LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08

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Probability-density averages and locally detectable upper essential values

Statement

Assume AC (The Axiom of Choice). Let $G$ be an arbitrary LCH group with fixed left Haar measure $μ$ , and put $P = {f \in L^{1} (G) : f \geq 0, ∥ f ∥_{1} = 1}$ . For a real global- $L^{\infty}$ class $h$ , choose a bounded Borel representative and define its locally detectable upper essential value by $β (h) : = \inf {t \in R : μ (K \cap {h > t}) = 0 for every compact K \subseteq G} .$ This is a finite real number independent of that representative, and $\sup_{f \in P} \int_{G} f h d μ = β (h) .$ For complex $h \in L^{\infty} (G)$ and $α \in C$ , the exact support-function formula is $\sup_{f \in P} Re (α \int_{G} f h d μ) = β (Re (α h)) .$ For every real bounded continuous $h$ , one has $β (h) = \sup_{x \in G} h (x)$ . All $L^{\infty}$ classes here retain the global-null convention of Complex $L^{\infty}$ space of a locally compact group; locally null functions are not identified with zero. The original global-norm formula and its naive global-essential-upper-value repair fail as explained in Remarks.

Facts & Assumptions

Given: AC; an LCH group $G$ with fixed Haar measure; and a real or complex bounded Borel representative $h$ .

[F1]

A Borel superlevel set contains finite-positive mass exactly when some compact intersection detects positive mass (Finite Haar mass, compact detection, and integrable pairings).

[F2]

Global $L^{\infty}$ classes have Borel representatives, can be clipped on a global null set to be bounded, and have the essential-supremum norm (Complex $L^{\infty}$ space of a locally compact group, The essential supremum of a measurable function with respect to a measure).

[F3]

Under AC, $C_{c} (G; C)$ is dense in $L^{1} (G)$ (Completeness of the complex Haar L1 and L2 spaces and density of Cc).

[F4]

The integral is linear on $L^{1}$ , monotone on nonnegative functions and satisfies the integral triangle inequality; a nonnegative function has zero integral exactly when it is zero a.e. (The Lebesgue integral is linear on $L^{1} (μ)$ , Monotonicity and nonnegative homogeneity of the nonnegative integral, The modulus of an integral is bounded by the integral of the modulus, A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[F5]

Nonempty open sets have positive Haar measure, compact sets have finite measure, and relatively compact open neighborhoods exist in an LCH space (Haar measure is positive on nonempty open sets and finite on compact sets, In a locally compact Hausdorff space every open set containing a point contains an open set containing it whose closure is compact and still inside; such a space is regular).

[F6]

Countable unions of measurable null sets are null by countable subadditivity (Finite and countable subadditivity of measures).

Proof

technique · optimize over finite-detectable superlevels, using compactly supported approximants for the upper bound

1.1F2F5F6givenconstruct

Global a.e. changes alter every compact superlevel intersection only on a null set, so the defining thresholds and $β$ are unchanged. The admissible thresholds form an upward-closed nonempty set, bounded below because $h$ is bounded and some relatively compact nonempty open set has positive finite measure by [F5]. Thus $β$ is finite. Every $t > β$ is admissible; on each fixed compact $K$ , apply this to $t = β + 1 / n$ and [F6] to obtain $h \leq β$ almost everywhere on $K$ . Also a normalized indicator of any relatively compact nonempty open set shows $P \neq \emptyset$ .

2.1F3F4step 1.1choosealgebra

Fix $f \in P$ . By [F3] choose $u_{n} \in C_{c} (G)$ tending to $f$ in $L^{1}$ and put $v_{n} = ∣ u_{n} ∣$ . Then $v_{n} \in C_{c} (G)$ , $v_{n} \geq 0$ , and $∥ v_{n} - f ∥_{1} \leq ∥ u_{n} - f ∥_{1} \to 0$ . On the compact support of $v_{n}$ , step 1.1 gives $h \leq β$ a.e., so $\int v_{n} h \leq β \int v_{n}$ . Since $h$ is bounded, [F4] gives $∣ \int (v_{n} - f) h ∣ \leq ∥ h ∥_{\infty} ∥ v_{n} - f ∥_{1}$ and $\int v_{n} \to \int f = 1$ . Passing to the limit proves $\int f h \leq β$ .

3.1F1F4step 2.1constructalgebra

If $t < β$ , it is not admissible, so some compact $K$ gives $E = K \cap {h > t}$ of positive finite measure. By [F1], $f_{E} = μ (E)^{- 1} 1_{E} \in P$ , and [F4] makes $\int f_{E} (h - t) > 0$ because $h - t$ is strictly positive on $E$ . Thus $\int f_{E} h > t$ . Letting $t ↑ β$ and combining with step 2.1 proves the signed formula. For complex $h$ and $α$ , linearity gives $Re (α \int f h) = \int f Re (α h)$ , so the same signed formula proves the complex support-function identity.

4.1F1F5step 1.1step 3.1∎

Let $h$ be real bounded continuous and put $s = \sup_{G} h$ . Since $h \leq s$ everywhere, $β \leq s$ . For $t < s$ , the superlevel ${h > t}$ is nonempty open. Choose a relatively compact nonempty open neighborhood $V$ inside it; [F5] gives $0 < μ (V) < \infty$ , and its compact closure detects this superlevel. Hence $t$ is not admissible and $β \geq t$ . Letting $t ↑ s$ gives $β = s$ , completing all claims.

Remarks

On $Z$ with counting Haar measure, $h = - 1$ has global norm one but every probability average is $- 1$ , so the original signed formula with $∥ h ∥_{\infty}$ was false. Even replacing that norm by the global signed essential supremum is insufficient in arbitrary LCH generality. In the torus-times-uncountable-discrete example retained in Finite Haar mass, compact detection, and integrable pairings, $A = {1} \times D$ is locally null and globally infinite. Thus $h = 1_{A}$ has global norm and global upper essential value one, while the finite-detectability lemma gives $\int f h = 0$ for every $f \in P$ and $β (h) = 0$ . Both original counterexamples remain; the formula now identifies the exact support value rather than silently imposing semifiniteness or changing global-null classes.

Depends on

Used by

Probability-density approximation of continuous tests and topological means Lemma

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Sources

Bachir Bekka, Pierre de la Harpe and Alain Valette, Kazhdan's Property (T) (standard reference, not scraped)
Anne Thomas, The Banach-Tarski Paradox and Amenability, Lecture 20: Invariant Mean implies Reiter's Property (standard reference, not scraped)