LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Tail sum integrability equivalence

Statement

For a measurable $X : Ω \to [0, \infty]$ on a probability space, $\sum_{n \geq 1} P (X > n) \leq E X \leq 1 + \sum_{n \geq 1} P (X > n)$ . Thus $E X < \infty$ if and only if the tail series is finite.

Facts & Assumptions

[F1]

Monotone convergence for the integral: Let $0 \leq f_{1} \leq f_{2} \leq \dots$ be measurable and suppose $f_{n} (x) ↑ f (x)$ for every $x$ . Then $\int f_{n} d μ ↑ \int f d μ .$

Proof

Given: The objects, hypotheses and definitions in the statement. Its conclusions are to be established below.

1.1givenalgebra

For $0 \leq t < \infty$ , let $N (t)$ count positive integers strictly less than $t$ . When $t = 0$ the count is zero; when $t = m \geq 1$ is an integer it is $m - 1$ ; between consecutive integers it is the lower integer. Thus $N (t) \leq t \leq 1 + N (t)$ . At $t = \infty$ both $N (t)$ and $t$ are infinite, and the extended inequalities remain valid.

2.1F1step 1.1

The functions $\sum_{n = 1}^{m} 1_{{X > n}}$ increase pointwise to $N (X)$ . F1 therefore gives $E N (X) = \sum_{n \geq 1} P (X > n)$ . Integrate both inequalities in step 1.1 and use $E 1 = 1$ to obtain the bracket.

3.1step 2.1∎

If $E X$ is finite, the left inequality in step 2.1 bounds the series. Conversely a finite series makes the right inequality finite. This proves both directions, including extended-valued X.

Depends on

Used by

Dependency tree · two levels

28 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

Durrett, §§2.4–2.5, pp. 76–87 (standard reference, not scraped)