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For $1 \leq p < \infty$ , every $L^{p} (μ)$ class has a sigma-finite essential support

Statement

Let $(X, A, μ)$ be a measure space and let $1 \leq p < \infty$ . For every element $[f] \in L^{p} (μ)$ there is a measurable sigma-finite set $S \subseteq X$ such that $f = 0$ almost everywhere on $X ∖ S$ .

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , an exponent $1 \leq p < \infty$ , and a class $[f] \in L^{p} (μ)$ .

[L1]

Elements of $L^{p} (μ)$ are almost-everywhere classes of measurable representatives (The space $L^{p} (μ)$ as the quotient by null functions).

[L2]

Chebyshev-Markov gives $μ ({∣ u ∣^{p} \geq t}) \leq t^{- 1} \int ∣ u ∣^{p} d μ (t > 0)$ (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 $n \geq 1$ , [L1, L2, given, choose, construct] set $E_{n} : = {∣ u ∣ \geq 1 / n} = {∣ u ∣^{p} \geq n^{- p}} .$ Since $u \in L^{p} (μ)$ , [L2] gives $μ (E_{n}) \leq n^{p} \int ∣ u ∣^{p} d μ < \infty .$ So every $E_{n}$ has finite measure.

2.1L3step 1.1

Put [L3, step 1.1] $S : = ⋃_{n = 1}^{\infty} E_{n} .$ By step 1.1 and [L3], the set $S$ is sigma-finite.

3.1step 2.1algebra∎

If $x \notin S$ , then $∣ u (x) ∣ < 1 / n$ for every $n \geq 1$ , 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 $S$ .

For $1 \leq p < \infty$ , every $L^{p} (μ)$ class has a sigma-finite essential support

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For $1 \leq p < \infty$ , every $L^{p} (μ)$ class has a sigma-finite essential support