Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11
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A bounded truncation function need not have an improper limit

Example

Define f(x)=(−1)k on [k,k+1) for every nonnegative integer k. Its truncation primitive is bounded, but ∫0∞f diverges.

Facts & Assumptions

Verification

technique · counterexample
1.1L1L2

For a positive integer N, additivity gives the displayed sum, whose parity values follow from the alternating sequence. [L1, L2] ∫0Nf=∑k=0N−1(−1)k, which equals one for odd N and zero for even N by [L2]. Thus the integer truncations are bounded but have no limit. [L1, L2] If N≤R<N+1, the remaining integral has absolute value at most one, so the full truncation primitive is bounded as asserted.

2.1L3step 1.1∎

An improper limit would restrict to the same limit along all integer truncations, contradicting step 1.1. The example changes sign, so it does not satisfy the nonnegativity hypothesis in [L3] and shows that hypothesis cannot be deleted. [L3, step 1.1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

37 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