Why is this function discontinuous exactly at $x .\in A$?

Question

I am studying Measure Theory using the book by Elstrodt and there is one thing I'm having trouble with in the section on contents on $\mathfrak{J} := \{]a,b]: a \leq b, a.b \in \mathbb{R} \}$. I'll first list a few things for reference.

Definition

Let $\mathfrak{H}$ be a semi-ring over the set $X$. A map $\mu: \mathfrak{H} \to \overline{\mathbb{R}}$ is called a content if it has the following properties:

(i) $\mu(\emptyset) = 0$

(ii) $\mu \geq 0$

(iii) For every finite sequence of disjoint sets $A_1,...,A_n \in \mathfrak{H}$ with $\bigcup \limits_{j=1}^{n} A_j \in \mathfrak{H}$:

$\mu \left(\bigcup \limits_{j=1}^{n} A_j \right) = \sum \limits_{j=1}^{n} \mu(A_j).$

If $\mu$ is defined on a ring $\mathfrak{R}$, then (iii) can be replaced by

(iii*) $\mu(A \cup B = \mu(A) + \mu(B)$ for all disjoint sets $A,B \in \mathfrak{R}.$

Theorem 1

Let $\mu: \mathfrak{R} \to \overline{\mathbb{R}}$ be a content on the ring $\mathfrak{R}$. Then:

a) $\mu$ is a premeasure

$\iff$ b) For every sequence $(A_n)_{n \geq 1}$ of sets in $\mathfrak{R}$ with $A_n \uparrow A \in \mathfrak{R}$, we have $\mu(A_n) \uparrow \mu(A)$.

$\implies$ c) For every sequence $(A_n)_{n \geq 1}$ of sets in $\mathfrak{R}$ with $A_n \downarrow A$ and $\mu(A_1) < \infty$, we have $\mu(A_n) \downarrow \mu(A)$.

$\iff$ d) For every sequence $(A_n)_{n \geq 1}$ of sets in $\mathfrak{R}$ with $A_n \downarrow \emptyset$ and $\mu(A_1) < \infty$, we have $\mu(A_n) \downarrow 0$.

If $\mu$ is finite, then a)-d) are equivalent.

Note: Since every measure on a semi-ring $\mathfrak{H}$ can be uniquely extended to the ring generated by $\mathfrak{H}$, we can also use this Theorem for semi-rings as long as all the sets involved are in $\mathfrak{H}$.

Theorem 2

a) If $F: \mathbb{R} \to \mathbb{R}$ is an increasing function, then $\mu_F: \mathfrak{I} \to \mathbb{R}, \mu_F(]a,b]) := F(b) - F(a) \quad (a \leq b)$ is a finitie content. For two increasing functions $F,G: \mathbb{R} \to \mathbb{R}$, $\mu_F = \mu_G$ if and only if $F - G$ is constant.

b) If $\mu: \mathfrak{I} \to \mathbb{R}$ is a finite content, then $F: \mathbb{R} \to \mathbb{R}$ defined by

$$F(x) := \begin{cases} \mu(]0,x]), & \text{for } x \geq 0\\ -\mu(]x,0]), & \text{for } x < 0 \end{cases},$$

is increasing and $\mu = \mu_F$.

$\mu_F$ is called the Stieltjes content corresponding to $F$.

Theorem 3

Let $F: \mathbb{R} \to \mathbb{R}$ be increasing and $\mu_F: \mathfrak{I} \to \mathbb{R}$ be the Stieltjes content corresponding to $F$. Then $\mu_F$ is a pre-measure if and only if $F$ is right-continuous.

Now here comes the part I am struggling with:

Every increasing, continuous function defines a finite pre-measure $\mu_F: \mathfrak{I} \to \mathbb{R}$. Such a pre-measure can be seen as a continuous mass distribution on $\mathbb{R}$. However, there are also completely different pre-measures:

Let $A \subset \mathbb{R}$ be a countable set and $p: A \to \mathbb{R}$ a strictly positive function such that

$$\sum \limits_{y \in A \cap [-n,n]} p(y)$$

converges for all $n \in \mathbb{N}$.

