TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Simple functions are dense in $L^{\infty} (μ)$ in the essential-supremum norm

Statement

Let $(X, A, μ)$ be a measure space. Simple functions are dense in $L^{\infty} (μ)$ for the essential-supremum norm.

Facts & Assumptions

Given: A measure space, $ε > 0$ , and $f \in L^{\infty} (μ)$ .

[L1]

Choose a representative of the $L^{\infty}$ class and interpret its norm by the least essential bound (The space $L^{\infty} (μ)$ of essentially bounded measurable functions, The essential supremum is attained as the least essential bound).

Proof

technique · direct

1.1L1givenchooseconstruct

Choose a measurable representative $u$ of $f$ and let $M : = ∥ f ∥_{\infty}$ . [L1, given, choose, construct] By [L1], $∣ u (x) ∣ \leq M$ almost everywhere. Partition the interval $[- M, M]$ into finitely many subintervals of length at most $ε$ , and on each strip $u^{- 1} (I_{j})$ choose one value $c_{j} \in I_{j}$ . The resulting function $s : = \sum_{j} c_{j} 1_{u^{- 1} (I_{j})}$ is simple.

2.1step 1.1algebra

On the full-measure set where $∣ u ∣ \leq M$ , the values $u (x)$ and $s (x)$ lie [step 1.1, algebra] in the same interval $I_{j}$ , so $∣ u (x) - s (x) ∣ \leq ε$ . Hence $∥ f - s ∥_{\infty} \leq ε .$

3.1step 2.1∎

Since $ε > 0$ was arbitrary, simple functions are dense in [step 2.1] $L^{\infty} (μ)$ for the essential-supremum norm.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

Simple functions are dense in L∞(μ)L∞(μ) in the essential-supremum norm