ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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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.
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The weights $2^{- (k + 1)}$ define a probability measure on $P (N)$

Example

For $E \subseteq N$ , define

$μ (E) : = \sum_{k \in E} 2^{- (k + 1)} = (\sum_{k = 0}^{\infty} 2^{- (k + 1)} δ_{k}) (E) .$

Then $μ$ is a probability measure on $P (N)$ . The shift by $1$ is essential: because $0 \in N$ , the unshifted weights $2^{- k}$ have total mass $2$ , not $1$ .

Facts & Assumptions

Given: The Dirac measures $δ_{k}$ on $N$ and the weights $c_{k} = 2^{- (k + 1)}$ .

[L1]

Nonnegative countable weighted sums of measures are measures (Nonnegative scalar multiples and countable weighted sums of measures are measures); $δ_{k} (E)$ is $1$ exactly when $k \in E$ and is $0$ otherwise (The Dirac set function at a point); and every $δ_{k}$ is a probability measure (A Dirac set function is a probability measure).

[L2]

A probability measure is a measure of total mass $1$ (Probability measures and probability spaces).

[L3]

Natural powers have initial value $r^{0} = 1$ (Integer powers $a^{m}$ ), and for $∣ r ∣ < 1$ , $\sum_{k = 0}^{\infty} r^{k} = 1 / (1 - r)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Verification

technique · direct

1.1

By [L1], $μ = \sum_{k} 2^{- (k + 1)} δ_{k}$ is a measure, and evaluating it on $E$ gives exactly the displayed subseries because $δ_{k} (E)$ is $1$ for $k \in E$ and $0$ otherwise.

givenL1

1.2

The geometric-series formula with $r = 1 / 2$ gives $μ (N) = \sum_{k = 0}^{\infty} 2^{- (k + 1)} = (1 / 2) \sum_{k = 0}^{\infty} (1 / 2)^{k} = 1$ .

givenL3algebra

1.3

The unshifted total is $\sum_{k = 0}^{\infty} 2^{- k} = 2$ , whose first term at $k = 0$ is $1$ ; thus those weights do not define a probability measure.

givenL3algebra

2.1

Steps 1.1 and 1.2 make $μ$ a probability measure by [L2], and step 1.3 verifies the index-zero boundary and rules out the unshifted construction.

step 1.1step 1.2step 1.3L2∎

Depends on

Used by

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The weights 2−(k+1)2−(k+1) define a probability measure on P(N)P(N)