Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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FALSE: a pointwise limit of continuous functions is continuous almost everywhere

Statement

False claim. A pointwise limit of continuous functions on R is continuous almost everywhere.

Facts & Assumptions

Given: The fat Cantor set S[0,1].

[L1]

The fat Cantor set is closed, nowhere dense, and not Lebesgue null. (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero)

Refutation

technique · direct
1.1

By [L2], there is a continuous function u:RR with [L2, choose] S={x:u(x)=0}. Put f0:=f1, and for n1 define

fn(x):=11+nu(x)2.

Each fn is continuous, because it is built from u by continuous algebraic operations and the denominator is everywhere positive. [L2, choose]

2.1

If xS, then u(x)=0 and fn(x)=1 for every n. If xS, [step 1.1, L1] then u(x)2>0, so 1+nu(x)2+ and fn(x)0. Thus the pointwise limit is 1S. Since S is closed and has empty interior by [L1], every point of S is a boundary point of S, and the indicator 1S is discontinuous at every such point. Because S is not Lebesgue null by [L1], the discontinuity set has positive measure.

step 1.1L1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

59 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources