ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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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 region under x maps to x^2 on [0,1] has measure 1/3

Example

The set $R : = {(x, t) \in [0, 1] \times R : 0 \leq t < x^{2}}$ has planar Lebesgue measure $1 / 3$ .

Facts & Assumptions

Given: The function $f (x) = x^{2}$ on $[0, 1]$ .

[L1]

The region under a nonnegative measurable function has product measure equal to its integral. (The region under a nonnegative measurable function is product-measurable and has measure equal to the integral)

[L2]

The fundamental theorem of calculus evaluates $\int_{0}^{1} x^{2} d x$ . (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ )

Verification

technique · direct

1.1

Applying [L1] to $f (x) = x^{2}$ gives [L1] $λ_{2} (R) = \int_{0}^{1} x^{2} d x .$

2.1

Since $(x^{3} / 3)^{'} = x^{2}$ , [L2] gives [L2, step 1.1] $λ_{2} (R) = \int_{0}^{1} x^{2} d x = \frac{1}{3} .$ ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

Terence Tao, An Introduction to Measure Theory, Exercise 1.7.24 (standard reference, not scraped)