Products, scalar Fubini and the polar integral

Written and self-checked by GPT-6 Astra (OpenAI), Ultra reasoning effort, 4 October 2026. Original exposition is CC0.

Read Scalar measure and integration: programme proofs first. Its Sections 1 and 2 construct one-dimensional Lebesgue measure and scalar integration and prove monotone and dominated convergence. Its Lemma 6.1 proves the fundamental theorem and oriented substitution for continuously differentiable functions. We use those written proofs below.

This lesson proves sigma-finite product integration in full, then the affine and polar substitutions and the integration-by-parts formulas used in the modular kernel. The general injective, merely pointwise differentiable Jacobian theorem is proved in The Jacobian theorem on a measurable source set, after its geometric foundations. The full fundamental theorem for an integrable derivative, including the non-C1 case, is proved in the following programme lesson, The fundamental theorem for an integrable derivative. The special substitutions below have direct proofs as well.

PI1. Generating classes and uniqueness of a measure

A pi-system is a family closed under finite intersections. A Dynkin class on a set $Z$ contains $Z$, is closed under complements in $Z$, and is closed under countable disjoint unions. Such a class is closed under differences $B\setminus A$ when $A\subseteq B$ are members: take the complement of the disjoint union of $A$ and $Z\setminus B$.

Lemma. If a Dynkin class contains a pi-system $\mathcal P$ containing $Z$, it contains the sigma-algebra generated by $\mathcal P$.

Proof. Let $\mathcal D$ be the intersection of all Dynkin classes containing $\mathcal P$; it is itself such a class. Fix $A\in\mathcal P$. The class of all subsets $B\subseteq Z$ such that $A\cap B\in\mathcal D$ is a Dynkin class: its whole-space condition is $A\in\mathcal D$, its complement condition is the nested-difference property just proved, and its disjoint-union condition is immediate. It contains $\mathcal P$ by finite intersection closure. Thus $A\cap B\in\mathcal D$ whenever $A\in\mathcal P$, $B\in\mathcal D$.

Now fix $B\in\mathcal D$ and repeat the argument with the class of $A$ for which $A\cap B\in\mathcal D$. It contains $\mathcal P$ by the preceding paragraph, so contains $\mathcal D$. Therefore $\mathcal D$ is closed under finite intersections and differences. Disjointifying any countable family by subtracting its preceding finite union shows that $\mathcal D$ is a sigma-algebra. This proves the lemma. $\square$

If two finite measures have equal values on $\mathcal P$, the sets on which their values agree form a Dynkin class: subtract from their equal finite total masses for complements, and use countable additivity for disjoint unions. They therefore agree on $\sigma(\mathcal P)$. For sigma-finite measures, it suffices that there be $C_n\in\mathcal P$, increasing to $Z$, on which both measures are finite and equal. Apply the finite assertion to the two measures $E\mapsto\mu(E\cap C_n)$, then let $n$ increase. Their values agree on $\mathcal P$ because $E\cap C_n\in\mathcal P$. This proves the required uniqueness statement, including its exhaustion hypothesis.

PI2. Construction of the sigma-finite product

Let $(X,\Sigma,\mu)$ and $(Y,\mathcal T,\nu)$ be sigma-finite measure spaces. Write $\Sigma\otimes\mathcal T$ for the sigma-algebra generated by the measurable rectangles $A\times B$. For $E\subseteq X\times Y$, set $E_x=\{y:(x,y)\in E\}$. If $E\in\Sigma\otimes\mathcal T$, then $E_x\in\mathcal T$ for every $x$: the sets with this property form a sigma-algebra and contain all rectangles.

First suppose $\nu(Y)<\infty$. The sets $E\in\Sigma\otimes\mathcal T$ for which $x\mapsto\nu(E_x)$ is measurable form a Dynkin class. The whole space gives a constant. Complements give $\nu(Y)-\nu(E_x)$, a difference of finite measurable functions. For a disjoint union the section measures are the pointwise sum of the section measures of its members, hence are measurable. Rectangles belong to the class because their section functions are $\nu(B)1_A$. PI1 proves the assertion for every product-measurable set.

