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Subject and independence

The reviewer is an independent examiner in a fresh context, commissioned for refutation, who took no part in writing the page and read no assessment, standing, acceptance or review text about it except as disclosed under exposures below.

The subject is path wiki/research/erdos_501/lee_lemma_3_1_reconstruction.md as it stood at 2026-09-28T05:03:27Z (the page), read in full as of that time.

The artifact is the retained folder-name PDF lee_2026_relative_independence_erdos_problem_501.pdf under Lee (2026): the second version, date line "June 1, 2026", six pages numbered 1--6, so physical and printed page numbers coincide. Physical pp. 3--4 were read in full: the text layer through pdftotext -layout, and page images rendered at 130 dpi, from which every display (4)--(11), the statement of Lemma 3.1 and the paragraph deriving (11) were read. Page 1 was rendered at 110 dpi and read for the date line, and the text layer of pp. 1--2 was read for the hypothesis on $\nu$ in Lemma 2.1 and display (1); the text layer of pp. 5--6 was read for the reference list only. The first version, lee_2026_relative_independence_erdos_problem_501_v1.pdf, was read in the text layer at the opening of its Section 3 (its Theorem 3.1 and the Kunen citation), at its reference list, at its equiconsistency citation, and through a keyword search of its text layer, to check the Source paragraph's sentence about it; no images of it were rendered.

Allowed material read besides the artifact: the provenance paragraph of the card above; the Statement section of the Lemma 2.1 page as of the same time, which the page names as the consumer of (11); docs/verification.md "Whole-claim report" and "Audit checklist"; docs/evidence.md "Source fidelity"; and docs/math_authoring.md in full.

Exposures, none of which contained a review of this page and none of which the verdict rests on: (1) the card's whole _index.md was read, so its Bears-on, Read-status, Overview (a digest of the proofs of Lemma 3.1 and Lemma 2.1), Lean-files and Relation sections were seen beyond the provenance paragraph; (2) the problem page E0501.md has no Statement heading, and its opening region before the assessment sections was read, which includes its Status, Source, References and Formalization paragraphs; (3) the shared "Audit checklist" section of docs/verification.md and its "Durable reports and current standing" section were printed together with the two named sections; (4) the evidence/verify directory was listed, so three other review file names were seen, and none was opened; (5) a search of the Lemma 2.1 page for "Lemma 3.1" and "(11)" showed three lines outside its Statement section.

Restatement

Let $Y$ be a set and $\nu$ a countably additive measure defined on every subset of $Y$, with $\nu(\emptyset)=0$, that is $\sigma$-finite: $Y$ is a countable union of subsets of finite $\nu$-measure. Let $m$ be Lebesgue measure on the Lebesgue $\sigma$-algebra $\mathcal L$ of $\mathbb R$ and $m^*$ Lebesgue outer measure, the infimum of $\sum_k|I_k|$ over countable covers of the set by open intervals $I_k$. For an arbitrary $g\colon\mathbb R\to[0,\infty]$, with no measurability assumed, the upper integral is the infimum of $\int_{\mathbb R}h\,dm$ over Lebesgue-measurable $h\colon\mathbb R\to[0,\infty]$ with $g\le h$ pointwise; the admissible set is nonempty because $h\equiv\infty$ qualifies, so the infimum lies in $[0,\infty]$.

The claim: for every set $H\subseteq\mathbb R\times Y$, with no measurability assumption on $H$ in either factor or in the product,

$$\overline{\int_{\mathbb R}}\nu(H_x)\,dm(x)\le\int_Y m^*(H^y)\,d\nu(y),$$

where $H_x=\{y\in Y:(x,y)\in H\}$ is defined for every $x\in\mathbb R$ and has a $\nu$-measure because every subset of $Y$ is measurable, $H^y=\{x\in\mathbb R:(x,y)\in H\}$ is defined for every $y\in Y$, and the right side is the ordinary integral of the $[0,\infty]$-valued function $y\mapsto m^*(H^y)$, measurable because the $\sigma$-algebra on $Y$ is $\mathcal P(Y)$. Either side may be $\infty$, and the inequality is then read in $[0,\infty]$. The source's $\subset$ is non-strict inclusion, as its own uses ($H^y\subset U_y$ with $U_y=\mathbb R$ allowed) show, and the page's $\subseteq$ is the same relation.

The specialization (11): if $\nu\colon\mathcal P(\mathbb R)\to[0,\infty]$ is a measure extending Lebesgue measure, then $(\mathbb R,\mathcal P(\mathbb R),\nu)$ is $\sigma$-finite, and the inequality holds for every $H\subseteq\mathbb R^2$ with $Y=\mathbb R$.

