In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational. ($M$ is a algebra).
Given a monotone function $\bar{\alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals. Let $\alpha(\bigcup_{j=1}^{l} [c_j,d_j) )=\sum_{j=1}^{l} (\bar{\alpha}(d_j)-\bar{\alpha}(c_j))$ finitely additive.
In $M$, let $\bar{\alpha}(x)=0$ if $x<1/2$ and $\bar{\alpha}(x)=1$ if $x\geq 1/2.$
Prove that the associated Lebesgue-Stieltjes measure $\mu$ has $\mu(A)=0$ if $1/2\not\in A$ and $\mu(A)=1$ if $1/2\in A$. For any Baire set $A.$
I have this: $\mu(A)=\int_{X} \mathcal{X}_{A}d(\bar{\alpha})=\lim_n \sum_{j=1}^{n-1} \mathcal{X}_{A}(x_j)(\bar{\alpha}(x_{j+1})-\bar{\alpha}(x_{j}))$
Why this sum is $0$ when $1/2\not \in A?$
In the partition $\bigcup_{j=1}^{n-1}[x_j,x_{j+1})$ of $[0,1]$, there is a specific $i$ with $\frac{1}{2}\in[x_i,x_{i+1})$. Consider $\frac{1}{2}\notin A$. Then there is a $N\in\mathbb{N}$, such that for all $n\ge N:$ $[x_i,x_{i+1})\subseteq [0,1]\backslash A$. This implies that $\chi_A(x_i)=0$, such that
$$\mu(A)=\lim_{n\rightarrow\infty} \sum_{j=1\\j\neq i}^{n-1} \mathcal{X}_{A}(x_j)(\bar{\alpha}(x_{j+1}-)-\bar{\alpha}(x_{j}-))$$
Then it follows for each $j\in\{1,\ldots,n-1\}$ with $ j\neq i$ that either $x_j,x_{j+1}\le\frac{1}{2}$ or $x_j,x_{j+1}>\frac{1}{2}$, such that
$$\bar{\alpha}(x_{j+1}-)-\bar{\alpha}(x_{j}-)=0,$$
such that $\mu(A)=0.$ Note, that you have to consider the left sided limits $\bar{\alpha}(x_{j+1}-)$ and $\bar{\alpha}(x_{j}-)$ as in the definition of Stieltjes integral.