smooth full density. For finite times the no-jump contribution is the positive measurewhere ${\delta }_z$ is the Dirac measure at $z$. Smooth jump destination probability density functions cannot remove this positive atom. The density equation above is therefore conditional on the extra density hypothesis, for example with a suitable random initial probability density function. For the specified fixed initial state, write ${\nu }_t$ for the whole probability measure and use insteadThis weak measure evolution includes both the atom and any continuous component. It becomes the displayed partial differential equation whenever ${\nu }_t(dx)=p_t(x)dx$ with the assumed regularity. It does not describe the literal hidden-reset construction without the correction in part (b).

An orthogonally invariant random vector has a probability measure determined by its radial pushforward measure. Here $\sigma$ is normalized surface measure on a sphere, and the formula holds for bounded measurable $f$ and for nonnegative measurable $f$. It includes an atom of a measure at zero: the inner integral there is simply $f(0)$.

To prove the formula first for continuous bounded $f$, average $f(Qx)$ over normalized Haar measure of the compact orthogonal group. Right invariance makes this average depend only on $|x|$; averaging also over surface measure on a sphere identifies it with the spherical average. Invariance of $\mu$ and Fubini's theorem then give the identity. Uniqueness of finite Borel measures and measurable approximation extend it to the stated classes of $f$. No extra sphere-area factor belongs in an average against normalized surface measure on a sphere.

For positive $u$, integrate $t-f$ from zero to $u$; for negative $u$, reverse the integral from $u$ to zero. This proves the identity without a positivity restriction on the data. For $u,f\in L^2$, the absolute integral is bounded by $u^2/2+|fu|$, so Fubini's theorem applies. The zero-function baseline removes the divergence from the negative-height tail.

For $1\le p<\infty$, an integer $k\ge 0$, and a normalized nonnegative mollifier $\rho \in C_c^{\infty }({\mathbb{R}}^n)$, put ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x/\epsilon )$. If $u\in W^{k,p}({\mathbb{R}}^n)$, its mollification is smooth and converges to $u$ in the Sobolev norm. The definition of a weak derivative and Fubini's theorem give ${\partial }^{\alpha }({\rho }_{\epsilon }\ast u)={\rho }_{\epsilon }\ast {\partial }^{\alpha }u$ for every multi-index with $|\alpha |\le k$. The approximate identity theorem in the Lp space $L^p$ applied to these finitely many weak derivatives proves the claim. On a domain with boundary this direct convolution argument applies only away from the boundary unless an extension is first supplied.

For a Sobolev space function, almost every line parallel to a coordinate axis gives an absolutely continuous restriction. An absolutely continuous function has derivative zero almost everywhere on each of its level sets. Slicing and Fubini's theorem give the assertion in higher dimension. If $w\in W^{2,p}$, apply it to $w$ and then to each component of its gradient to get $D^{2}w=0$ almost everywhere on $\{w=0\}$. This supplies the contact differentiation used in noncontact of ROF level boundaries.

For a real function in the $L^{\infty }$ unit ball, push uniform measure on $t\in [0,1]$ forward by the displayed sign-valued functions. In the weak-star topology this is a Borel probability concentrated on the extreme points, and Fubini's theorem shows that its barycenter is $f$. Weak-star compactness and separability of the predual $L^1$ place the example within Choquet theorem.

A triple integral is a multiple integral over a three-dimensional region. For a mass density $\rho$, ${\iiint }_D\rho dV$ gives the total mass, while ${\iiint }_{D}1dV$ gives its volume. Iterated one-dimensional integration is justified by Fubini's theorem under absolute integrability, and a change of variables includes the absolute Jacobian determinant in the volume element.

For an independent private values model, let $X_i(v)$ and $P_i(v)$ be interim allocation and payment. The interim payment identity gives $P_i(v)=v{X}_i(v)-U_i({\underline{v}}_i)-{\int }_{{\underline{v}}_i}^v{X}_i(t)dt$. Applying Fubini's theorem to reverse the order of integration yields $\mathbb{E}{\int }_{{\underline{v}}_i}^{V_i}{X}_i(t)dt=\int X_i(t)(1-F_i(t))dt$. Subtracting this term from $\int v{X}_i(v)f_i(v)dv$ proves the identity. Interim individual rationality bounds the lowest-type utilities below by zero.

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.