For general sigma-finite $\nu$, choose a disjoint measurable partition $Y=\bigcup_jY_j$ with $\nu(Y_j)<\infty$, by disjointifying a finite-measure cover. Apply the finite argument to each measure $B\mapsto\nu(B\cap Y_j)$. Then $\nu(E_x)=\sum_j\nu(E_x\cap Y_j) \tag{PI2.1}$ is measurable. We can therefore define $\lambda(E)=\int_X\nu(E_x)\,d\mu(x), \qquad E\in\Sigma\otimes\mathcal T. \tag{PI2.2}$ This is a measure: the empty set has value zero, and countable additivity on disjoint sets follows from countable additivity of $\nu$ followed by monotone convergence on $X$. On a rectangle it has the value $\lambda(A\times B)=\mu(A)\nu(B), \tag{PI2.3}$ with $0\cdot\infty=0$. This formula also holds if $\nu(B)=\infty$, by increasing the finite-measure portions of $B$.

Choose increasing finite-measure sets $X_n\uparrow X$, $Y_n\uparrow Y$. The rectangles $X_n\times Y_n$ are a finite-measure exhaustion, so $\lambda$ is sigma-finite. PI1 shows it is the unique measure on $\Sigma\otimes\mathcal T$ with the rectangle values (PI2.3). Reversing the construction gives the measure $E\mapsto\int_Y\mu(E^y)\,d\nu(y)$, where $E^y=\{x:(x,y)\in E\}$. Its rectangle values are the same, so uniqueness proves that both constructions agree. This establishes symmetry without assuming an interchange theorem in advance. We write $\mu\otimes\nu=\lambda$.

PI3. Tonelli, Fubini and exceptional sections

Theorem. For every nonnegative product-measurable function $f$, its section integrals are measurable and $\int_{X\times Y}f\,d(\mu\otimes\nu) =\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) =\int_Y\left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y). \tag{PI3.1}$ The values may be infinite. If $f$ is real or complex and $\int|f|<\infty$, almost every section is integrable; defining its integral to be zero on the exceptional set gives an integrable measurable function, and (PI3.1) holds with finite scalar values.

Proof. The assertion for indicator functions is PI2, and finite nonnegative linear combinations give the assertion for simple functions. Choose increasing simple functions converging to $f$, as constructed in Section 2 of the preceding lesson. For every section, monotone convergence identifies the limit of the section integrals with its integral. These limits are measurable. Applying monotone convergence once more to the outer integral proves (PI3.1). The reversed construction in PI2 proves its other equality.

For scalar $f$, apply this result to $|f|$. The function $g(x)=\int_Y|f(x,y)|\,d\nu(y)$ has finite integral, so is finite almost everywhere. Indeed, if $\mu\{g=\infty\}>0$, the inequality $g\geq n1_{\{g=\infty\}}$ for every $n$ contradicts finiteness of its integral. Outside this measurable null set, define the section integral by the positive and negative parts of the real and imaginary parts. Their nonnegative section integrals are measurable by the first part; subtract them on $\{g<\infty\}$ and put zero on its complement. This proves measurability, and the bound $\left|\int_Y f(x,y)\,d\nu(y)\right|\leq g(x) \tag{PI3.2}$ proves integrability. Applying (PI3.1) separately to those four nonnegative parts and using scalar linearity gives the claimed equality. Reverse the factors for the other order. $\square$

In particular, if $u\in L^1(\mu)$ and $v\in L^1(\nu)$, then $f(x,y)=u(x)v(y)$ is measurable and $\int|u(x)v(y)|\,d(\mu\otimes\nu) =\left(\int|u|\,d\mu\right)\left(\int|v|\,d\nu\right)<\infty. \tag{PI3.3}$ The scalar Fubini identity then gives $\int f=(\int u)(\int v)$. This also justifies integrable product majorants in the modular kernel.

