The common submeasure of two finite densities

Put $h=\min \{f,g\}$. It is measurable and nonnegative by [F5]. Thus $\eta (A)={\int }_{A}hd\lambda$ is a measure by [F3]. Since $h\le f,g$, monotonicity [F4] gives $\eta (A)\le \rho (A)$ and $\eta (A)\le \sigma (A)$ for every $A$. In particular $\eta (E)\le \rho (E)<\mathrm{\infty }$, so $\eta$ is finite.

Let $\theta$ be any measure with $\theta \le \rho$ and $\theta \le \sigma$. Then $\theta (E)\le \rho (E)<\mathrm{\infty }$, so $\theta$ is finite. If $\lambda (B)=0$, then $\rho (B)={\int }_{B}fd\lambda =0$; hence $\theta (B)\le \rho (B)=0$. Therefore $\theta \ll \lambda$. It is sigma-finite because it is finite [F1]. By [A1] and [F2], choose a finite-valued nonnegative measurable density $k$ with $\theta (A)={\int }_{A}kd\lambda$ for every $A$.

For each integer $n\ge 1$, set $E_n=\{k\ge f+1\mathrm{/}n\}$. This set is measurable by [F5]. Monotonicity, additivity, and homogeneity [F4] give $\theta (E_n)={\int }_{E_n}kd\lambda \ge {\int }_{E_n}fd\lambda +\frac{\lambda (E_n)}{n}=\rho (E_n)+\frac{\lambda (E_n)}{n}\mathrm{.}$ But $\theta (E_n)\le \rho (E_n)<\mathrm{\infty }$, so $\lambda (E_n)=0$. The same argument with $g$ and $\sigma$ shows that each $G_n=\{k\ge g+1\mathrm{/}n\}$ is $\lambda$-null. Hence $N:={\bigcup }_{n\ge 1}(E_n\cup G_n)$ is $\lambda$-null, and $k\le f$ and $k\le g$, thus $k\le h$, on $E\setminus N$.

Absolute continuity gives $\theta (A)=\theta (A\setminus N)$ for every $A$, and [F6] gives $\eta (A)=\eta (A\setminus N)$. On $A\setminus N$, $k\le h$, so monotonicity [F4] yields $\theta (A)={\int }_{A\setminus N}kd\lambda \le {\int }_{A\setminus N}hd\lambda =\eta (A)\mathrm{.}$ Thus every common lower measure is dominated by $\eta$.

Step 1.1 shows that $\eta$ is dominated by both $\rho$ and $\sigma$; step 3.1 shows every measure dominated by both is itself dominated by $\eta$. Therefore $\eta$ is the greatest common submeasure.