What is so special about the Lebesgue-Stieltjes measure 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? 
 A: $\mu$ 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


*

*meaningless, and

*surely not the same as $1$, which is the actual measure of $(0,2]$.


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 $F$.
A: If you have a measure $\mu$ in $\mathbb{R}$ that is finite in compact sets then we define the function 
$$
  F(x) = \begin{cases}
     \mu((0,x])&\text{if $x > 0$} \\
    -\mu((x,0]) &\text{if $x < 0 $} \\
    0 &\text{if $x=0$}
  \end{cases}
$$
Then $F$ is increasing càdlag function and $\mu = \mu_{F}$.
This also shows that every measure in $\mathbb{R}$ which are finite in compact sets is regular.