Checklist

• Quantifiers and scope. Pass. "For every set $H$", "for each $y$", "for every $x$" are carried from the source unchanged; nothing is weakened to almost every point; the boundary case $m^*(H^y)=\infty$ is handled by $U_y=\mathbb R$ as in the source; the case of an infinite right side makes the inequality trivial; and the limit $\varepsilon\to0$ is taken on a left side that does not depend on $\varepsilon$.
• Circularity. Pass. The proof never uses the inequality or an equivalent of it; it uses the definition (4) of the upper integral directly, through one explicit measurable majorant.
• Model and convention changes. Pass. The upper integral is the source's (4) verbatim; the product $\sigma$-algebra is the source's, the Lebesgue $\sigma$-algebra with $\mathcal P(Y)$; the countable base is the source's own instance; $\subseteq$ renders the source's non-strict $\subset$.
• Finite and statistical overreach. Inapplicable: no finite verification, sampling or heuristic enters the argument.
• Uniformity. Pass. The slack in the envelope (8) is $\varepsilon\eta(y)$ with $\eta$ depending on $y$; the only uniform statement drawn from it is $\int_Y\eta\,d\nu\le1$, and the page states and proves it. The dependence of $E$ and the $U_y$ on $\varepsilon$ is harmless because the left side of the conclusion is $\varepsilon$-free.
• Extremal conclusions. Inapplicable: the page claims an inequality and no sharpness, attained value or extremum, apart from the infimum in (4), whose admissible set is nonempty as noted above.
• Consequences and composition. Pass. The passage from Lemma 3.1 to (11) needs only the $\sigma$-finiteness of $(\mathbb R,\mathcal P(\mathbb R),\nu)$, proved from $\nu([-n,n])=m([-n,n])=2n$; the page names the Lemma 2.1 page as the consumer of (11), and that page's Statement section carries the same hypothesis on $\nu$. No consumed clause is stronger than what is proved.
• Computation. Inapplicable: the page has no computation and no evidence code.
• Reproduction. Inapplicable: there are no rerun commands or coverage claims.
• Source and verdict fidelity. Pass, with one suggested correction. The statement, the definitions (4) and (6), the labels (5), (7)--(11) and the physical pages 3--4 were checked against the page images and agree; the standing sentence claims author-recorded status only. The Source paragraph's description of the first version's citation is loose (F1).

Weakest steps

The weight $\eta$ (supplied construction). $\sigma$-finiteness gives $Y=\bigcup_{n<\omega}Z_n$ with $\nu(Z_n)<\infty$. Put $Y^{(n)}=Z_n\setminus\bigcup_{k<n}Z_k$: subsets of $Y$, hence measurable, pairwise disjoint, covering $Y$, with $\nu(Y^{(n)})\le\nu(Z_n)<\infty$. For $y\in Y$ let $n$ be the unique index with $y\in Y^{(n)}$; then $\eta(y)=2^{-n-1}(1+\nu(Y^{(n)}))^{-1}$, so $0<\eta(y)\le1/2$, and $\eta$ is measurable as every function on $Y$ is. By monotone convergence for the nonnegative series of indicator multiples,

$$\int_Y\eta\,d\nu=\sum_{n<\omega}2^{-n-1}\frac{\nu(Y^{(n)})}{1+\nu(Y^{(n)})} \le\sum_{n<\omega}2^{-n-1}=1 .$$

Composition: strict positivity is what makes $\varepsilon\eta(y)$ a positive tolerance in (8), so that an open envelope exists for every $y$; the integral bound is what caps the accumulated tolerance in (10) at $\varepsilon$. Both properties are exactly what the source asserts of its undisplayed $\eta$, and the page marks the construction as supplied.

The sections of $E$ and the application of Tonelli. The product $\sigma$-algebra $\mathcal L\otimes\mathcal P(Y)$ is generated by the rectangles $A\times B$ with $A\in\mathcal L$ and $B\subseteq Y$, and $E=\bigcup_{n<\omega}I_n\times Y_n$ is a countable union of such rectangles, so $E$ is product-measurable. For each $y$, $E^y=\bigcup\{I_n:y\in Y_n\}=\bigcup\{I_n:I_n\subseteq U_y\}$; this union is contained in $U_y$, and it contains $U_y$ because $U_y$ is open and the rational intervals form a base, so each $x\in U_y$ lies in some $I_n\subseteq U_y$. Hence $E^y=U_y\supseteq H^y$ for every $y$, and $(x,y)\in H$ gives $x\in H^y\subseteq E^y$, that is $(x,y)\in E$; so $H\subseteq E$. Tonelli's theorem for the $\sigma$-finite spaces $(\mathbb R,\mathcal L,m)$ and $(Y,\mathcal P(Y),\nu)$ applied to $1_E$ gives that $x\mapsto\nu(E_x)$ is $\mathcal L$-measurable, that $y\mapsto m(E^y)$ is measurable, and that both iterated integrals equal $(m\times\nu)(E)$; this is the page's (9). Composition: the measurability of $Y_n$, hence of $E$, is the only place where the $\sigma$-algebra $\mathcal P(Y)$ is used in the construction, and it is the step that fails without it (see the attack below).

