Lebesgue integration and smoothing — selected programme proofs

These are selected foundation sections from Banach estimates, quotient spaces and compact parameter arguments, programme lesson AN03-P004. They retain the section and equation numbers of that earlier lesson. The selection and introductory note were prepared by GPT-6 Astra (OpenAI), Ultra reasoning effort, October 2026.

The original course-writing task and OpenAI Codex credits, dedication and history are retained in the original title page, history and rights notice. This independently written programme selection carries CC0 1.0 Universal, to the extent rights are held. The CC0 dedication identifies the terms.

The original notice files describe the original course edition. Selection history describes these excerpts.

The mathematical entry is natural-number arithmetic with induction, ordinary set theory and countable choice. The real-field construction is in the scalar foundations; complex arithmetic and finite algebra are in the algebra foundations; integration is in the measure foundations.

This selection supplies: Lebesgue measure, convergence, Euclidean Fubini and affine changes, L1 density and translations, mollification, disk area and the generating-class argument.

15.0. Constructing the measure without importing a convergence theorem

We now construct the measure input used in Section 15.1. Fix the original coordinates on , . The covering boxes are bounded half-open coordinate boxes , , with original geometric volume . The empty box has volume zero. For every subset , put Finite covers are allowed by appending empty boxes. The empty cover gives . Every set has a cover, since a countable collection of bounded coordinate boxes covers the whole original space. Enlarging a set only reduces the choices of covers, so is monotone. For a countable collection , if is finite, choose a cover of with cost below , for . Concatenating these original covers proves . Let . If that sum is infinite, the inequality already holds. Thus This proof uses only countable sums of nonnegative real numbers, not an integral or a convergence theorem for one.

The box volume in (LM1) is exact. One box covering itself gives . To prove the reverse inequality, take any countable box cover. If its cost is infinite there is nothing to show. Otherwise enlarge each covering box to an open box with added geometric volume less than . This is possible by continuity of its full finite product in its endpoints. For , the closed coordinate box lies in the original and has a finite subcover by these enlarged open boxes. The finite-subcover fact has a direct proof here. If a closed box had no finite subcover by an open cover, bisect all its original coordinate intervals. At least one of its closed subboxes still has no finite subcover, because otherwise the union of the finite subcovers would cover the original box. Repeat inside that chosen subbox. Real completeness gives a point in all the resulting nested boxes: each coordinate's increasing lower and decreasing upper endpoints have the same limit, since their difference is its original side length times . The diameter at step is the full , which tends to zero. One covering open set contains the limiting point and a small ball about it. Eventually the entire chosen box lies in that one set, contradicting its lack of a finite subcover. This proves exactly the compactness needed for .

For a finite family of open boxes covering , include every covering endpoint and every endpoint of in a finite coordinate grid, inside a bounded box containing them all. The product distributive law writes the full volume of each box as the sum of the grid-cell products contained in it. On the interior of each grid cell, membership in a covering open box is constant, because all its boundary coordinates are grid endpoints. Every grid cell inside is therefore contained, in its interior, in at least one covering box. Summing the full products, with their actual multiplicities, gives Grid faces have a zero side length in this geometric calculation; no assertion about an already constructed measure is being used. First let , then . Every covering cost is at least the original full . Taking its infimum proves .

Define the measurable sets by the exact splitting condition The reverse inequality to this equality always holds by (LM2), so only its other direction needs proof. Complements preserve (LM4). If , split first at and then split both pieces at . The resulting four original pieces have sum of outer measures . The three pieces outside cover its complement in ; (LM2) therefore proves the required inequality for . Thus finite intersections and finite unions preserve measurability.

For disjoint measurable , repeated finite splitting gives, for every positive integer , Let . The full sum bounds from below by (LM2), proving (LM4) for their union. For an arbitrary countable union of measurable sets, replace its -th set by its difference from the preceding finite union; finite intersections and complements have already proved these disjoint differences measurable. Their union is the original union, so is a sigma-algebra. Taking in (LM5) and using (LM2) proves countable additivity of . No subtraction of two infinite numbers occurs.

Every covering box is in . Indeed, subdivide any covering box at all coordinate endpoints of . The half-open grid pieces partition exactly. Each is either inside or outside it, and the sum of their full volume products is , by the distributive law. The inside pieces cover , and the outside pieces cover , for any original cover of . Thus each covering cost is at least . Taking its infimum proves (LM4). These boxes generate the Borel sigma-algebra: every open set is a countable union of coordinate boxes with rational endpoints whose closures are contained in it, since each of its points has an interior ball and rational endpoints can be chosen on both sides of each original coordinate. Hence every Borel set is measurable, with the exact box volumes proved in (LM3).

