Question with Lebesgue-Stieltges outer measure Let $g$ and $h$ be two increasing functions and $\theta_g$, $\theta_h$ be the associated Lebesgue-Stieltges outer measures on $R$ (the set of real numbers). We can also associate to $g+h$ the L-S outer measure $\theta_{g+h}$
Is it possible to show that $\theta_g + \theta_h = \theta_{g+h}$? There is one easy inequality to do $\theta_{g+h}$ is greater than $\theta_g+\theta_h$. 
Thank you so much 
 A: Let $A\subset \mathbb{R}$. As you mentioned, by definition:
$$
\theta_g(A) = \inf \{ \sum_{i=1}^{\infty} (g(b_i)-g(a_i)): A\subset \cup_{i=1}^{\infty} (a_i,b_i) \},
$$
$$
\theta_h(A) = \inf \{ \sum_{i=1}^{\infty} (h(b_i)-h(a_i)): A\subset \cup_{i=1}^{\infty} (a_i,b_i) \},
$$
We have $\theta_g(A)+\theta_h(A) \leq \theta_{g+h}(A)$. 
For the reverse inequality, let $\epsilon>0$. There exist   open subsets $U$, $V$ of $\mathbb{R}$ such that $A\subset U\cap V$ and 
$$
\theta_g(A)\leq \theta_g(U) < \theta_g(A)+\epsilon, $$
$$
\theta_h(A)\leq \theta_h(V) < \theta_h(A)+\epsilon. $$
Then by monotonicity of the measures, 
$$
\theta_g(A)\leq \theta_g(U\cap V) < \theta_g(A)+\epsilon, $$
$$
\theta_h(A)\leq \theta_h(U\cap V) < \theta_h(A)+\epsilon.$$
Since $\theta_{g+h}=\theta_g+\theta_h$ for open sets, in particular for $U\cap V$, we have
$$
\theta_g(A)+\theta_h(A)\leq \theta_{g+h}(U\cap V) < \theta_g(A)+\theta_h(A)+2\epsilon.$$
Then we have $\theta_{g+h}(A)\leq \theta_{g+h}(U\cap V) < \theta_g(A)+\theta_h(A)+2\epsilon.$
Since $\epsilon>0$ is arbitrary, we have
$$
\theta_{g+h}(A)\leq \theta_g(A)+\theta_h(A).
$$
