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Source. S. Korsky, A resolution of the de Bruijn--Erdős consecutive-gap problem, arXiv:2609.07196v2, Section 7: Theorem 7.1 (p. 13) and Lemma 7.2 (pp. 13--14) of the retained PDF, read in the canonical conversion and checked against the text layer at the displayed constants; held by its library card, Korsky 2026, resolution.

Standing. Author-recorded reconstruction of Lemma 7.2; not an independent review; changes no status and assigns no tier. Theorem 7.1 is an external input, cited by the source to G. Halász, On Roth's method in the theory of irregularities of point distributions, in Recent Progress in Analytic Number Theory, vol. 2 (Academic Press, 1981), 79--94; that paper is not held and the statement is not checked against it here.

Definitions

For a finite set $\mathcal P\subset[0,1]^2$ of $M$ points put

$$D_{\mathcal P}(u,v)=\#\bigl(\mathcal P\cap((0,u]\times(0,v])\bigr)-Muv \qquad(0\le u,v\le1);$$

endpoint conventions do not affect the integrals below. Points, $P_n$ and $N_n(\cdot)$ are as on the Lemma 2.1 page; only integer times occur here.

The imported input (Theorem 7.1, Halász)

There is an absolute constant $c_H>0$ such that every set $\mathcal P$ of $M\ge2$ points in $[0,1]^2$ satisfies

$$\int_0^1\!\!\int_0^1\bigl|D_{\mathcal P}(u,v)\bigr|\,du\,dv\ \ge\ c_H\sqrt{\log M}.$$

This is the unnormalized form of Halász's planar $L^1$ discrepancy theorem, as the source states it; the source gives no further derivation and none is supplied here.

Statement (Lemma 7.2, p. 13)

There are absolute constants $c_4>0$ and $S_0$ with the following property. Suppose $S\ge S_0$, $B\ge1$, and, for all sufficiently large integers $n$,

$$\int_{\mathbb T}\Bigl|N_n\bigl((x,x+D/n]\bigr)-D\Bigr|\,dx\ \le\ B \qquad(0\le D\le S). \tag{7.1}$$

Then $B\ge c_4\sqrt{\log S}$.

Proof

Put $L=\lfloor S\rfloor$, let $n_0$ be a threshold for (7.1), choose an integer $N>\max(L,n_0)$, and put $w=L/N$. For $a\in\mathbb T$ let $M_a=N_N((a,a+w])$ be the number of the first $N$ points in the arc of length $w=L/N$ starting at $a$. Applying (7.1) at $n=N$ with $D=L$,

$$\int_{\mathbb T}|M_a-L|\,da\ \le\ B . \tag{7.2}$$

The planar point sets. View the oriented arc $(a,a+w]$ as a copy of $(0,1]$. For each $x_i\in P_N\cap(a,a+w]$ form the point

$$\Bigl(\frac{x_i-a}w,\ \frac iN\Bigr)\in(0,1]^2 ,$$

the first coordinate measured along the arc. Let $\mathcal P_a$ be the resulting set of $M_a$ points, and define

$$G_a(u,v)=N_{\lfloor Nv\rfloor}\bigl((a,a+uw]\bigr)-Luv .$$

A point $x_i$ of the arc lies in the box $(0,u]\times(0,v]$ exactly when $x_i\in(a,a+uw]$ and $i\le Nv$, that is, $i\le\lfloor Nv\rfloor$; so, off a null set of $(u,v)$ caused by endpoints,

$$D_{\mathcal P_a}(u,v)=N_{\lfloor Nv\rfloor}\bigl((a,a+uw]\bigr)-M_auv =G_a(u,v)+(L-M_a)uv .$$

Lower bound from Halász. Since $\int_0^1\int_0^1uv\,du\,dv=1/4$, Theorem 7.1 gives, whenever $M_a\ge2$,

$$\|G_a\|_{L^1([0,1]^2)}\ \ge\ c_H\sqrt{\log M_a}-\tfrac14|M_a-L| . \tag{7.3}$$

If $B\ge L/4$, then $B\ge c_4\sqrt{\log S}$ for any fixed $c_4$ once $S_0$ is large, because $L\ge S-1$ grows faster than $\sqrt{\log S}$; so assume $B<L/4$. By Markov's inequality and (7.2), the set of $a$ with $|M_a-L|>L/2$ has measure at most $(2/L)B<1/2$, so

$$\bigl|\{a:\ L/2\le M_a\le3L/2\}\bigr|\ \ge\ \tfrac12 .$$

On this set $M_a\ge L/2\ge2$ for $L\ge4$, and $\sqrt{\log M_a}\ge\sqrt{\log(L/2)}$; integrating (7.3) over it and using (7.2) for the subtracted term,

$$\int_{\mathbb T}\|G_a\|_1\,da\ \ge\ \frac{c_H}2\sqrt{\log(L/2)}-\frac B4 . \tag{7.4}$$

Upper bound from (7.1). Fix $(u,v)$ and put $n=\lfloor Nv\rfloor$. If $n\ge n_0$, set $D=nuw$; then $D\le Nuw=Lu\le L\le S$, the arc $(a,a+uw]$ has length $uw=D/n$, and

$$G_a(u,v)=\Bigl(N_n\bigl((a,a+uw]\bigr)-nuw\Bigr)+Lu\Bigl(\frac nN-v\Bigr),$$

because $nuw=Lu\cdot n/N$. The second term has absolute value at most $Lu/N\le w$, since $0\le v-n/N<1/N$; the integral over $a$ of the absolute value of the first term is at most $B$ by (7.1). On the strip $0\le v<n_0/N$, where $n<n_0$, trivially $|G_a(u,v)|\le n_0+L$. Integrating over $(u,v)$ and then $a$,

$$\int_{\mathbb T}\|G_a\|_1\,da\ \le\ B+\frac LN+\frac{n_0(n_0+L)}N =B+o_{N\to\infty}(1), \tag{7.5}$$

with $L$ and $n_0$ fixed.

Conclusion. Combining (7.4) and (7.5) and letting $N\to\infty$,

$$\frac54B\ \ge\ \frac{c_H}2\sqrt{\log(L/2)} .$$

Since $L=\lfloor S\rfloor$ and $S\ge S_0$ is large, $\log(L/2)\ge\frac12\log S$, say, and $B\ge c_4\sqrt{\log S}$ with an absolute $c_4>0$ after adjusting the constants; the case $B\ge L/4$ treated above is covered by the same $c_4$ once $S_0$ is large enough.

Role in the argument

Proposition 6.4 supplies (7.1) with $B=C_3A$ and $S=\sqrt{Ar}/\Lambda^2$; the Section 8 proof takes $A=c\sqrt{\log r}$ and compares $C_3A$ with $c_4\sqrt{\log S}\ge c_4\sqrt{(\log r)/3}$.