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

  • 1
    $\begingroup$ @ChristianRemling : Since the question is, not about Lebesgue--Stieltjes measures, but about the corresponding outer measures, it seems to me not altogether trivial. $\endgroup$ – Iosif Pinelis Aug 1 '16 at 20:25

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

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