Every subset of an outer-measure-zero set is also measurable. For such a subset , , and (LM2) and monotonicity give . This is (LM4), with its first term zero. Therefore is complete. A coordinate hyperplane has outer measure zero: its bounded part with other coordinates in is covered by one box of thickness in its fixed coordinate and side lengths in the others. Its full cost is The whole hyperplane is a countable union of these bounded parts, so (LM2) applies. Consequently every open, closed or half-open choice of endpoints on a bounded coordinate box differs only by a measurable null subset of its faces and retains the full original volume.

The completion is exactly the completed Borel measure, rather than a larger unexplained collection. If has finite measure, choose covers with costs tending to , enlarge their boxes to open boxes with additional total cost tending to zero, and denote their open unions by . Additivity gives . Thus is Borel, contains , and . Every measurable null set has a Borel null superset: make the same open-cover construction with total cost below , and intersect the resulting open sets. Apply this to . It follows that differs from a Borel set by a subset of a Borel null set. For infinite , apply the finite construction to , where bounded coordinate boxes increase to the whole original space. Each has finite measure by (LM3). The countable union of the resulting Borel supersets is a Borel superset of , and its difference from lies in the countable union of the Borel null supersets just constructed. This again has measure zero. Conversely, completeness has already shown that every Borel set changed on a subset of a Borel null set belongs to .

Finally, this construction is the unique completed Borel measure with the stated coordinate-box volumes. On a bounded coordinate box , suppose two finite Borel measures have these volumes. The sets on which they agree form a Dynkin class: agreement holds on , relative complements subtract from its finite original volume, and disjoint countable unions use countable additivity. The coordinate rectangles form an intersection-closed generating class. The independent Dynkin-class argument in Section 16.2, which uses only these set operations, makes agreement extend to all Borel subsets of . Exhausting by bounded and writing their increasing union as successive disjoint differences proves agreement on all Borel sets. Both completions contain exactly the same Borel null sets and all their subsets, as proved above; their completed measures therefore agree as well. Thus (LM1)–(LM6) construct precisely the countably additive completed Lebesgue measure used in (LP1)–(LP19), with its actual original coordinates, full products, face contributions and outer-measure covers.

15.1. Convergence, product integration and the full linear Jacobian

Countable additivity gives continuity of measure on increasing sets: write their union as the disjoint union of the first set and successive differences. For , the definition of the nonnegative integral first gives . If is a nonnegative simple function below and , the sets increase to a set containing . Integrating the finitely many values of and using measure continuity gives , including infinite values. Thus . First let tend to infinity, then tend to one, and then take the supremum over . This proves The zero values of cause no problem, and no finiteness of the ambient measure is needed. Every nonnegative measurable has increasing simple approximations: truncate at and round down to multiples of . These approximations increase and tend to , with infinite values treated by the same truncation. Additivity for nonnegative simple functions follows by taking their common finite measurable partition and adding the coefficient of each part. Apply (LP1) to increasing simple approximations of , of , and to their sums, which increase to . This proves , including infinite values; positive scalar multiplication follows by the same argument. Decomposition into the positive and negative parts of real and imaginary components then gives linearity on absolutely integrable complex functions, where each of those component integrals is finite.

For arbitrary nonnegative , apply (LP1) to . Since , this proves Fatou's bound . If complex measurable almost everywhere and with , then almost everywhere. Fatou applied to gives Indeed its lower bound is , so the upper limit of the subtracted nonnegative integral is zero. Values on the original null exceptional set can be set to zero for this calculation; integrals and equivalence classes remain unchanged.

Here is product integration with its actual measurability requirement. In a fixed finite coordinate box , let consist of the Borel sets whose sections are measurable, whose section measure is a measurable function of , and which satisfy Coordinate rectangles belong to , by the full box-volume formula. The whole box belongs. Relative complementation preserves membership because every section measure is at most the finite , so both sides subtract from . Disjoint countable unions preserve membership by countable additivity and (LP1), which also proves measurability of the sum of the section measures. Thus is a Dynkin class containing the rectangles.

