A measure $\lambda: B(\mathbb{R}^n) \rightarrow \overline{{\mathbb{R_{\ge 0}}}}$ that is associated with a monotone increasing and right-side continuous function $F$ is called a Lebesgue-Stieltjes measure. But I am wondering, why it is not true that every measure $\lambda: B(\mathbb{R}^n) \rightarrow \overline{{\mathbb{R_{\ge 0}}}}$ is a Lebesgue-Stieltjes measure?
[Math] so special about the Lebesgue-Stieltjes measure
lebesgue-integrallebesgue-measuremeasure-theoryreal-analysis
Related Solutions
I don't think my understanding is completely correct and should have posted as a comment, but it has length restrictions.
It seems to me that when $g$ is BV and right continuous, and $f$ is Borel measurable, $f$ does not have to be bounded. There are unbounded Lebesgue integrable functions, so the same should be true for Lebesgue-Stieltjes integral, which is just Lebesgue integral w.r.t. the signed measure $\mu_g$ on $\mathcal{B}(\mathbb{R})$ induced by $g$.
It should be OK for function $g$ with bounded variation. We can write $g$ as the difference of two non-decreasing functions.
I don't think Riemann-Stieltjes integral requires the integrator to be non-decreasing. It might be BV or possibly an even broader class of functions. I guess the author of Wikipedia entry mentions only nondecreasing functions because s/he has CDF of a random variable in mind and wants to discuss its application in probability theory. I also have the impression that these two integrals agree whenever the Riemann-Stieltjes integral exists. (Just like the relation between the Lebesgue integral and the Riemann integral.)
If these two notions agree, there's no danger of using the same notation. Otherwise, I've seen authors using prefix to distinguish different types of integrals, e.g., $(R)\int_a^b...$ for Riemann(-Stieltjes) integrals.
The general notion at work here is the completion of a measure.
$\newcommand{\R}{\mathbb{R}} \newcommand{\B}{\mathcal{B}}$ Let's write $\B$ for the Borel $\sigma$-algebra on $\R$. If $\mu$ is a positive Borel measure on $\R$ (i.e. a countably additive set function $\mu : \B \to [0,\infty]$), let $\B_\mu$ be the $\sigma$-algebra generated by $\B$ together with the sets $\{A \subset B : B \in \B, \mu(B) = 0\}$ (i.e. throw in all subsets of sets with measure zero). This is called "completing $\B$ with respect to $\mu$", and of course $\mu$ has a natural extension to $\B_\mu$. When we take $\mu$ to be Lebesgue measure $m$, $\B_m$ is precisely the Lebesgue $\sigma$-algebra.
In this notation, I think your questions are as follows:
Is $\B_m \subset \B_\mu$ for every $\mu$?
If not, is there a measure $\mu$ with $\B_\mu = \B$?
For 1, the answer is no. As you suspect, the Cantor measure $\mu_C$ is a counterexample. If $C$ is the Cantor set and $f : C\to [0,1]$ is the Cantor function, then we can write $\mu_C(B) = m(f(B))$. If $A \notin \B_m$ is a non-Lebesgue measurable set, then $f^{-1}(A) \notin \B_{\mu_C}$. But $f^{-1}(A) \subset C$ and $m(C) = 0$, so $f^{-1}(A) \in \B_m$.
For the second question, the answer is yes, sort of. One example is counting measure $\mu$ which assigns measure 1 to every point (hence measure $\infty$ to every infinite set). Here the only set of measure 0 is the empty set, which is already in $\B$, so $\B_\mu = \B$. Another example is a measure which assigns measure 0 to every countable (i.e. finite or countably infinite) set, and measure $\infty$ to every uncountable set. Now the measure zero sets are all countable, hence so are all their subsets, but all countable sets are already Borel. Note these are not Lebesgue-Stieltjes measures, because they give infinite measure to every nontrivial interval.
In fact, suppose $\mu$ is a measure such that $\B_\mu = \B$. Then I claim every uncountable Borel set $B$ has $\mu(B) = \infty$. Suppose $B$ is an uncountable Borel set. It is known that such $B$ must have a subset $A$ which is not Borel. If $\mu(B) = 0$, then $A \in \B_\mu \backslash \B$, which we want to avoid. So we have to have $\mu(B) > 0$. On the other hand, it is also known that an uncountable Borel set can be written as an uncountable disjoint union of uncountable Borel sets. Each of these must have nonzero measure, so this forces $\mu(B) = \infty$. In particular $\mu$ is not Lebesgue-Stieltjes.
So in fact, any Lebesgue-Stieltjes measure has measure-zero sets with non-Borel subsets, and hence can be properly extended by taking the completion.
A closely related idea is that of universally measurable sets, which are those sets $B$ which are in $\B_\mu$ for every finite (or, equivalently, every $\sigma$-finite) measure $\mu$. There do exist universally measurable sets which are not Borel. On the other hand, the above example with Cantor measure shows that there are Lebesgue measurable sets which are not universally measurable.
Best Answer
$\mu$ being a lebesgue-stiltjes measure with corresponding function $F$ implies that $$ \mu\left((a,b]\right) = F(b) - F(a) \text{.} $$
Now take the (rather silly) measure $$ \mu(X) = \begin{cases} \infty &\text{if $0 \in X$} \\ 1 &\text{if $0 \notin X$, $1 \in X$} \\ 0 &\text{otherwise.} \end{cases} $$
We'd need to have $F(x) = \infty$ for $x \geq 0$ and $F(x) = 0$ for $x < 0$ to have $\mu\left((a,b]\right) = F(b) - F(a)$ for $a < 0$, $b \geq 0$. But then $$ \mu\left((0,2]\right) = F(2) - F(0) = \infty - \infty $$ which is
Note that a measure doesn't necessarily need to have infinite point weights (i.e., $x$ for which $\mu(\{x\}) = \infty$ to cause trouble. Here's another measure on $B(\mathbb{R})$ which isn't a lebesgue-stiltjes measure $$ \mu(X) = \sum_{n \in \mathbb{N}, \frac{1}{n} \in X} \frac{1}{n} \text{.} $$ For every $\epsilon > 0$, $\mu\left((0,\epsilon]\right) = \infty$, which again would require $F(x) = \infty$ for $x > 0$, and again that conflicts the requirement that $\mu((a,b]) = F(b) - F(a) < \infty$ for $0 < a \leq b$. Note that this measure $\mu$ is even $\sigma$-finite! You can write $\mathbb{R}$ as the countable union $$ \mathbb{R} = \underbrace{(-\infty,0]}_{=A} \cup \underbrace{(1,\infty)}_{=B} \cup \bigcup_{n \in \mathbb{N}} \underbrace{(\tfrac{1}{n+1},\tfrac{1}{n}]}_{=C_n} $$ and all the sets have finite measure ($\mu(A)=\mu(B) = 0$, $\mu(C_n) = \frac{1}{n}$).
You do have that all finite (i.e., not just $\sigma$-finite, but fully finite) measures on $B(\mathbb{R})$ are lebesgue-stiltjes measures, however. This is important, for example, for probability theory, because it allows you to assume that every random variable on $\mathbb{R}$ has a cumulative distribution function (CDF), which is simply the function $F$.