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Continuous functions are dense in Lp for compactly supported finite measures
Statement
Let be a finite regular Borel measure on concentrated on a compact set , and let . Form the real space using the Borel -algebra. Then restrictions to of real-valued compactly supported continuous functions on are dense in .
Statement references
Facts & Assumptions
Given: A finite regular Borel measure concentrated on compact , an exponent , a real-valued Borel class , and .
consists of measurable functions with finite , with functions identified when equal -almost everywhere (The function space for , The space as the quotient by null functions).
A measure is a nonnegative countably additive set function on a sigma-algebra, and regularity here means Borel sets can be approximated from inside by compact sets and outside by open sets with arbitrarily small measure excess (Measures on sigma-algebras); regularity is the hypothesis on .
If is compact and is open with , there is a smooth compactly supported cutoff with , near , and (Test function cutoffs and euclidean localization).
Dominated convergence applies to and almost everywhere (Dominated convergence).
For , the quotient has a norm whose triangle inequality is Minkowski's inequality (The norm descends to the quotient and makes a normed space for ).
Proof
Choose a measurable representative of . Truncate it to . Since and pointwise wherever is finite, [F4] makes for some large . Partition into finitely many intervals of length at most and replace each value of by a fixed endpoint of its interval. If , choose with ; if , any suffices. The resulting finite-valued Borel simple function satisfies pointwise, so . By [F5] and the triangle inequality, . Since is concentrated on , replace by : this finite Borel simple function represents the same class and admits a representation with every set . Thus such supported simple functions are dense in .
Write such a simple function as with Borel . For any , regularity [F2] gives compact and open with . By [F3] choose with and on . Outside the functions and agree; everywhere their difference has magnitude at most one. Therefore .
The triangle inequality [F5] gives . Choose small enough that this is below , and choose from step 1.1 within of . Then the compactly supported continuous function is within of in .
Depends on
- Measures on sigma-algebras
- The function space $\mathcal{L}^p(\mu)$ for $0 < p < \infty$
- The space $L^p(\mu)$ as the quotient by null functions
- The $L^p$ norm descends to the quotient and makes $L^p$ a normed space for $1 \le p \le \infty$
- Dominated convergence
- Test function cutoffs and euclidean localization
Used by
Dependency tree · two levels
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Sources
- Leshno, Lin, Pinkus, and Schocken, Multilayer Feedforward Networks With a Non-Polynomial Activation Function Can Approximate Any Function, proof of Theorem 1, Lp part (standard reference, not scraped)
- Kurt Hornik, Approximation Capabilities of Multilayer Feedforward Networks, §2 (standard reference, not scraped)