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Source. S. Lee, Relative independence of Erdős problem #501, second version dated 2026-06-01 (the retained folder-name PDF), Lemma 3.1 and the display (11) that specializes it, physical pp. 3--4, in the six-page PDF held by its library source card, Lee (2026). The first version, also held, took the inequality from Kunen's theorem as stated in Fremlin's Measure Theory, Volume 5, Chapter 54, result 543C (its Theorem 3.1, citing its reference [3]) instead of proving it; the labels here are the second version's.

Standing. This is an author-recorded reconstruction. It is not an independent review and changes no status and assigns no tier. Imported: Tonelli's theorem for the product of two $\sigma$-finite measure spaces, and the definition of Lebesgue outer measure as the infimum of the total lengths of countable open-interval covers, which gives, for every $S\subseteq\mathbb R$ and $\delta>0$, an open $U\supseteq S$ with $m(U)\le m^*(S)+\delta$.

Definitions

$m$ and $m^*$ are Lebesgue measure and Lebesgue outer measure on $\mathbb R$. For $g\colon\mathbb R\to[0,\infty]$ the Lebesgue upper integral is

$$\overline{\int_{\mathbb R}}g\,dm =\inf\Bigl\{\int_{\mathbb R}h\,dm:\ g\le h,\ h\text{ Lebesgue measurable}\Bigr\}$$

(the source's (4)); it is monotone in $g$. For a set $H\subseteq\mathbb R\times Y$ write

$$H_x=\{y\in Y:(x,y)\in H\},\qquad H^y=\{x\in\mathbb R:(x,y)\in H\}$$

(the source's (6)).

Statement

Let $(Y,\mathcal P(Y),\nu)$ be a $\sigma$-finite measure space, so that every subset of $Y$ is $\nu$-measurable. For every set $H\subseteq\mathbb R\times Y$,

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

(the source's (5)). Both integrands take values in $[0,\infty]$; the right-hand integrand is $\nu$-measurable because every function on $Y$ is.

Proof

A weight. Since $\nu$ is $\sigma$-finite, write $Y=\bigcup_nY^{(n)}$ with $\nu(Y^{(n)})<\infty$ and the $Y^{(n)}$ pairwise disjoint, and put $\eta=\sum_n2^{-n-1}(1+\nu(Y^{(n)}))^{-1}1_{Y^{(n)}}$. Then $\eta\colon Y\to(0,\infty)$ and

$$\int_Y\eta\,d\nu\le\sum_n2^{-n-1}\le1$$

(the source's (7); the source asserts the existence of such an $\eta$ without displaying one).

Open envelopes. Fix $\varepsilon>0$. For each $y\in Y$ choose an open $U_y\subseteq\mathbb R$ with

$$H^y\subseteq U_y\quad\text{and}\quad m(U_y)\le m^*(H^y)+\varepsilon\eta(y)$$

(the source's (8)), taking $U_y=\mathbb R$ when $m^*(H^y)=\infty$.

A measurable majorant. Let $(I_n)_{n<\omega}$ enumerate the open intervals with rational endpoints, a base of $\mathbb R$. For $n<\omega$ put $Y_n=\{y\in Y:I_n\subseteq U_y\}$, a subset of $Y$ and hence $\nu$-measurable, and set

$$E=\bigcup_{n<\omega}(I_n\times Y_n),$$

a countable union of measurable rectangles, so $E$ is measurable for the product of the Lebesgue $\sigma$-algebra with $\mathcal P(Y)$. For every $y\in Y$, $E^y=U_y$: if $x\in E^y$ then $x\in I_n$ for some $n$ with $y\in Y_n$, so $x\in I_n\subseteq U_y$; conversely, if $x\in U_y$, then since $U_y$ is open some basic interval satisfies $x\in I_n\subseteq U_y$, so $y\in Y_n$ and $(x,y)\in I_n\times Y_n\subseteq E$. Since $H^y\subseteq U_y=E^y$ for every $y$, $H\subseteq E$.

Tonelli. Both $(\mathbb R,m)$ and $(Y,\nu)$ are $\sigma$-finite, so Tonelli's theorem applied to the measurable set $E$ gives that $x\mapsto\nu(E_x)$ is Lebesgue measurable and

$$\int_{\mathbb R}\nu(E_x)\,dm(x)=\int_Ym(E^y)\,d\nu(y)=\int_Ym(U_y)\,d\nu(y)$$

(the source's (9)). By the envelope bound and the weight,

$$\int_Ym(U_y)\,d\nu(y)\le\int_Ym^*(H^y)\,d\nu(y)+\varepsilon\int_Y\eta\,d\nu \le\int_Ym^*(H^y)\,d\nu(y)+\varepsilon$$

(the source's (10)).

Conclusion. For every $x$, $H_x\subseteq E_x$, so $\nu(H_x)\le\nu(E_x)$; thus $x\mapsto\nu(E_x)$ is a Lebesgue-measurable majorant of $x\mapsto\nu(H_x)$, and by the definition of the upper integral

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

Letting $\varepsilon\to0$ gives the statement.

The specialization used later

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, because $\mathbb R=\bigcup_{n\ge1}[-n,n]$ and $\nu([-n,n])=2n<\infty$. Taking $Y=\mathbb R$ gives, for every $H\subseteq\mathbb R^2$,

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

(the source's (11)), which Lemma 2.1 applies.

Boundary. The measure $\nu$ on the second factor must be defined on all subsets: this is what makes $Y_n$ measurable and the right-hand integrand measurable. The Lebesgue side carries no measurability assumption on $H$; the upper integral absorbs it.