The citations are just for reference. The question here is not rooted in some details in this proof but in the intuition behind the Stieltjes measures.
(3.3.1) $\mathcal{A}$ is dyadic half-open algebra.
Definition (Stieltjes premeasures on $\mathcal{A}$ ) Let $F$ be an monotone nondecreasing function from $\mathbb{R}$ to $\mathbb{R}$ that is $\textbf{right continuous}$; i.e., for all $x \in \mathbb{R}$$$\begin{equation*}F(x)=\lim _{h \downarrow 0} F(x+h) \tag{3.3.4}\end{equation*}$$Let $F(-\infty)=\lim _{x \rightarrow-\infty} F(x)$ and $F(\infty)=\lim _{x \rightarrow \infty} F(x)$.For any set $A$ in $\mathcal{A}$ given by (3.3.1), define$$\begin{equation*}m_{F}(A)=\sum_{j=1}^{m}\left[F\left(b_{j}\right)-F\left(a_{j}\right)\right] \tag{3.3.5}\end{equation*}$$It is readily verified, using only the monotonicity of $F$, that $m_{F}$ is a premeasure on $\mathcal{A}$. The set of premeasures obtained in this way is the set of $\textit{Stieltjes premeasures}$.
Theorem 3.17 (Continuity at the empty set of Stieltjes premeasures). Let $F$ be a monotone non-decreasing and right continuous function from $\mathbb{R}$ to $\mathbb{R}$. Then Stieltjes premeasure $m_{F}$ on the dyadic interval algebra $\mathcal{A}$ in $\mathbb{R}$ is continuous at the empty set as well as semifinite, and thus extends to a countably additive measure $\mu$ that is defined on a $\mathcal{B}_{\mathbb{R}}=\sigma(\mathcal{A})$.
Proof. Suppose that $\left\{A_{j}\right\}$ is a decreasing sequence of sets in $\mathcal{A}$; i.e., $A_{j+1} \subset$$A_{j}$ for all $j$, and suppose that$$\begin{equation*}\lim _{j \rightarrow \infty} m_{F}\left(A_{j}\right)=c_{0} \neq 0 \tag{3.3.6}\end{equation*}$$and also $m\left(A_{1}\right)<\infty$. Then we must show that $\bigcap_{j=1}^{\infty} A_{j} \neq \emptyset$.If $F$ is such that $F(-\infty)=-\infty$ and $F(\infty)=\infty$, then since $m_{F}\left(A_{1}\right)<\infty$$A_{1}$ contains no infinite intervals, and hence is bounded. If both $F(\infty)$ and $F(-\infty)$ are finite, then then we can choose $k \in N$ such that $\left(F\left(-2^{k}\right)-\right.$$F(-\infty))+\left(F(\infty)-F\left(2^{k}\right)\right)<\frac{1}{2} c_{0}$. Then for all $j$,$$A_{j} \backslash\left(A_{j} \cap\left(-2^{k}, 2^{k}\right]\right) \subset\left(-\infty,-2^{k}\right] \cup\left(2^{k}, \infty\right]$$and hence $\left.m_{F}\left(A_{j} \cap\left(-2^{k}, 2^{k}\right]\right)\right) \geq \frac{1}{2} c_{0}$. Hence replacing each $A_{j}$ by $A_{j} \cap$$\left(-2^{k}, 2^{k}\right] \in \mathcal{A}$, we may assume that $A_{1}$ is bounded. The case in which only one of $F(-\infty)$ or $F(\infty)$ is finite is handled similarly.Thus, we may assume that for some $k \in \mathbb{N}$ and all $\ell$,$$\begin{equation*}A_{\ell} \subset A_{1} \subset I_{0}:=\left(-2^{k}, 2^{k}\right] \tag{3.3.7}\end{equation*}$$We shall produce, by a successive bisection procedure a nested sequence of half open dyadic intervals$$I_{0} \supset I_{1} \supset I_{2} \supset I_{3} \ldots$$in which $I_{j+1}$ is obtained by bisecting $I_{j}$ and keeping one of the two halves: Write $I_{0}$ as the union of its left and right halves$$I_{0}=J_{0}^{\text {left }} \cup J_{0}^{\text {right }}$$bisecting as above. Since $A_{\ell} \subset I_{0}$ for all $\ell$, we have from (3.3.6) that$$\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{0}^{\text {left }}\right)+\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{0}^{\text {right }}\right)=\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell}\right)=c_{0}>0$$If $\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{0}^{\text {right }}\right)>0$, define this number to be $c_{1}$, and chose $I_{1}=J_{0}^{\text {right }}$. Otherwise, $\lim _{k \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{0}^{\text {left }}\right)=c_{0}$, and we define $c_{1}=c_{0}$, and $I_{1}=J_{0}^{\text {left }}$. Either way, we have$$\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap I_{1}\right)=c_{1}>0$$We proceed inductively as follows: At the $n$th stage, with $I_{n}$ and $c_{n}>0$ defined, we write $I_{n}=J_{n}^{\text {left }} \cup J_{n}^{\text {right }}$, and note that$$\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{n}^{\text {left }}\right)+\lim _{k \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{n}^{\text {right }}\right)=\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap I_{n}\right)=c_{n}>0$$If $\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{n}^{\text {right }}\right)>0$, define this number to be $c_{n}$, and chose $I_{n+1}=$$J_{n}^{\text {right }}$. Otherwise, $\lim _{\ell \rightarrow \infty} m_{F}\left(A_{\ell} \cap J_{n}^{\text {left }}\right)=c_{n}$, and we define $c_{n+1}=c_{n}$, and $I_{n+1}=J_{n}^{\text {left }}$.In this way, we produce a nested sequences of intervals $\left\{I_{n}\right\}$ such that $m_{F}\left(A_{\ell} \cap I_{n}\right) \geq c_{n}>0$ for each $\ell$, and since $m_{F}(\emptyset)=0$, this means that $A_{\ell} \cap I_{n} \neq \emptyset$ for any $\ell$ and $n$. But then by the inclusion-exclusion lemma, there is an integer $n_{\ell}$ such that $I_{n} \subset A_{\ell}$ for all $n \geq n_{\ell}$.For each $n$, let $a_{n}, b_{n}$ be such that $I_{n}=\left(a_{n}, b_{n}\right]$. By construction, $\left\{b_{n}\right\}$ is a decreasing sequence that is bounded, and hence $b:=\lim _{n \rightarrow \infty}$ exists. Moreover, by construction, $\left\{a_{n}\right\}$ is an increasing sequence, and for any $m, n$, $b_{n} \geq a_{m}$. Hence if we can find an $n_{\ell}$ such that$$\begin{equation*}\left[a_{n_{\ell}+1}, b_{n_{\ell}+1}\right] \subset A_{\ell} \tag{3.3.8}\end{equation*}$$then $b \in A_{\ell}$, and since $\ell$ is arbitrary, $b \in \cap_{\ell=1}^{\infty} A_{\ell}$.To find an $n_{\ell}$ such that (3.3.8) is satisfied, we first show that the rule leads us to choose the right interval infinitely often. If this were not the case, then for some $n_{0}$, we would always choose the left interval in each later step. We would then have, for all $\ell$, and all $n>n_{0}$,$$\begin{equation*}m_{F}\left(A_{\ell} \cap I_{n}\right)=c_{n_{0}}>0 \tag{3.3.9}\end{equation*}$$and $I_{n}=\left(a_{n_{0}}, b_{n}\right]$. Therefore, for all sufficiently large $n$,$$m_{F}\left(A_{\ell} \cap I_{n}\right) \leq m_{F}\left(I_{n}\right)=F\left(b_{n}\right)-F\left(a_{n_{0}}\right)$$$\color{red}{\text{Since }F\text{ is right continuous, }\lim _{n \rightarrow \infty}\left(F\left(b_{n}\right)-F\left(a_{n_{0}}\right)\right)=0,\text{ which contradicts (3.3.9).}}$$\color{red}{\text{Therefore, the the right half gets picked infinitely often.}}$ Then for all sufficiently large $n$ (depending on $\ell),\left(a_{n}, b_{n}\right] \subset A_{\ell}$. Since the right half gets picked infinitely often, we can find an $n_{\ell}$ so that$$\left[a_{n_{\ell}+1}, b_{n_{\ell}+1}\right] \subset\left(a_{n_{\ell}}, b_{n_{\ell}}\right] \subset A_{\ell}$$so that (3.3.8) is satisfied.
To my understanding, the left portion of the bisection can't be picked infinitely many times is because picking left-hand side leads to a descending sequence converging to $a_{n_0}$ from right. Since $F$ is right continuous, the measure goes to zero, contradicting to the nonzero assumption.
My question is rather simple, what will happen if $F$ is just continuous? Does this proof fail? Or Stieltjes premeasures is by its nature not both sides continuous? If the latter is the case, then what's the intuition behind letting a measure be only one-sided continuous?