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Existence and uniqueness of the equilibrium measure

Statement

Assume the Axiom of Choice. Let K⊆C be nonempty and compact with cap⁡(K)>0. Then there is exactly one Borel probability measure μK on K with

I(μK)=VK=inf⁡μ∈P(K)I(μ)<+∞.

The measure μK is the equilibrium measure of K. If cap⁡(K)=0 then VK=+∞, every μ∈P(K) has I(μ)=+∞, and no equilibrium measure is asserted.

The Axiom of Choice is spent twice: through Countable Choice for the minimizing sequence, and through the weak sequential compactness of probability laws of Probability laws on a compact metric space have weakly convergent subsequences. The energy lower semicontinuity used below is choice-free (Lower semicontinuity of logarithmic potential and energy).

Facts & Assumptions

Given: a nonempty compact set K⊆C with cap⁡(K)>0, the probability measures P(K) on K, the Robin constant VK and the logarithmic energy I of Robin constant and logarithmic capacity of a compact set and Logarithmic potential and energy of a positive compactly supported measure, and the Axiom of Choice (The Axiom of Choice).

[F1]

P(K) is the set of Borel probability measures on K (Probability measures and probability spaces), each of compact support contained in K; VK=inf⁡μ∈P(K)I(μ)∈(−∞,+∞] and cap⁡(K)=exp⁡(−VK) when VK<+∞ and cap⁡(K)=0 when VK=+∞, so cap⁡(K)>0 is equivalent to VK<+∞ (Robin constant and logarithmic capacity of a compact set).

[F2]

For finite positive Borel measures μ,ν of compact support the mixed energy I(μ,ν)=∬k dμ dν∈(−∞,+∞] is symmetric, I(μ,ν)=I(ν,μ), and it is computed from the shifted nonnegative kernel kR=k+log⁡R with R>diam⁡(supp⁡μ∪supp⁡ν) by I(μ,ν)=∬kR dμ dν−μ(C)ν(C)log⁡R (Logarithmic potential and energy of a positive compactly supported measure).

[F3]

If μn,μ∈P(K) with μn⇒μ in the sense of Weak convergence of borel probability measures, then Uμ(z)≤lim inf⁡nUμn(z) for every z∈C and I(μ)≤lim inf⁡nI(μn) (Lower semicontinuity of logarithmic potential and energy).

[F4]

Assume the Axiom of Choice: every sequence in P(K) has a subsequence converging weakly to some element of P(K) (Probability laws on a compact metric space have weakly convergent subsequences).

[F5]

Assume Countable Choice, and let μ,ν be finite positive Borel measures on C with compact support, equal total mass and finite energy. Then I(μ,ν) is finite, I(μ−ν):=I(μ)−2I(μ,ν)+I(ν) is a real number, I(μ−ν)≥0, and I(μ−ν)=0 if and only if μ=ν (Strict positivity of logarithmic energy for a zero-mass signed charge).

[F6]

The Axiom of Choice implies Dependent Choice, which implies Countable Choice (AC implies DC implies countable choice, The Axiom of Countable Choice (ACω)).

[F7]

If S⊆R is nonempty and bounded below with infimum L, then for every ε>0 there is s∈S with s<L+ε (Epsilon characterisation of the infimum).

[F8]

Finite nonnegative weighted sums of measures are measures, and a convex combination (1−t)μ+tν with 0≤t≤1 of probability measures is again a probability measure (Nonnegative scalar multiples and countable weighted sums of measures are measures, Probability measures and probability spaces).

Proof

technique · direct
1.1F1F7given

Since cap⁡(K)>0, [F1] and the finite-energy construction in Robin constant and logarithmic capacity of a compact set give the nonempty real set EK:={I(μ):μ∈P(K), I(μ)<+∞} with inf⁡EK=VK∈R. It is bounded below by −log⁡diam⁡K. The infinite energies do not alter its real lower bounds. Applying [F7] to EK, for each n≥1 there is a finite energy below VK+1/n, so {μ∈P(K):I(μ)<VK+1/n} is nonempty.

2.1step 1.1F6

By [F6] the Axiom of Choice yields Countable Choice, so there is a sequence (μn)n≥1 in P(K) with I(μn)<VK+1/n for every n.

3.1step 2.1F4

By [F4], which assumes the Axiom of Choice of the hypothesis, the sequence (μn) has a subsequence (μnk)k≥1 and a limit μ∈P(K) with μnk⇒μ.

4.1step 3.1F3F1

Applying [F3] to the weakly convergent subsequence of step 3.1 and using step 2.1 along it gives I(μ)≤lim inf⁡kI(μnk)≤lim inf⁡k(VK+1/nk)=VK, while VK≤I(μ) holds because VK is an infimum over P(K)∋μ; hence I(μ)=VK<+∞.

5.1step 4.1F2F5F8given

For uniqueness let μ,ν∈P(K) satisfy I(μ)=I(ν)=VK; both have compact support in K, finite energy and total mass 1, so [F5] applies to the pair and the average σ:=12μ+12ν lies in P(K) by [F8], whence I(σ)≥VK. Expanding the double integral of σ⊗σ=14μ⊗μ+14μ⊗ν+14ν⊗μ+14ν⊗ν and using I(μ,ν)=I(ν,μ) from [F2] gives the finite value I(σ)=14I(μ)+12I(μ,ν)+14I(ν); comparing with the companion expansion I(μ−ν)=I(μ)−2I(μ,ν)+I(ν) of the same bilinear form from [F5], this says I(12(μ+ν))=12I(μ)+12I(ν)−14I(μ−ν). Substituting I(μ)=I(ν)=VK yields VK≤12VK+12I(μ,ν), that is, I(μ,ν)≥VK.

6.1step 5.1F5

With σ as in step 5.1 the signed measure μ−ν also meets the hypotheses of [F5], so I(μ−ν)=I(μ)−2I(μ,ν)+I(ν)≤VK−2VK+VK=0 by step 5.1, while [F5] gives I(μ−ν)≥0; hence I(μ−ν)=0 and [F5] gives μ−ν=0, that is, μ=ν, so the minimizer of step 4.1 is the only minimizer and is the stated equilibrium measure μK.

7.1step 4.1step 6.1F1∎

The zero-capacity case is [F1] verbatim: if cap⁡(K)=0 then VK=+∞, so an element μ∈P(K) with I(μ)=VK would have I(μ)=+∞ by definition of the extended infimum, and the statement asserts nothing about the existence of such a μ, which completes the proof.

Remarks

The equilibrium measure is a probability on the conductor. The minimizer μK of the theorem is carried by K, since P(K) consists of the probability measures on K; this is used by every later item that integrates against μK over K.

Uniqueness is strict convexity of the energy. The proof shows more than the statement needs: any two finite-energy probabilities of equal mass on a common compact carrier satisfy I(12(μ+ν))=14I(μ)+12I(μ,ν)+14I(ν) and μ≠ν forces I(μ−ν)>0 by Strict positivity of logarithmic energy for a zero-mass signed charge.

Where the two choice uses sit. Countable Choice selects one measure per level n in step 2.1; the Axiom of Choice itself is the hypothesis of the weak-compactness statement [F4] used in step 3.1. The lower semicontinuity [F3] and the infimum characterization [F7] are choice-free.

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