关于满足紧有限性的ℝ上测度在勒贝格-斯蒂尔杰斯可测集上是否与对应L-S测度一致的技术问询

Great question—this cuts to the core of how measure extensions and completions behave, so let's break this down clearly:

直接结论 #

Yes, μ restricted to 𝒜' is exactly equal to μ*. Here's the step-by-step reasoning:

先回顾已知背景 #

First, let's recap the setup to align on definitions:

• You have a measure μ on some σ-algebra 𝒜 over ℝ, where μ is finite on all compact sets.
• The function α is defined using μ: α(x) = μ((0,x]) for x ≥ 0, and α(x) = -μ((x,0]) for x < 0.
• μ* is the Lebesgue-Stieltjes outer measure generated by α, which extends the interval measure ℓ((a,b]) = α(b)-α(a) to every subset of ℝ.
• 𝒜' is the collection of all μ*-measurable sets (via Carathéodory's criterion), which includes the Borel σ-algebra plus all subsets of μ*-null sets (this is the completion of the Borel σ-algebra with respect to μ*'s Borel restriction).
• The key condition here is that every μ*-measurable set is in 𝒜, so μ is defined on all of 𝒜'.

核心论证 #

1. μ and μ agree on Borel sets*: You already proved this using the Carathéodory extension theorem. The set function ℓ on half-open intervals (a,b] is a σ-finite pre-measure, and both μ and μ* extend ℓ to the Borel σ-algebra. Since the Carathéodory extension is unique for σ-finite pre-measures, their restrictions to the Borel σ-algebra must be identical.

2. μ assigns 0 to all μ-null sets*: Suppose N is a μ*-null set (so μ*(N) = 0). By definition of outer measure, for any ε > 0, we can cover N with a countable collection of half-open intervals {(aₙ, bₙ]} such that the sum of ℓ((aₙ, bₙ]) is less than ε. Since μ agrees with ℓ on these intervals, we have:

   μ(N) ≤ μ(∪(aₙ, bₙ]) ≤ Σμ((aₙ, bₙ]) = Σℓ((aₙ, bₙ]) < ε
   

   Since ε can be made arbitrarily small, μ(N) must be 0, which matches μ*(N).

3. μ and μ agree on all of 𝒜'*: Every set in 𝒜' can be written as either B ∪ N or B \ N, where B is a Borel set and N is a subset of a μ*-null set (this is a standard property of measure completions). For either case:

   • If A = B ∪ N (with B and N disjoint), then μ(A) = μ(B) + μ(N) = μ*(B) + μ*(N) = μ*(B ∪ N) = μ*(A) (using additivity of both measures, and the fact we already established agreement on Borel sets and null sets).
   • If A = B \ N, then μ(A) = μ(B) - μ(N) = μ*(B) - μ*(N) = μ*(B \ N) = μ*(A) (again, using agreement on B and N, and that μ*(N)=0 doesn't affect the measure of B).

为什么你的反例直觉不成立 #

You wondered if there could be a measure that agrees with μ* on Borel sets but diverges on non-Borel μ*-measurable sets. But the condition 𝒜' ⊆ 𝒜 forces μ to be defined on all those non-Borel sets, and our argument shows μ can't assign anything other than 0 to the null sets that distinguish these sets from Borel ones. There's no room for disagreement here—any deviation would break the subadditivity of μ or contradict the outer measure approximation.

总结 #

To wrap up: Since μ matches μ* on the Borel σ-algebra (via uniqueness of extension) and on all μ*-null sets (via outer measure covering), it must match μ* on every μ*-measurable set that's in its domain.

备注：内容来源于stack exchange，提问作者이희원