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For 1≤p<∞, every Lp(μ) class has a sigma-finite essential support

Statement

Let (X,A,μ) be a measure space and let 1≤p<∞. For every element [f]∈Lp(μ) there is a measurable sigma-finite set S⊆X such that f=0 almost everywhere on X∖S.

Facts & Assumptions

Given: A measure space (X,A,μ), an exponent 1≤p<∞, and a class [f]∈Lp(μ).

[L1]

Elements of Lp(μ) are almost-everywhere classes of measurable representatives (The space Lp(μ) as the quotient by null functions).

[L2]

Chebyshev-Markov gives μ({∣u∣p≥t})≤t−1∫∣u∣p dμ(t>0) (Chebyshev-Markov inequality for the integral).

[L3]

Sigma-finiteness means a countable union of finite-measure measurable sets (Finite, sigma-finite, and semifinite measures).

Proof

technique · Take a representative $u$ and use the level sets $\{|u|\ge1/n\}$. Chebyshev-Markov makes each level set finite-measure, and their union contains every point where $u\ne0$
1.1L1L2givenchooseconstruct

Choose a measurable representative u of [f] and, for each n≥1, [L1, L2, given, choose, construct] set En:={∣u∣≥1/n}={∣u∣p≥n−p}. Since u∈Lp(μ), [L2] gives μ(En)≤np∫∣u∣p dμ<∞. So every En has finite measure.

2.1L3step 1.1

Put [L3, step 1.1] S:=⋃n=1∞En. By step 1.1 and [L3], the set S is sigma-finite.

3.1step 2.1algebra∎

If x∉S, then ∣u(x)∣<1/n for every n≥1, hence u(x)=0. [step 2.1, algebra] Therefore u=0 on X∖S, so the class [f] vanishes almost everywhere outside the sigma-finite set S.

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Sources