real analysis - 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?

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

1. meaningless, and


2. 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 $F$.