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Simple functions with finite-measure support are dense in $L^{p} (μ)$ for $1 \leq p < \infty$

Statement

Let $(X, A, μ)$ be a measure space and let $1 \leq p < \infty$ . Then every element of $L^{p} (μ)$ can be approximated in $L^{p}$ by simple functions whose supports have finite measure.

Facts & Assumptions

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

[L1]

Every measurable function admits dominated simple approximations (Every measurable function admits simple approximations dominated by its absolute value).

[L2]

Dominated convergence applies in $L^{1}$ (Dominated convergence).

[L3]

Elements of $L^{p} (μ)$ are almost-everywhere classes, so one may choose a measurable representative when making pointwise constructions (The space $L^{p} (μ)$ as the quotient by null functions).

Proof

technique · direct

1.1

Choose a measurable representative $u$ of $f$ by [L3]. For each [L1, L3, given, choose, construct] $m \in N_{\geq 1}$ , [L1] gives a simple function $s_{m}$ with $∣ s_{m} ∣ \leq ∣ u ∣$ and $s_{m} (x) \to u (x)$ for every $x$ . Define

t_{m} : = s_{m} 1_{{∣ u ∣ \geq 1 / m}} .

Then each $t_{m}$ is simple and $supp (t_{m}) \subseteq {∣ u ∣ \geq 1 / m}$ .

L1L3givenchooseconstruct

2.1

Since $∣ u ∣^{p}$ is integrable, [step 1.1, algebra]

μ ({∣ u ∣ \geq 1 / m}) \leq m^{p} \int ∣ u ∣^{p} d μ < \infty,

so every $t_{m}$ has finite-measure support. Also $t_{m} (x) \to u (x)$ for every $x$ : if $u (x) = 0$ then eventually both sides are $0$ , and if $u (x) \neq 0$ then $1_{{∣ u (x) ∣ \geq 1 / m}} = 1$ for all large $m$ . Finally

∣ u - t_{m} ∣^{p} \leq (2 ∣ u ∣)^{p},

whose right-hand side is integrable.

step 1.1algebra

3.1

Applying [L2] to $∣ u - t_{m} ∣^{p}$ gives [L2, step 2.1]

∥ f - t_{m} ∥_{p}^{p} = \int ∣ u - t_{m} ∣^{p} d μ ⟶ 0.

So the simple functions with finite-measure support are dense in $L^{p} (μ)$ .

L2step 2.1∎

Depends on

Used by

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Sources

John K. Hunter, Measure Theory (standard reference, not scraped)

Simple functions with finite-measure support are dense in Lp(μ)Lp(μ) for 1≤p<∞1≤p<∞