For completeness, the needed class argument is finite and exact. Let be the smallest Dynkin class containing the rectangles. A Dynkin class is closed under differences of nested members: if , use the disjoint union and complement. For a rectangle , the sets for which form a Dynkin class and contain the rectangles, since intersections of rectangles are rectangles or empty. Hence they contain . Fixing now any and making the same argument in shows that is closed under all finite intersections. Complements give finite unions. Disjointifying a countable union by subtracting its finitely many preceding members then shows closure under countable unions. Therefore is a sigma-algebra containing the rectangles, which generate the Borel sets of . This proves (LP3) for every Borel set in the box.

Increasing simple approximation and (LP1) now prove nonnegative Tonelli on . Exhausting both original spaces by expanding coordinate boxes proves it on . A Lebesgue measurable set differs from a Borel set by a subset of a Borel null set. For that null set, (LP3) says that its sections are null outside a null set of 's. Completeness of the section measure makes every subset section measurable there. On the exceptional null set one may choose the value zero for the inner integral. Thus completed Lebesgue measurability gives the same iterated identity almost everywhere; it does not assert measurable sections at every exceptional point. Applying this argument to simple approximations proves the completed version for nonnegative functions. Finally, apply it to when , and to the positive and negative parts of the real and imaginary components. All these integrals are finite. We obtain the full identities for nonnegative , with possibly infinite integrals, and for absolutely integrable complex . Sections and inner integrals have their almost-everywhere meanings in the completed case.

Translations, coordinate reflections, coordinate permutations and positive diagonal scalings preserve the box identities, with the full product of the diagonal scale factors retained. The same Dynkin argument compares the two measures on Borel sets; completion extends it to Lebesgue sets because null sets remain null. A shear , , preserves measure: all other coordinates are fixed, and each one-dimensional section is translated by the fixed section parameter ; (LP4) and one-dimensional translation give equality. Every invertible real matrix is a finite product of swaps, nonzero diagonal scalings and shears, by Gaussian elimination on its actual rows. Each shear has determinant one, a swap has determinant minus one, and a diagonal scaling retains its actual signed factor. Multiplying the absolute factors therefore proves on the nonnegative and absolutely integrable domains. The transformations carry Borel null supersets to null supersets, so the completed measure and its exceptional sets are respected. This is a comparison of the original integrals with their complete Jacobian; no determinant is removed from the working formula.

15.2. Hölder, Minkowski and all Young endpoints

For , put . The scalar inequality , , follows by maximizing as a function of : its derivative is , with maximum at ; the zero case follows directly. Apply it to and when both norms are nonzero and finite. The original functions and norms are retained, and multiplication restores both norm factors. If a norm is zero, its function vanishes almost everywhere and the conclusion follows directly. This proves The second inequality is the pointwise essential bound. It also covers its reversed ordering. The first inequality includes integral Cauchy–Schwarz at . Repeated application proves Hölder for any finite number of factors whose reciprocal exponents sum to one, including infinite exponents by their essential bounds.

For , the convexity of gives , so the sum is integrable. Using and (LP6) gives . Divide when that norm is positive; if it is zero the result is immediate. The integral triangle inequality gives the case , and the essential-supremum triangle inequality gives . Thus every original norm satisfies Minkowski.

Let satisfy . Convolution uses the original density . When , the relation forces and . Split the absolute integrand into the three factors If an exponent is zero, its factor is omitted; this convention also applies where its base vanishes. Hölder uses exponents , and for the retained factors. A factor with zero exponent requires no division by zero. Their reciprocal sum is . Hence, outside any exceptional set where the first integral is infinite, Zero input norms give zero convolution. For positive norms, integrating (LP8), using (LP4) and the translation case of (LP5), shows that the first integral is finite for almost every , and proves If , then , so (LP6) proves (LP9) directly for each defined section and its essential supremum. This includes and . For , (LP8) is simply the integrated triangle bound; the two omitted factors have exponent zero. These cases exhaust the displayed exponent relation, and no strong estimate with an invalid exponent is being asserted.

15.3. Completeness and compact smooth density

For , let be norm Cauchy. Choose a subsequence with . The increasing finite sums have norms at most , by Minkowski. Equation (LP1), applied to , proves that their pointwise limit is in and finite almost everywhere. Thus the series of original differences converges absolutely almost everywhere, giving a measurable limit . The same argument for the tail gives One obtains the first bound by dominating the pointwise difference by that nonnegative tail and passing its finite partial sums through (LP1). The Cauchy property and the triangle inequality then give convergence of the entire sequence to . For , use the same subsequence, remove the countable union of exceptional null sets for the initial essential bound and all difference bounds, and sum the uniformly convergent difference series on the complement. Its limit is essentially bounded and satisfies (LP10) in the essential-supremum norm. Again the full Cauchy sequence converges. These arguments use only measure additivity, (LP1) and the norm inequalities; they prove completeness on any measure space with these definitions, not only Euclidean Lebesgue space.

