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Continuous compactly supported functions are translation-continuous in
Statement
Assume the Axiom of Countable Choice.
Let and let . Then
Facts & Assumptions
Given: The Axiom of Countable Choice, , and .
Continuous functions on compact metric spaces are uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).
Bounded sets have finite Lebesgue measure, and translation preserves Lebesgue measure (Lebesgue measure is sigma-finite, and every metrically bounded subset of has finite outer measure, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).
Translation is the convention of Translation of a function on , and is defined in The spaces and .
Proof
Let , and choose so that [L1, L3, given, choose] . Then for , the support of lies in the compact set . By [L1], is uniformly continuous on that compact set.
Let . Uniform continuity gives such that [L1, L2, step 1.1, choose, algebra] whenever and . Hence for , The right-hand side tends to with , and [L2] makes the measure finite.
Therefore as .
Depends on
- Translation of a function on $\mathbb{R}^n$
- The spaces $C_c(\mathbb{R}^n)$ and $C_c^\infty(\mathbb{R}^n)$
- Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous
- Lebesgue measure is sigma-finite, and every metrically bounded subset of $\mathbb{R}^n$ has finite outer measure
- Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation
Used by
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)