Stieltjes Integral Let $\alpha$ be Stieltjes weight function (monotone increasing), and $\alpha = \alpha_1+\alpha_2$, where $\alpha_1,\alpha_2$ are also monotone increasing. suppose $f$ is Stieltjes-integrable on $[a,b]$
$$\int_a^bf d\alpha = \int_a^b f d\alpha_1 + \int_a^b f d\alpha_2$$
Under what conditions is this statement true? How can I prove it easily using basic knowledge of upper and lower Stieltjes integral? If it is not true all the time counter examples?
 A: $\newcommand\uint{\overline{\int_a^b}}\newcommand\lint{\underline{\int_a^b}}$
Notations
Suppose $\alpha$ is monotone increasing on $[a,b]$, $P=\{\,x_0,x_1,\dotsc,x_n\,\}$ is a partition of $[a,b]$, where $a=x_0\le x_1\le\dotsb\le x_n=b$, $f$ is a bounded function on $[a,b]$, we'll use notations here:
\begin{align}
&\Delta\alpha_k=\alpha(x_k)-\alpha(x_{k-1})\\\\
&M_k=\sup_{x_{k-1}\le x\le x_k}f(x)
&m_k=\inf_{x_{k-1}\le x\le x_k}f(x)\\
&U(P,f,\alpha)=\sum_{k=1}^nM_k\Delta\alpha_k
&L(P,f,\alpha)=\sum_{k=1}^nm_k\Delta\alpha_k\\
&\uint fd\alpha=\inf_P U(P,f,\alpha)
&\lint fd\alpha=\sup_P U(P,f,\alpha)
\end{align}
Stronger results
\begin{align}
\uint fd(\alpha_1+\alpha_2)&=\uint fd\alpha_1+\uint fd\alpha_2\\
\lint fd(\alpha_1+\alpha_2)&=\lint fd\alpha_1+\lint fd\alpha_2
\end{align}
Proof
We'll only prove the first one. For each partition $P$, it's easier to check that
$$U(P,f,\alpha_1+\alpha_2)=U(P,f,\alpha_1)+U(P,f,\alpha_2)$$
Take $\inf_P$ on both sides, we have
$$\uint fd(\alpha_1+\alpha_2)\ge\uint fd\alpha_1+\uint fd\alpha_2$$
Next, for each $u_1>\uint fd\alpha_1$, $u_2>\uint fd\alpha_2$, there's some partitons $P_1$ and $P_2$ such that $U(P_1,f,\alpha_1)<u_1$ and $U(P_2,f,\alpha_2)<u_2$.
Consider $P=P_1\cup P_2$, we have $U(P,f,\alpha_1)\le U(P_1,f,\alpha_1)<u_1$ and $U(P,f,\alpha_2)<u_2$, then
$$U(P,f,\alpha_1+\alpha_2)=U(P,f,\alpha_1)+U(P,f,\alpha_2)<u_1+u_2$$
therefore
$$\uint fd(\alpha_1+\alpha_2)<u_1+u_2$$
Hence
$$\uint fd(\alpha_1+\alpha_2)\le\uint fd\alpha_1+\uint fd\alpha_2$$
Proved.
