Alphabeta Math
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→R be measurable on a measure space (X,A,μ), and suppose ∥f∥∞<∞. Then

∣f∣≤∥f∥∞μ-almost everywhere.

Moreover, if M≥0 and ∣f∣≤M almost everywhere, then

∥f∥∞≤M.

So ∥f∥∞ is the least essential bound of ∣f∣.

Facts & Assumptions

Given: A measurable real-valued function f with finite essential supremum s:=∥f∥∞.

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

1.2L1

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

2.1step 1.1L3algebra

Put E:=⋃n=1∞En. Then E is measurable and μ(E)≤∑n=1∞μ(En)=0. If x∉E, then ∣f(x)∣≤s+1/n for every n, hence ∣f(x)∣≤s. Therefore ∣f∣≤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∥∞ is the least essential bound. [step 2.1, step 1.2].

Depends on

Used by

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Sources