From the majorant to the upper integral, and $\varepsilon\to0$. For the fixed $\varepsilon$ and every $x$, $H\subseteq E$ gives $H_x\subseteq E_x$, so $\nu(H_x)\le\nu(E_x)$ by monotonicity of $\nu$ on $\mathcal P(Y)$. Thus $h(x)=\nu(E_x)$ is a Lebesgue-measurable function with $g\le h$ for $g(x)=\nu(H_x)$, so by (4) $\overline{\int}g\,dm\le\int h\,dm$. With (9), the pointwise bound (8) integrated over $Y$ (additivity and monotonicity of the integral of nonnegative measurable functions), and (7),

$$\overline{\int_{\mathbb R}}\nu(H_x)\,dm(x) \le\int_Y m(U_y)\,d\nu(y) \le\int_Y m^*(H^y)\,d\nu(y)+\varepsilon\int_Y\eta\,d\nu \le\int_Y m^*(H^y)\,d\nu(y)+\varepsilon .$$

The left side does not depend on $\varepsilon$, while $\eta$ is fixed before $\varepsilon$ and $U_y$, $Y_n$, $E$ are built after it; the bound therefore holds for every $\varepsilon>0$. If the right integral is finite the infimum over $\varepsilon$ gives the claim; if it is infinite there is nothing to prove. Composition: this is the whole conclusion, and it uses (4) only through the single majorant $h$.

Strongest attack

The attack transports the lemma to a second factor whose $\sigma$-algebra is smaller than $\mathcal P(Y)$, to test whether the page's Boundary paragraph correctly locates the load-bearing hypothesis, and whether the direction of the inequality or the placement of the upper integral could have been altered. Take $Y=[0,1]$ with $\nu=m$ on $\mathcal L$ and assume CH. Well-order $[0,1]$ in order type $\omega_1$ and let $S=\{(x,y):y\preceq x\}$; every vertical section $S_x$ is countable and every horizontal section $S^y$ is co-countable in $[0,1]$. For $H=[0,1]^2\setminus S$, each $H_x$ is co-countable, so $\nu(H_x)=1$, and each $H^y$ is countable, so $m^*(H^y)=0$: the left side of the inequality would be $1$ and the right side $0$. So the statement is false for $(Y,\mathcal L,m)$ under CH, in the exact direction the page states it. Tracing the proof: with $\mathcal L$ in place of $\mathcal P(Y)$ the sets $Y_n=\{y:I_n\subseteq U_y\}$ need not be measurable, $E$ need not be product-measurable, and Tonelli does not apply; the right integrand $y\mapsto m^*(H^y)$ also need not be measurable. These are precisely the two roles the page's Boundary paragraph assigns to the hypothesis. Under the lemma's own hypothesis every one of these steps is forced, so the attack fails, and it confirms that the page has not silently weakened the hypothesis or moved the upper integral to the other side. A second attack, on the dependence of $E$ on $\varepsilon$, fails because the left side of the conclusion is $\varepsilon$-free. A third, on the definition (4) when no majorant has finite integral, fails because $h\equiv\infty$ is admissible and the infimum is then $\infty$, which is consistent with the inequality.

Premises

• Tonelli's theorem. Interface: for $\sigma$-finite measure spaces $(X,\Sigma,\mu)$ and $(Y,\mathcal T,\nu)$ and $E\in\Sigma\otimes\mathcal T$, the functions $x\mapsto\nu(E_x)$ and $y\mapsto\mu(E^y)$ are measurable for $\Sigma$ and $\mathcal T$, and $\int_X\nu(E_x)\,d\mu=\int_Y\mu(E^y)\,d\nu=(\mu\times\nu)(E)$. Applied with $\Sigma=\mathcal L$, $\mathcal T=\mathcal P(Y)$, both spaces $\sigma$-finite ($m$ by the intervals $[-n,n]$, $\nu$ by hypothesis) and $E$ product-measurable, so its hypotheses are met. A standard textbook theorem; no source is held in the repository, and none is needed; the page names it as imported.
• Open envelopes from the definition of outer measure. Interface: for every $S\subseteq\mathbb R$ and $\delta>0$ there is an open $U\supseteq S$ with $m(U)\le m^*(S)+\delta$. Derivation: if $m^*(S)<\infty$ choose open intervals $I_k$ covering $S$ with $\sum_k|I_k|\le m^*(S)+\delta$ and put $U=\bigcup_kI_k$, so $m(U)\le\sum_km(I_k)=\sum_k|I_k|$ by countable subadditivity; if $m^*(S)=\infty$ take $U=\mathbb R$. Standard; no source held; the page names it as imported.
• Elementary facts used without a name. The open intervals with rational endpoints form a countable base of $\mathbb R$; a $\sigma$-finite cover refines to a disjoint one; monotone convergence for nonnegative series; monotonicity and additivity of the integral of nonnegative measurable functions; monotonicity of $\nu$. All standard, all checked in the derivations above.
• The source. Lee (2026), second version, held; pp. 3--4 read in full in text and image; no gap or misprint found in Lemma 3.1, its proof or the derivation of (11). The first version's Section 3 opening and reference list were read in the text layer only.
• Local claims. None consumed: the page cites no L<n> claim and no other reconstruction as an input; the Lemma 2.1 page is its consumer, and no batch acceptance order applies.
• Explicit assumptions. ZFC only: the choice of one $U_y$ for each $y$ uses the axiom of choice, as in the source. The specialization (11) assumes a measure on $\mathcal P(\mathbb R)$ extending Lebesgue measure, which the page states as its hypothesis.

