PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31

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The essential supremum is attained as the least essential bound

Statement

Let $f : X \to R$ be measurable on a measure space $(X, A, μ)$ , and suppose $∥ f ∥_{\infty} < \infty$ . Then

$∣ f ∣ \leq ∥ f ∥_{\infty} μ -almost everywhere .$

Moreover, if $M \geq 0$ and $∣ f ∣ \leq M$ almost everywhere, then

$∥ f ∥_{\infty} \leq M .$

So $∥ f ∥_{\infty}$ is the least essential bound of $∣ f ∣$ .

Facts & Assumptions

Given: A measurable real-valued function $f$ with finite essential supremum $s : = ∥ f ∥_{\infty}$ .

[L1]

The essential supremum is the infimum of the essential bounds (The essential supremum of a measurable function with respect to a measure).

[L2]

Absolute values and threshold sets of measurable functions are measurable (Closure properties of measurable functions used by the integral).

[L3]

Countable subadditivity bounds the measure of a countable union (Finite and countable subadditivity of measures).

Proof

technique · Take the countable family of bad sets $\{|f|>\|f\|_\infty+1/n\}$; each is null by minimality of the infimum. Their union is null, giving $|f|\le\|f\|_\infty$ almost everywhere, and leastness is built into the definition

1.1L1L2given

For each $n \geq 1$ , the number $s + 1 / n$ is strictly larger than the infimum in [L1], so it is an essential bound. Therefore the measurable set $E_{n} : = {∣ f ∣ > s + 1 / n}$ has measure $0$ .

1.2L1

If $M \geq 0$ and $∣ f ∣ \leq M$ almost everywhere, then $M$ is one of the essential bounds in [L1], so the infimum $s$ satisfies $s \leq M$ . [L1].

2.1step 1.1L3algebra

Put $E : = ⋃_{n = 1}^{\infty} E_{n}$ . Then $E$ is measurable and $μ (E) \leq \sum_{n = 1}^{\infty} μ (E_{n}) = 0.$ If $x \notin E$ , then $∣ f (x) ∣ \leq s + 1 / n$ for every $n$ , hence $∣ f (x) ∣ \leq s$ . Therefore $∣ f ∣ \leq s$ almost everywhere.

3.1step 2.1step 1.2∎

Step 2.1 proves that $s$ itself is an essential bound, and step 1.2 proves that no smaller essential bound exists. Thus $s = ∥ f ∥_{\infty}$ is the least essential bound. [step 2.1, step 1.2].

The essential supremum is attained as the least essential bound

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources