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Statement

Notation as in Theorem 2: $S$ is a set of positive reals and $\mu_S(\varrho)$ is the measure of $S\cap(0,\varrho)$.

Theorem 3 (printed p. 138, quoted). "Suppose that

$$\limsup_{\varrho\to\infty}\mu_S(\varrho)/\varrho=c>0 \tag{9}$$

(„$S$ has positive upper density"). Then for almost every $\alpha$, infinitely many multiples $m\alpha$ lie in $S$."

Remark 3 on the same page introduces it: the sets of Theorem 2 have $\mu_S(\varrho)=o(\varrho)$, and Theorem 3 shows this is necessary for a set with the properties of Theorem 2. The $\alpha$ range over the positive reals, as the proof (p. 143) states.

Source. W. M. Schmidt, Disproof of some conjectures on Diophantine approximations, Studia Sci. Math. Hungar. 4 (1969), 137--144; Theorem 3 on printed p. 138, the proof in Section 5 on printed pp. 143--144, read on the page images. The edition read is identified on the source card.

Read depth. Claims checked: the statement and Remark 3 were read clause by clause on the page image of p. 138; the proof (pp. 143--144) was read for its structure and not checked. Nothing here is independently reviewed.

Proof pointer

Section 5 (pp. 143--144). The set $T$ of $\alpha>0$ with infinitely many multiples in $S$ is measurable, and it suffices to show (24) $\mu(I\cap T)\ge c\mu(I)$ for every interval $I$ of positive reals, since a complement of positive measure would have density points contradicting (24). Lemma 3 (p. 143): for each $\varepsilon>0$ there are infinitely many positive integers $m$ with (25) $\mu(mI\cap S)\ge(c-\varepsilon)\mu(mI)$, proved (p. 144) by packing disjoint dilates $n_kI,\ldots,n_1I$ into $[0,\varrho]$ at a scale $\varrho$ where $\mu_S(\varrho)\ge(c-\varepsilon/2)\varrho$. Taking $m_1<m_2<\cdots$ from the lemma, the sets $S_k=\{\alpha\in I:m_k\alpha\in S\}$ have $\mu(S_k)\ge(c-\varepsilon)\mu(I)$, and $T\cap I$ contains their limit superior. Not reconstructed here.

Dependencies

Lemma 3 of the paper (p. 143); otherwise self-contained.

Bears on

No problem page is reached by this result.