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I'm having difficulty convincing myself the Lebesgue-Stieltjes measure is indeed a measure. The Lebesgue-Stieltjes measure is defined as such:

Given a nondecreasing, right-continuous function $g$, let $\mathcal{H}_1$ denote the algebra of half-open intervals in $\mathbb{R}$. We define the Lebesgue-Stieltjes integral to be $\lambda: \mathcal{H}_1 \rightarrow[0,\infty]$, with $\lambda(I)=0$ if $I=\emptyset$, $ \lambda(I)=g(b)-g(a)$ if $I=(a,b]$, $-\infty\leq a < b < \infty$ and $\lambda(I)=g(\infty)-g(a)$ if $I=(a,\infty)$, $-\infty\leq a < \infty$.

Showing countable subadditivity is done by a careful application of the $\epsilon2^{-n}$ trick, and while I couldn't do this on my own, this can be found in most analysis textbooks. What about the other inequality to show $\sigma$-additivity? Does anyone know of a resource that proves this or could share how to do this? I suspect this requires quite a bit more trickery than the proof of subadditivity.

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  • $\begingroup$ en.m.wikipedia.org/wiki/… is relevant, any reference there works $\endgroup$
    – Evan
    Aug 16 '13 at 0:03
  • $\begingroup$ Folland's book does a good job constructing measures from pre-measures and the Caratheodory condition. The Lebesgue-Stieltjes measure then falls out as a special case. $\endgroup$ Aug 16 '13 at 8:44
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Firstly, note that the measure defined here is a Radon measure (that is $\lambda(B)<\infty$ for any bounded borel set $B$). hence it is also $\sigma$-finite (Because $\mathbb{R}=\bigcup_{n\in\mathbb{Z}}(n,n+1]$). So if I can only show that the measure $\lambda$ is $\sigma$-additive on the semifield $\{(a,b]:-\infty\leq a\leq b\leq\infty\}$ (showing this is trivial), then it would be so over $\mathcal{B}$ by Caratheodory Extension Theorem.

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To show superadditivity of the Lebesgue-Stieltjes premeasure, use finite additivity (on pairwise disjoint semiring elements) and monotonicity of the premeasure - take a limit (supremum) when bounded above. Otherwise, the result is trivial. Superadditivity is actually easier to verify than subadditivity.

Subadditivity is shown by a compactness argument. It is not that tricky, because you "follow your nose" with the semiring elements involved by "adding" and "subtracting" endpoints as needed to use compactness followed by right continuity.

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