Then

$$\mu(]a,b]) := \sum \limits_{y \in A \cap ]a,b]} p(y) \quad (a \leq b)$$

is a finite pre-measure on $\mathfrak{I}$, and so

$$G(x) := \begin{cases} \sum \limits_{y \in A \cap ]0,x]} p(y), & \text{for } x \geq 0\\ -\sum \limits_{y \in A \cap ]x,0]} p(y), & \text{for } x < 0 \end{cases}$$

is right-continuous. Using Theorem 1 it can be shown that $G$ is discontinuous exactly at the points in $A$.

I don't see how to prove the statement in bold. Since $G$ is right-continuous, I have to show that $A$ is not left-continuous exactly at the points in $A$. Intuitively, $G$ jumps by $p(y)$ at $y \in A$, so the natural way to prove this that came to my mind would be (for $y \geq 0$)

\begin{align*} \lim \limits_{\substack{h \to 0 \\ h>0}} (G(y) - G(y-h)) & = \lim \limits_{\substack{h \to 0 \\ h>0}} \mu(]0,y]) - \mu(]0,y-h]) \\ & = \lim \limits_{\substack{h \to 0 \\ h>0}} \mu(]y-h,y]) \\ & = \mu(\{y\}) \\ & = p(y) \end{align*}

for $y \in A$.

But the set $\{y\}$ is not in $\mathfrak{J}$, so I cannot use Theorem 1 this way.

Can anybody explain how this follows from Theorem 1 or give me a hint?

Thanks a lot!

Answer 1

Let us prove that $G$ is discontinuous exactly at the points in $A$.

Remark: As discussed in the comments, there is no direct way to apply Theorem $1$, because, for instance, the sets $\{y\}$ (singletons) are not in $\mathfrak{J}$.

Proof: Given any $x, y \in \Bbb R$ such that $x<y$. Then, if $0 \leqslant x <y$ , then $$ G(y) - G(x)= \mu(]0,y]) - \mu(]0,x] = \mu(]x,y]) $$ If $x < 0 \leqslant y$, then $$ G(y) - G(x)= \mu(]0,y]) - (- \mu(]x,0]) = \mu(]x,y]) $$ If $x < y <0$ then $$ G(y) - G(x)= (-\mu(]y,0])) - (- \mu(]x,0]) = \mu(]x,y]) $$ So, in the three cases, we have
$$ G(y) - G(x)= \mu(]x,y]) $$ Now, given any $y\in \Bbb R$, if $y \in A$, let $\{x_n\}_n$ be a sequence such that, for all $n$, $x_n < y$ and $x_n \to y$. Then $$ \lim_n (G(y) - G(x_n)) = \lim_n \mu(]x_n,y]) \geqslant p(y) > 0 $$ So, $G$ is not continuous at $y \in A$.

Now, if $y \notin A$, then take $N \in \Bbb N$ such that $N> |y|$. Let $\{a_k\}_k$ be an enumeration of $A \cap [-N,N]$. Since $\sum \limits_{y \in A \cap [-N,N]} p(y)$ converges, for all $\varepsilon > 0$, there is $K \in \Bbb N$ such that $\sum_{k=K+1}^{\infty} p(a_k)< \varepsilon$. Let us define $$ \delta = \frac{1}{2} \min ( \{y-(-N)\} \cup\{y - a_k: k \in \{0,\dots, K\} \text{ and } a_k < y\} ) $$ Note that $\delta >0$, and, for any $x \leqslant y$, if $y-x <\delta$ we have $$ G(y) - G(x) = \mu(]x,y]) \leqslant \mu(]y-\delta, y]) \leqslant \sum_{k=K+1}^{\infty} p(a_k)< \varepsilon $$ So, for all $\varepsilon >0$, there is $\delta>0$ such that if $0 \leqslant y-x <\delta$, then $G(y) - G(x)< \varepsilon$. Since we already know tha $G$ is a non-decreasing right-continuous function, we have that $G$ is continuous at points not in $A$.