Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-21
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FALSE: continuity from above needs no finiteness hypothesis

Statement

False claim. For every decreasing sequence (Ek) of measurable sets, one has μ(⋂kEk)=inf⁡kμ(Ek), without requiring any Ek to have finite measure. The valid theorem Continuity from above when one set has finite measure includes precisely that missing hypothesis.

Facts & Assumptions

Given: Counting measure # on N and the tails Ek:={n∈N:k≤n}.

[L1]

Counting measure assigns +∞ to every infinite set (Counting measure on an arbitrary set) and is a measure (Counting measure is a measure).

[L2]

Continuity from above is proved when one member of the decreasing sequence has finite measure (Continuity from above when one set has finite measure).

[L3]

The natural order is defined by m≤n exactly when m+k=n for some natural k (Order on the natural numbers), natural addition is cancellative (Addition is cancellative), and k<k+1 (Discreteness: σ(n) is the immediate successor).

Refutation

technique · direct
1.1givenL3

The tails decrease, Ek+1⊆Ek, and E0=N.

1.2givenL1L3

Every Ek is infinite, because n↦k+n injects N into it; hence #(Ek)=+∞ for every k.

1.3givenL3

The intersection is empty: if n belonged to every tail, it would belong to En+1, which would say n+1≤n, contrary to discreteness of the natural order.

2.1step 1.2step 1.3L1L2∎

Thus #(⋂kEk)=#(∅)=0 but inf⁡k#(Ek)=+∞. This refutes the claim and shows why [L2] cannot be applied: no tail has finite counting measure.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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