How can we prove that Lebesgue-Stieltjes measures are regular Borel measures?

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I need to show that any lebesgue stieltjes measure is a regular borel measure. I'm really clueless , can anyone help??
We know the definition and facts about the distribution function , how can we conclude approximation by compact or closed sets??
Regards, hermanni.
 
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First, can you give your definition of Lebesque-Stieltjes measure??

Let \mu be the Lebesque-Stieltjes measure, we need to show that for every Borel set A holds that

\mu(A)=\inf\{\mu(G)~\vert~G~\text{open and}~A\subseteq G\}

and something analogous in the case of closed sets.
Can you prove this for A=]a,b]? Can you find a sequence of closed sets G_n containing ]a,b] such that \lim_{n\rightarrow +\infty}{\mu(G_n)}=\mu(]a,b])??


Of you've proven this first case, then you'll need to use that the intervals of the form ]a,b] form a semiring (i.e. apply an approximation theorem.)
 
Ok , here's our course's definiton : Let F be a right-continuos and nondecreasing function.
Then lebesgue - stieltjes measure associated to F is u and:
u(a, b] = F(b) - F(a)
For the compact sets , we do approximation from inside .The thing that bothers me is extension from a semiring to the ring.Any way , I'll try your suggestions , thanks :))
 
Hi,
I showed approximations for intervals. Can you give me an idea how I can show it for any set??
Regards, hermanni.
 
You'll need an approximation theorem.
Have you seen the following?

If \mathcal{A} is a semiring and if A is a Borel set. Then there exist A_1,...,A_n such that

A\subseteq \bigcup A_i~\text{and}~\mu\left(\bigcup{A_i}\setminus A\right)<\epsilon

Or did you see any other theorem that looks like it?
 
Actually no , in the course we only saw that if we have a premasure on a semiring , then we can extend it to a measure on the ring.
Also we noted down

If E \in S and F \in S then there exists a finite number of mutually disjoint sets C_i \in S for i=1,\ldots,n such that E \setminus F = \cup_{i=1}^n C_i without proof , it looks like what you said.Can you explain how the result will follow from your lemma? We also did something similar in the class at characterization of the measurable sets: If A is any lebesgue measurable set , then what you said follows and Ai's are open sets.
 
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