Completion and almost-everywhere strong measurability. These assertions extend to a function equal off a product-null set to a product-measurable representative. To see why the exceptional sections are harmless, enclose the exceptional set in a product-measurable null set $N$. Formula (PI3.1) for $1_N$ gives $\nu(N_x)=0$ for almost every $x$, and $\mu(N^y)=0$ for almost every $y$. The representatives therefore give the same section integrals almost everywhere. On incomplete marginal spaces this means integrability of the equivalence classes, or integrability for the completed marginal measures; an arbitrary pointwise null-set modification need not be literally measurable.

Here are the representative details. A completion-measurable scalar function has increasing dyadic simple approximations for each nonnegative component. Every set occurring in these countably many approximations differs from a set in the original sigma-algebra by a subset of a measurable null set, by the definition of completion. Replace those sets, take the countable union of the null sets, and take the limits off that union, assigning zero on it. The result is an original-measurable representative. Likewise, an almost-everywhere limit of measurable scalar simple functions has a measurable representative: take its pointwise limit where the sequence converges and zero elsewhere. The Cauchy criterion writes this convergence set using countably many measurable inequalities. These arguments cover the almost-everywhere strong measurability formulation of the Fubini contract. Ordinary measurable complex functions themselves have such simple approximations, by truncating and rounding their real and imaginary parts.

PI4. Euclidean products and null rays

Let $m$ be the Borel Lebesgue measure constructed in Section 1 of the preceding lesson. It is sigma-finite because $\mathbb R=\bigcup_n[-n,n]$. Iterate PI2 to define $m_d$ on $\mathbb R^d$. The product Borel sigma-algebra is exactly the Euclidean Borel sigma-algebra: coordinate projections are continuous, giving one inclusion; every Euclidean open set is a union of the rational open boxes it contains, a countable family, giving the other. Repeated use of (PI2.3) gives $m_d\left(\prod_{j=1}^d I_j\right) =\prod_{j=1}^d\operatorname{length}(I_j) \tag{PI4.1}$ for bounded intervals of any endpoint type. The case of a degenerate interval follows from zero singleton mass. These box values uniquely determine $m_d$: use the pi-system of boxes, the exhaustion $[-n,n]^d$, and PI1. In particular, grouping or permuting factors leaves the resulting measure unchanged. Its completion is the completed Euclidean Lebesgue measure used here. Scalar linearity, monotonicity, almost-everywhere invariance and the absolute integral bound are proved for every measure space in the preceding lesson, so apply to all these products.

For later use, the ray $\{(x,0):x\leq0\}$ is null. It is the countable union of $[-n,0]\times\{0\}$, each having measure zero by (PI4.1). This is the actual exceptional set for the polar coordinates below.

PI5. Affine substitutions and rotations

Section 1 of the preceding lesson proves directly from interval covers that $m(aE+b)=|a|m(E)$ for $a\ne0$. Affine maps and their inverses are continuous, so preserve Borel measurability; the same formula for outer measure preserves null sets and hence also applies to the completed measure. Indicator functions, simple approximations and monotone convergence now give, for nonnegative measurable $h$, $\int_{\mathbb R}h(ax+b)\,dx =\frac1{|a|}\int_{\mathbb R}h(u)\,du. \tag{PI5.1}$ For example, for an indicator $h=1_E$, this is the measure formula for $(E-b)/a$. Applying it to $|h|$ gives the equivalence of integrability; taking positive and negative real and imaginary parts proves (PI5.1) for integrable scalar $h$. This supplies translations, reflections and Gaussian dilations, with the absolute value in the full-line measure formula.

By PI3, the planar shear $S_c(x,y)=(x+cy,y)$ preserves the integral of every nonnegative Borel function: integrate first in $x$, apply translation invariance for each fixed $y$, then integrate in $y$. The shear $T_c(x,y)=(x,y+cx)$ does the same in the reverse order. The coordinate swap preserves integrals by (PI3.1), and $(x,y)\mapsto(ax,by)$, $ab\ne0$, multiplies them by $|ab|^{-1}$ by two applications of (PI5.1).

