Seeking a property about Lebesgue-Stieltjes outer measure I am a graduate student and this is not something related to my work but I was just wondering and did not find an answer on the Internet. I asked this on the other math site two weeks ago and no one responded. Question:
Let $g$ and $h: R \rightarrow R$, where $R$ is the set of all real numbers, be two increasing functions and $\theta_g$, $\theta_h$ be the associated Lebesgue--Stieltjes outer measures on $R$. We can also associate to $g+h$ the Lebesgue--Stieltjes outer measure $\theta_{g+h}$. 
Is it possible to show that $\theta_g + \theta_h = \theta_{g+h}$?
Thank you so much 
 A: A more general question is as follows. Let $\mu$ and $\nu$ be two measures on a measurable space $(S,\Sigma)$. Does it then always follow that $(\mu+\nu)^*=\mu^*+\nu^*$, where ${}^*$ denotes the corresponding outer measure? 
The answer to this question is yes. Indeed, for any $E\subseteq S$, let $\mathcal{A}_E$ denote the set of all sequences $(A_j)$ of pairwise disjoint members of $\Sigma$ such that $\bigcup_j A_j\supseteq E$. Then 
\begin{equation}
\mu^*(E)+\nu^*(E)=\inf_{(A_j)\in\mathcal{A}_E}\sum_j\mu(A_j)
+\inf_{(A_j)\in\mathcal{A}_E}\sum_j\nu(A_j)
\end{equation}
\begin{equation}
\le\inf_{(A_j)\in\mathcal{A}_E}\Big(\sum_j\mu(A_j)+\sum_j\nu(A_j)\Big)
=\inf_{(A_j)\in\mathcal{A}_E}\sum_j(\mu+\nu)(A_j)
= (\mu+\nu)^*(E). 
\end{equation}
To prove the reverse inequality, $(\mu+\nu)^*(E)\le\mu^*(E)+\nu^*(E)$, take any real numbers $a$ and $b$ (if they exist) such that $\mu^*(E)<a$ and $\nu^*(E)<b$. Then for some sequences $(A_i)$ and $(B_j)$ in $\mathcal{A}_E$ one has $\sum_i\mu(A_i)<a$ and $\sum_j\mu(B_j)<b$. Let $C_{i,j}:=A_i\cap B_j$. Somehow enumerating the pairs $(i,j)$, we may assume that the double-sequence $(C_{i,j})$ is in $\mathcal{A}_E$. So, 
\begin{equation}
 (\mu+\nu)^*(E)\le\sum_{i,j}(\mu+\nu)(C_{i,j})
 =\sum_i\sum_j\mu(C_{i,j})+\sum_j\sum_i\nu(C_{i,j})
 \end{equation}
    \begin{equation}
 =\sum_i\mu(A_i)+\sum_j\nu(B_j)<a+b, 
\end{equation}
for any real $a$ and $b$ such that $\mu^*(E)<a$ and $\nu^*(E)<b$. 
Hence, $(\mu+\nu)^*(E)\le\mu^*(E)+\nu^*(E)$. QED
