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Statement

Write $R(A)=\sum_{n\in A}1/n$. For sufficiently large $X$, $A\subseteq[1,X]$ with $R(A)\ge\eta$ and fixed $\alpha\in(0,3/4)$, there are $1\le M\le N\le X$ such that

$$M\ge N\exp(-(\log N)^{1-\alpha}),\qquad R(A\cap[M,N])\gg_\alpha \frac{\eta}{(\log X)^\alpha\log\log X}.$$

Source. Liu–Sawhney, arXiv:2404.07113v1, Lemma 6.1, p. 19. This is the intended interval inequality used immediately before and after that lemma. Its printed statement omits the minus sign and would require $M>N$ for $N>1$. The correction is explicit here.

Rewritten proof

Starting with $u_1=\log X$, set $u_{i+1}=\max(u_i-u_i^{1-\alpha},2)$ while $u_i>2$. Let $N_i=e^{u_i}$. Each resulting interval $[N_{i+1},N_i]$ has lower endpoint at least $N_i\exp(-(\log N_i)^{1-\alpha})$.

To bound the number of intervals, consider a band $u_i\in[U,2U]$, where $U\ge2$. Each step before leaving this band decreases $u_i$ by at least $U^{1-\alpha}$, so at most $O(U^\alpha+1)$ steps occur in it. There are $O(\log\log X)$ such bands and each has $U\le\log X$. Thus there are $O_\alpha((\log X)^\alpha\log\log X)$ intervals altogether. Cover the remaining integers below $e^2$ by singleton intervals, which also satisfy the stated endpoint inequality.

These intervals cover $[1,X]\cap\mathbb N$. Their reciprocal masses sum to at least $R(A)$; overlap of endpoints does not invalidate this inequality. One interval therefore has the claimed mass. This proves the localization statement with the displayed correction.

The printed recursion has terminal value $\log N_i=1$ but then asserts $N_i=1$. The explicit terminal treatment above avoids that additional endpoint typo.

Dependencies and verification

Only the pigeonhole principle. This rewritten proof passed independent blind review on 2026-09-18, retained as the fresh main-proof review with its distinct grade; the earlier main-proof review was ruled on 2026-09-18 a coordinated compilation check, not an independent review.

Bears on

• Problem 298
• Problem 299