Every invertible real two-by-two matrix is a product of these elementary matrices. Here is the needed elimination argument: swap rows if necessary to make the first entry nonzero, scale that row to make the entry one, subtract a multiple of it from the second row, scale the second diagonal entry to one (it cannot be zero by invertibility), and subtract the upper off-diagonal entry times the second row from the first. These are precisely swaps, nonzero row scalings and row shears; taking their inverses gives the factorization. The determinant formula for a two-by-two matrix shows respectively absolute determinants one, the absolute scaling factor, and one. Multiplying the elementary integral identities therefore proves $\int_{\mathbb R^2}h(Az)\,dz =|\det A|^{-1}\int_{\mathbb R^2}h(z)\,dz. \tag{PI5.2}$ Rotations have determinant one and thus preserve planar measure. Images of null sets under these homeomorphisms are null, so the completed-measure and scalar-integrable versions follow as in (PI5.1).

PI6. Polar integration with its exceptional set

Use the usual trigonometric parametrization of the unit circle, with angle measured in radians. We use its elementary identities, derivatives, and its one-to-one parametrization away from one point. These are the real exponential and trigonometric conventions already used for the scalar complex-variable programme proofs; no general Jacobian theorem is assumed.

First compute the area of the open disk of radius $R>0$ by its vertical sections and PI3: $m_2\{x^2+y^2<R^2\} =\int_{-R}^{R}2\sqrt{R^2-x^2}\,dx =2R^2\int_{-\pi/2}^{\pi/2}\cos^2t\,dt =\pi R^2. \tag{PI6.1}$ The middle equality is Lemma 6.1 applied to $x=R\sin t$, whose derivative is $R\cos t$; cosine is nonnegative on this interval. The last equality follows by integrating $\cos^2t=(1+\cos(2t))/2$, again by Lemma 6.1. The boundary circle has measure zero: it is contained, for every $0<\varepsilon<R$, in the difference between the disks of radii $R+\varepsilon$ and $R-\varepsilon$. Their area difference tends to zero. The centre has zero measure by PI4.

A radial segment is null because it is a rotation of a subset of the horizontal axis, and PI5 preserves null sets. Partition a disk into $n$ sectors of equal angle $2\pi/n$. The sectors are congruent under rotations; the only overlaps and omissions lie on the finitely many null radial segments and the centre. Thus each has area $\pi R^2/n$, and a sector of angle $2\pi k/n$ has area $k\pi R^2/n$. Increasing rational fractions of a turn to any angle $\alpha\in[0,2\pi]$, continuity from below of measure proves the area formula $\alpha R^2/2$. Rotation invariance makes it independent of the initial angle. Removing the smaller concentric sector gives $m_2\{a<|z|\leq b,\ c<\arg z\leq d\} =\frac{b^2-a^2}{2}(d-c) \tag{PI6.2}$ when $0<a<b<\infty$ and $-\pi<c<d<\pi$. The boundary choices make no difference because the circles and radial segments are null.

Consider $\Phi:(0,\infty)\times(-\pi,\pi)\longrightarrow \mathbb R^2\setminus\{(x,0):x\leq0\}, \qquad \Phi(r,\theta)=(r\cos\theta,r\sin\theta). \tag{PI6.3}$ This is a homeomorphism. The inverse has radius $\sqrt{x^2+y^2}$ and angle $2\arctan\bigl(y/(\sqrt{x^2+y^2}+x)\bigr)$; its denominator is positive exactly on the indicated target, and the elementary half-angle identities verify the inverse formula. In particular images of Borel sets are Borel.

Pulling planar measure back by $\Phi$ gives a Borel measure on the parameter domain. Another Borel measure there is $\rho(E)=\int_E r\,d(m\otimes m)(r,\theta). \tag{PI6.4}$ Countable additivity of this weighted measure follows from monotone convergence. On a half-open parameter rectangle $(a,b]\times(c,d]$, PI3 and Lemma 6.1 give $\rho=(b^2-a^2)(d-c)/2$. This equals the pullback measure by (PI6.2). Such rectangles, with the empty set and whole domain adjoined, form a generating pi-system. Both measures are infinite on the whole domain and have a common increasing finite-measure exhaustion by rectangles bounded away from $r=0$ and the two angular endpoints. PI1 therefore identifies them on every Borel set.

