Lebesgue–Stieltjes Measures and Atoms

Setup.

Let $G:\mathbb{R}\to \mathbb{R}$ be a non-decreasing and right-continuous function. The Lebesgue–Stieltjes measure ${\mu }_G$ on the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ is defined by ${\mu }_G((a,b]):=G(b)-G(a)$ for any interval $(a,b]$. For any $x\in \mathbb{R}$, we define the left-hand limit and the jump size of $G$ at $x$ as follows:

$$G(x-):=\underset{t<x}{\sup }G(t)=\underset{t\uparrow x}{\lim }G(t),{\Delta }_G(x):=G(x)-G(x-)$$

Since $G$ is non-decreasing, ${\Delta }_G(x)\ge 0$.

Theorem (Atoms and Continuity in Lebesgue–Stieltjes Measures).

Let $G$ and ${\mu }_G$ be as defined above.

1. The function $G$ is discontinuous at a point $x\in \mathbb{R}$ if and only if ${\Delta }_G(x)>0$.
2. For any $x\in \mathbb{R}$, the measure of the singleton set $\{x\}$ is equal to the jump size of $G$ at $x$, i.e., ${\mu }_G(\{x\})={\Delta }_G(x)$.
3. The measure ${\mu }_G$ has no atoms (i.e., ${\mu }_G(\{x\})=0$ for all $x\in \mathbb{R}$) if and only if the function $G$ is continuous on $\mathbb{R}$.

Proof.

(1) Since $G$ is non-decreasing, the left-hand limit $G(x-)={\lim }_{t\uparrow x}G(t)$ exists and satisfies $G(x-)\le G(x)$. By hypothesis, $G$ is right-continuous, so ${\lim }_{t\downarrow x}G(t)=G(x)$. A discontinuity at $x$ occurs if and only if the left and right limits are unequal, which for a right-continuous function means $G(x-)\ne G(x)$. Monotonicity implies this is equivalent to $G(x)>G(x-)$. This chain of equivalences establishes the result: $G$ is discontinuous at $x\iff G(x-)\ne G(x)\iff G(x)>G(x-)\iff {\Delta }_G(x)>0$.

(2) For any $x\in \mathbb{R}$, the singleton set $\{x\}$ can be expressed as the intersection of a decreasing sequence of intervals: $\{x\}={\bigcap }_{n=1}^{\infty }(x-1/n,x]$. Since $G$ is real-valued, ${\mu }_G((x-1,x])=G(x)-G(x-1)$ is finite. By the continuity from above property of measures, we have:

$$\begin{array}{rl}{\mu }_G(\{x\})& ={\mu }_G\left(\bigcap _{n=1}^{\infty }(x-1/n,x]\right)\\ & =\underset{n\to \infty }{\lim }{\mu }_G((x-1/n,x])\\ & =\underset{n\to \infty }{\lim }(G(x)-G(x-1/n))\\ & =G(x)-\underset{n\to \infty }{\lim }G(x-1/n)\\ & =G(x)-G(x-)={\Delta }_G(x).\end{array}$$

(3) The measure ${\mu }_G$ has no atoms if and only if ${\mu }_G(\{x\})=0$ for all $x\in \mathbb{R}$. By part (2), this is equivalent to ${\Delta }_G(x)=0$ for all $x\in \mathbb{R}$. By part (1), this is equivalent to $G$ being continuous at every $x\in \mathbb{R}$. $\square$