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From http://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration:

The Lebesgue–Stieltjes measure is a regular Borel measure, and conversely every regular Borel measure on the real line is of this kind.

As far as I know, a Lebesgue-Stieltjes measure is a regular Borel measure. But I can't find any proof for the reversed side. Could anyone help me?

Martin Sleziak
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Du Phan
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  • What about $\mu(\emptyset) := 0$, $\mu(A) := \infty$ for all $A \in \mathcal{B}(\mathbb{R}),, A \neq \emptyset$? – Thomas Apr 17 '13 at 09:52
  • @Thomas: your measure is not regular – Du Phan May 17 '13 at 10:34
  • Why not? If $A \subseteq \mathbb{R}$ is measurable, then either $A=\emptyset$, so it is already compact and open, or it contains a point $x$, then ${x} \subseteq A$ is compact and $\mu({x}) = \infty = \mu(A)$, and any open superset $O$ of $A$ satisfies $\mu(O) = \infty = \mu(A)$. – Thomas May 17 '13 at 14:58
  • @Thomas: I think that a regular Borel measure on the real line is outer regular and finite for every compact set (according to Folland's book, page 99). The definition of regular measure in link is not this kind of measure, as your example. Thank you for your simple but elegant example. – Du Phan May 18 '13 at 02:51
  • After re-reading Folland's book, I find the answer is exactly Theorem 1.16 of the book which mentions that a measure on $\mathbb{R}$ that is finite on all bounded Borel sets is a Lebesgue-Stieltjes measure (after completion). As Thomas mentioned, the statement in this question is not true when using the definition of a regular Borel measure on wikipedia. – Du Phan May 18 '13 at 09:45

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