ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-09-08 (Codex)

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Mollifying $1_{[0, 1]}$ at two scales

Example

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $φ \in C_{c}^{\infty} (R)$ satisfy $\int φ = 1$ and $supp (φ) \subseteq [- 1, 1]$ . For $f : = 1_{[0, 1]}$ and $ε > 0$ ,

$(f * φ_{ε}) (x) = \int_{0}^{1} φ_{ε} (x - y) d y = \int_{(x - 1) / ε}^{x / ε} φ (u) d u .$

For $ε = 1 / 4$ and $ε = 1 / 8$ , the graph is $0$ outside $[- ε, 1 + ε]$ , equals $1$ on $[ε, 1 - ε]$ , and can differ from these constant pieces only in $[- ε, ε]$ and $[1 - ε, 1 + ε]$ , each of width $2 ε$ . Thus the possible transition intervals have width $1 / 2$ at scale $1 / 4$ and width $1 / 4$ at scale $1 / 8$ . Actual transitions may be narrower if the kernel has smaller support. A signed kernel need not produce monotone transitions or values between zero and one.

Facts & Assumptions

Given: Countable Choice, a unit-mass smooth bump $φ$ supported in $[- 1, 1]$ , and $f = 1_{[0, 1]}$ .

[L1]

Mollifier families are approximate identities (A unit-mass smooth bump generates an $L^{1}$ approximate identity).

[L2]

Convolution with a mollifier is smooth (Convolution with a mollifier is smooth, and derivatives pass under the integral sign).

[L3]

Linear change of variables and translation invariance give the affine one-dimensional substitution used below (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

Verification

technique · direct

1.1L1L3givenalgebra

The substitution $u = (x - y) / ε$ has absolute Jacobian $1 / ε$ and reverses the interval endpoints, giving the displayed integral by [L3]. Since the scaled kernel is supported in $[- ε, ε]$ , the original integral is zero when $x \notin [- ε, 1 + ε]$ .

2.1L1step 1.1algebra

If $x \in [ε, 1 - ε]$ , then $x / ε \geq 1$ and $(x - 1) / ε \leq - 1$ . Hence the integration interval contains the whole support of $φ$ , and the value is exactly $1$ . For $0 < ε < 1 / 2$ , all remaining possible variation is confined to the two disjoint intervals $[- ε, ε]$ and $[1 - ε, 1 + ε]$ , each of width $2 ε$ .

3.1L2step 2.1givenalgebra∎

By [L2] the convolution is smooth. Substituting $ε = 1 / 4$ and $ε = 1 / 8$ into step 2.1 gives the claimed widths $1 / 2$ and $1 / 4$ . Support containment need not be equality, and no sign condition on $φ$ was imposed, so the calculation asserts neither exact transition width nor monotonicity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)

Mollifying 1[0,1]1[0,1]​ at two scales