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Source. E. Glazer, Erdős Problem 501 after adding $\omega_2$ random reals, draft rev10, Lemma 2.1 (positive-measure selection), physical p. 2, in the eight-page PDF held by its library source card, Glazer (2026). The printed page numbers of that PDF coincide with its physical pages.

Standing. This is an author-recorded reconstruction. It is not an independent review and changes no status and assigns no tier. The only external input is Tonelli's theorem for $\sigma$-finite product measures, used for the measurability of section-measure functions and for the interchange of the two integrals.

Definitions

Let $(S,\Sigma,\mu)$ be a $\sigma$-finite measure space. For a set $E\subseteq S^2$ measurable for the product $\sigma$-algebra $\Sigma\otimes\Sigma$, and for $t,s\in S$, write

$$E_t=\{s\in S:(t,s)\in E\},\qquad E^s=\{t\in S:(t,s)\in E\}.$$

Both sections lie in $\Sigma$. The source reads $(t,s)\in E$ as "the point represented by $t$ is forbidden by the envelope represented by $s$": $E_t$ is the set of envelopes that forbid $t$, and $E^s$ the set of points that $s$ forbids.

Statement

The following is provable in ZFC. Suppose $\mu(S)=\infty$, $K<\infty$, $E\subseteq S^2$ is measurable, and the column bound

$$\mu(E^s)\le K\qquad(s\in S)$$

holds (the source's (2.1)). If $C\subseteq S$ is measurable with $\mu(C)=\infty$, then

$$Q(C)=\{t\in C:\mu(C\setminus E_t)=\infty\}$$

(the source's (2.2)) is measurable and $\mu(Q(C))>0$.

The hypothesis $\mu(S)=\infty$ is implied by $\mu(C)=\infty$ and is not used separately.

Proof

Measurability. For $t\in S$,

$$\mu(C\setminus E_t)=\int_S 1_C(s)\,1_{S^2\setminus E}(t,s)\,d\mu(s).$$

The integrand is $\Sigma\otimes\Sigma$-measurable in $(t,s)$ and nonnegative, so Tonelli's theorem makes $t\mapsto\mu(C\setminus E_t)$ a $\Sigma$-measurable map into $[0,\infty]$. Hence $Q(C)=C\cap\{t:\mu(C\setminus E_t)=\infty\}$ lies in $\Sigma$.

Positivity. Suppose $\mu(Q(C))=0$. Then $\mu(C\setminus Q(C))=\infty$. Since $\mu$ is $\sigma$-finite, $C\setminus Q(C)$ is an increasing union of measurable sets of finite measure whose measures tend to $\infty$; fix one of them, $D\subseteq C\setminus Q(C)$, with

$$K<\mu(D)<\infty.$$

Every $t\in D$ lies outside $Q(C)$, so $\mu(C\setminus E_t)<\infty$. For $k\in\mathbb N$ put

$$D_k=\{t\in D:\mu(C\setminus E_t)\le k\},$$

measurable by the first paragraph. The $D_k$ increase with $k$ and their union is $D$, so continuity from below gives some $k$ with

$$d:=\mu(D_k)>K;$$

also $d\le\mu(D)<\infty$.

By $\sigma$-finiteness again, choose measurable $C_0\subseteq C_1\subseteq\cdots$ with union $C$, each of finite measure, and $M_n:=\mu(C_n)\to\infty$. Because $K<d<\infty$, the inequality

$$(M_n-k)\,d>KM_n$$

(the source's (2.3)) is equivalent to $M_n(d-K)>kd$ and so holds for all large $n$; fix such an $n$.

For $t\in D_k$, since $C_n\setminus E_t\subseteq C\setminus E_t$ and $\mu(C_n)<\infty$,

$$\mu(E_t\cap C_n)=M_n-\mu(C_n\setminus E_t)\ge M_n-k.$$

Apply Tonelli's theorem to the measurable set $E\cap(D_k\times C_n)$: its $t$-section is $E_t\cap C_n$ for $t\in D_k$ and empty otherwise, and its $s$-section is $E^s\cap D_k$ for $s\in C_n$ and empty otherwise. Hence

$$(M_n-k)\,d \le\int_{D_k}\mu(E_t\cap C_n)\,d\mu(t) =\int_{C_n}\mu(E^s\cap D_k)\,d\mu(s) \le\int_{C_n}K\,d\mu(s) =KM_n,$$

the last inequality by the column bound. This contradicts the choice of $n$. Therefore $\mu(Q(C))>0$.

Boundary. Nothing here concerns the family $(A_y)$; the lemma is applied in Theorem 3.2 to a Borel graph on $\mathbb Z\times\Omega$ with $K=1$.