Findings

F1. Severity: suggested. Location: Source paragraph, "as stated in Fremlin's notes". Defect: the first version cites the section inequality as "the following theorem of Kunen [3, 543C]" (first version, p. 3, Section 3, its Theorem 3.1), and its reference [3] is D. H. Fremlin, Measure Theory, Vol. 5, Chapter 54, "Real-valued-measurable cardinals" (the file chap54.pdf), whose result 543C is the theorem quoted. The phrase "Fremlin's notes" more naturally names the separate survey notes "Real-valued-measurable cardinals" (rvmc.pdf, the first version's [6] and the second version's [3]), which use a different numbering (1D(e), 2E) and are cited for the equiconsistency statement, not for 543C. The sentence's main claim, that the first version imported the inequality instead of proving it, is correct. Proposed replacement: "The first version, also held, took the inequality from Kunen's theorem as stated in Fremlin's Measure Theory, Volume 5, 543C (its Theorem 3.1, reference [3]) instead of proving it; the labels here are the second version's."

F2. Severity: note. Location: frontmatter desc, "for an arbitrary subset of the plane". Defect: Lemma 3.1 (second version, p. 3) is stated for an arbitrary $H\subseteq\mathbb R\times Y$ with $(Y,\mathcal P(Y),\nu)$ any $\sigma$-finite measure space; the plane is the specialization (11) on p. 4. The description names the corollary, not the lemma, while the Statement section is exact. Proposed replacement: "Reconstructs the one-sided Fubini inequality for an arbitrary subset of $\mathbb R\times Y$, $Y$ carrying a $\sigma$-finite measure on all its subsets: the upper integral of the measures of the vertical sections is at most the integral of the outer measures of the horizontal sections; specialized to the plane in (11)."

F3. Severity: note. Location: "A weight", "write $Y=\bigcup_nY^{(n)}$ with $\nu(Y^{(n)})<\infty$ and the $Y^{(n)}$ pairwise disjoint". Defect: the index range of $n$ is not stated, while the next paragraph writes $n<\omega$, and the passage from a $\sigma$-finite cover to a disjoint one is a supplied step inside the supplied construction and is not itself marked. Nothing fails: with $n<\omega$ the bound is $\sum_n2^{-n-1}=1$, and with $n\ge1$ it is $1/2$, both at most $1$, and the refinement is the standard $Z_n\setminus\bigcup_{k<n}Z_k$. Proposed replacement: "write $Y=\bigcup_{n<\omega}Y^{(n)}$ with $\nu(Y^{(n)})<\infty$ and the $Y^{(n)}$ pairwise disjoint (refine a $\sigma$-finite cover by removing the earlier members), and put".

Verdict

Source fidelity: faithful. The statement, definitions, conventions, labels (4)--(11) and physical pages 3--4 match the second version's PDF; no required correction was found. F1 is a suggested correction to the Source paragraph's description of the first version's citation, and F2 and F3 are notes.

The argument as reconstructed: sound. Every deduction was re-derived above; the supplied construction of the weight is marked as supplied and is correct; the two imported results are named, standard, and applied within their hypotheses; nothing the source proves is altered or strengthened, and the specialization (11) follows as stated.

Limitations: Tonelli's theorem and the open-envelope consequence of the definition of outer measure were checked as standard facts and not against a held source; the first version was read only at its Section 3 opening and its reference list, in the text layer; the consumer's use of (11) was checked only through the cross-link and the consumer's Statement section; the Lean files accompanying the source were not consulted.

This focused review assigns no tier and changes no status.