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Let be a meaurable subset with Lebesgue measure . Is it true that, for almost all ,
for all ?
Let be a meaurable subset with Lebesgue measure . Is it true that, for all ,
for almost all ?
Source: erdosproblems.com/994
An accepted solution exists. The statement is false.
DISPROVED (LEAN), the label the community database that the site displays lists as of its last update on 2026-09-16 (the site's label on 2026-09-04 was DISPROVED): Khintchine's question of 1923 [Kh23], which Erdős's 1964 problem paper [Er64b] calls a conjecture, was refuted by Marstrand [Ma70], so the precise Statement is false, and the site's wording is false in either order of its two quantifiers, as the accepted claim Marstrand 1970 explains. The Lean behind the qualifier is a third-party formalization of Marstrand's disproof in the fixed-set order, not built or audited in this corpus.
The site's wording places "for all " after "for almost all
", so it also admits a simultaneous reading, in which one set of
of full measure serves every measurable at once. That reading fails
for every : the orbit is countable, so its
complement in is measurable, has measure and is never visited, and
its visit frequency is ; the argument is elementary and is set out on the
claim page. The change swaps the two phrases "for almost all " and "for
all ", so that the null set of exceptional may depend on ;
nothing else changes. The evidence is the poser's own text as the site credits
it: the site's commentary calls the problem a conjecture of Khintchine [Kh23],
and Khintchine's question (§ 5, "Ein neues Problem", pp. 303–304) fixes the set
first and asks whether his relations (6) and (7) hold "für alle mit
Ausnahme höchstens einer Menge vom Maße Null". The ambiguity is already in
Erdős's statement in [Er64b] (Part II, p. 57), which the site's wording follows:
"Then for almost all and every ". Erdős's own words there point to
the same reading, since he credits the conjecture to Khintchine with the locator
"see p. 303–304" and calls it "very deep", which the simultaneous reading is
not. The choice does not change the answer: Marstrand's refutation [Ma70] of the
precise Statement refutes the simultaneous reading as well. Results about the
simultaneous reading alone, credited here and not counted: the variant
erdos_994.variants.simultaneous of the formal-conjectures statement file
(added on 2026-09-22, pinned under Formalization) and the theorem
not_erdos_994 of the file
Erdos994.lean
in Boris Alexeev's lean-proofs collection (in the repository since 2026-08-17;
formal authors Codex and GPT-5.6 Sol, as the file names them), both proving the
orbit argument in Lean.