Use indicator functions, then simple approximations and monotone convergence to conclude $\int_{\mathbb R^2}h(x,y)\,dx\,dy =\int_{-\pi}^{\pi}\int_0^\infty h(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta \tag{PI6.5}$ for every nonnegative Borel $h$. The omitted ray contributes zero by PI4. For scalar $h$, applying the identity first to $|h|$ proves equivalence of absolute integrability, and applying it to the four real nonnegative parts proves the scalar identity. It also extends to completed-measurable functions: a planar null exceptional set has $\rho$-null preimage; because $r>0$, its intersection with $\{r\geq1/n\}$ has product measure at most $n$ times its $\rho$-measure. The preimage is consequently product-null. Thus the representative and completion argument of PI3 applies on both sides.

As a check of normalization, $I=\int_{\mathbb R}e^{-x^2}\,dx$ is finite: on $|x|\geq1$, $e^{-x^2}\leq e^{-|x|}$, and the latter has finite integral by finite-interval calculus and monotone convergence. The same calculus gives $\int_0^R r e^{-r^2}\,dr=(1-e^{-R^2})/2$, and hence $I^2=\int_{\mathbb R^2}e^{-(x^2+y^2)}\,dx\,dy =2\pi\int_0^\infty r e^{-r^2}\,dr=\pi. \tag{PI6.6}$ Here the first equality is PI3.3 and the second is PI6.5. Since the integrand is positive and bounded below on $[-1,1]$, $I>0$, so $I=\sqrt\pi$. This supplies the polar step in the modular Gaussian calculation with no imported area formula.

PI7. Finite and improper integration by parts

For complex $u,v\in C^1([a,b])$, the product rule and Lemma 6.1 give $\int_a^b u'v=u(b)v(b)-u(a)v(a)-\int_a^b uv'. \tag{PI7.1}$ The product rule follows directly by subtracting $u(t)v(t)$ from $u(t+h)v(t+h)$, dividing by $h$, and using continuity and differentiability of the factors. Thus this use of finite calculus is covered by the preceding written proof, componentwise over the real scalars.

Suppose now $u,v\in C^1(\mathbb R)$, both $u'v$ and $uv'$ are integrable, and $u(t)v(t)\to L_\pm$ as $t\to\pm\infty$. Apply (PI7.1) to $[-n,n]$. Dominated convergence with majorants $|u'v|$ and $|uv'|$ gives $\int_{\mathbb R}u'v=L_+-L_--\int_{\mathbb R}uv'. \tag{PI7.2}$ In particular, if both endpoint limits are zero, the two integrals are negatives. Separate endpoints $-A,B$ may tend independently to infinity: the omitted absolute integrals on either tail tend to zero by dominated convergence. On a half-line the same proof retains its finite endpoint term. No endpoint limit is inferred merely from boundedness.

The finite oriented substitution in Lemma 6.1 keeps $\phi'$, including its sign, because it differentiates a primitive. The full-line measure formula PI5.1 instead has $|a|$. These are distinct identities with the orientation and hypotheses displayed. For differentiation under a scalar parameter integral, pointwise convergence of the difference quotients and a common integrable majorant give the derivative by dominated convergence; a limit along every real sequence tending to the parameter yields the usual derivative. The modular Gaussian arguments must, and do, check their specific majorants and endpoint limits separately.

Proof scope and free references

The preceding programme proofs plus PI1–PI4 supply the entire declared Lebesgue-measure and sigma-finite scalar-Fubini interfaces. PI5–PI7 supply the affine, polar and continuously differentiable calculus cases used in the modular kernel. The following integrable-derivative lesson, FT1–FT5, supplies the larger FTC interface. The later relative Jacobian lesson, GV1–GV5 and JC1–JC7, supplies the full general change-of-variables interface. These prerequisite proofs do not certify the whole modular course.

For free human references, see D. H. Fremlin, Measure Theory, Chapter 25, Sections 251–252, and Sheldon Axler, Measure, Integration & Real Analysis, Chapter 5. The local comparison inspected Fremlin 251A–251E, pages 2–3; it did not inspect the whole chapter. The proofs needed here are written above and in the preceding programme lesson; the links are reading references, not substitutes for a proof.