The smooth density assertion has a different domain: on . First truncate the original by and . Dominated convergence applied to proves convergence of these truncations as grow. On the resulting bounded support and bounded complex range, a finite square grid of mesh gives a measurable simple function , with . Each has finite measure, so the error is at most times the support measure to the power .

For a measurable with finite measure, its outer-measure definition gives a countable coordinate-box cover with total volume less than . Enlarge its boxes to open boxes, choosing the extra volume of box less than . Their open union contains and has . The finite unions of the first boxes increase to . Since , (LP1) gives . Therefore , proving approximation in measure by a finite union of bounded coordinate boxes.

Write one such union as . For smaller than half every side length, choose one-dimensional smooth functions between zero and one, equal to one on and supported in . Their product is , and is compact smooth, between zero and one. The difference from is supported in the union of the full box layers. Hence Coordinate faces have measure zero: cover a bounded face by a box with arbitrarily small thickness and use the full volume product; unbounded faces are countable unions of bounded ones. Thus open or closed endpoint choices do not alter this estimate. Combining the finite-union approximation with (LP11) proves compact smooth approximation of each original indicator . Minkowski gives the full coefficient bound . Keeping these finitely many coefficients and then taking the preceding truncation and grid errors to zero proves density of in .

Translations are isometries by (LP5). For a fixed compact smooth , all supports with lie in a fixed bounded coordinate box . The fundamental theorem of calculus on the actual segment gives The same supremum bound holds at , without a volume factor. For general finite- , choose the compact smooth just proved dense. The triangle inequality gives . First let , then the approximation error tend to zero. This proves translation continuity for every finite . The general assertion would be false: and each translate with differ by one on a set of positive measure. Their essential-supremum distance is one. The finite- density and continuity results have not been extended to that endpoint.

15.4. The original scaled approximate identity

Let have integral one; it may be signed or complex. Set , retaining its actual scale and density. The scalar change of variables gives . For , Young's bound proves , including . For finite , the full difference formula is The original factor and its Jacobian are both included in this substitution. Hölder for the weighted scalar integral, or its triangle inequality when , followed by Tonelli gives The norm is positive because its integral is one. To verify Hölder's weighted bound for , write the integrand as times a factor of modulus ; the zero values of contribute zero. Translation continuity makes the inner norm tend to zero for each , and its bound is integrable against . Equation (LP2) proves the stated limit, with the entire factor retained.

The convolution is smooth even before the approximation limit. On a compact output neighborhood, the input variable in lies in one compact set. Hölder makes integrable on that set. Difference quotients of the kernel and all their subsequent derivatives converge uniformly there by the scalar fundamental theorem, with a common bound. Integration against this locally integrable therefore proves Every kernel derivative and scale factor remains. The general convergence conclusion is again false. The convolutions of are continuous; a continuous function has essential-supremum distance at least from that indicator, since values arbitrarily near an endpoint from its two sides would otherwise force incompatible limits. Uniform boundedness at that endpoint is the bound already proved, not convergence for all inputs.

15.6. The full polar formula on the original plane

The remaining change of variables has a singular point and an angular endpoint. We prove it from the measure in Section 15.0 and the product and linear formulas (LP4)–(LP5), without assuming a nonlinear change-of-variables theorem. The finite scalar calculus used here includes the original sine and cosine, their derivatives and addition formulas, and their period . Define The angular inverse, including its endpoints. We use the original scalar functions constructed in Arctangent, circular parameters and the original pi. There , , , , , and on , with , the full addition laws and . If , those laws give . Thus throughout . The mean value theorem makes strictly decreasing on , including comparisons with either endpoint. Its endpoint values are and . Continuity and intermediate values therefore give a bijection .

Define as this exact inverse. Both compositions are proved by bijectivity: for , and for . In particular and . To prove continuity at every , including , take . If failed to tend to , some subsequence would stay at distance at least . Compactness of supplies a convergent further subsequence with limit . Continuity of gives ; uniqueness forces , a contradiction. Sequential continuity is continuity in these original real intervals.

For a nonzero point , its radius is strictly positive. If , then , so belongs to . We have and ; the full square identity gives . If , put . Periodicity, evenness of and oddness of give , , and . If , the positive ray has angle zero and the negative ray has angle , with the stated coordinates in both cases.