The common submeasure and its bounded density

The measure $\lambda =\rho +\sigma$ is finite, and $\rho ,\sigma \ll \lambda$. By [A1], [F1]–[F3], choose finite-valued nonnegative measurable representatives $f=d\rho \mathrm{/}d\lambda$ and $g=d\sigma \mathrm{/}d\lambda$. For every $B\in \mathcal{F}$, ${\int }_B(f+g)d\lambda =\rho (B)+\sigma (B)=\lambda (B)={\int }_{B}1d\lambda \mathrm{.}$ Uniqueness in [F2] gives $f+g=1$ $\lambda$-almost everywhere. On the measurable null set where this fails, replace $f$ by $0$ and $g$ by $1$; the modified functions remain representatives and now satisfy $f+g=1$ everywhere.

The function $h=\min \{f,g\}$ is measurable and nonnegative by [F7]. By [F4], $\eta (A)={\int }_{A}hd\lambda$ defines a measure. Since $h\le f,g$, monotonicity in [F5] and the density identities in step 1.1 give $\eta (A)\le \rho (A)$ 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 $\theta \le \sigma$. It is finite because $\theta (E)\le \rho (E)<\mathrm{\infty }$, and $\theta \ll \lambda$: if $\lambda (B)=0$, nonnegativity gives $\rho (B)=0$, and hence $\theta (B)=0$. By [A1], [F1]–[F3], choose a finite-valued nonnegative density $k=d\theta \mathrm{/}d\lambda$. For each integer $n\ge 1$, the measurable set $E_n=\{k\ge f+1\mathrm{/}n\}$ satisfies $\theta (E_n)={\int }_{E_n}kd\lambda \ge {\int }_{E_n}fd\lambda +\lambda (E_n)\mathrm{/}n=\rho (E_n)+\lambda (E_n)\mathrm{/}n\mathrm{.}$ Since $\theta (E_n)\le \rho (E_n)<\mathrm{\infty }$, this forces $\lambda (E_n)=0$. Thus $\{k>f\}={\bigcup }_{n\ge 1}{E}_n$ is null; the same argument using $\theta \le \sigma$ gives $k\le g$ almost everywhere. Therefore $k\le h$ outside a measurable $\lambda$-null set. Integrals on that null set vanish by simple approximation and [F5]–[F6], so monotonicity yields $\theta (A)\le \eta (A)$ for all $A$. Together with step 2.1, this proves that $\eta$ is the greatest common submeasure.

For each real $c<0$, $\{a>c\}=E$; for $0\le c<1$, $\{a>c\}=\{f=0\}\cup (\{f>0\}\cap \{g-cf>0\});$ and for $c\ge 1$ the set is empty. By [F7], each such set is measurable, so $a$ is measurable. It lies in $[0,1]$, and $af=\min \{f,g\}$ everywhere: on $\{f>0\}$ this follows by multiplying the ratio by $f$, while on $\{f=0\}$ both sides are zero. For a fixed $A$, approximate $a{\mathbf{1}}_A$ increasingly by simple functions. [F6] gives equality of their integrals against $\rho$ and against $f\lambda$; applying [F5] to both limits gives ${\int }_{A}ad\rho ={\int }_{A}afd\lambda ={\int }_A\min \{f,g\}d\lambda =\eta (A)\mathrm{.}$ Thus $a$ is a nonnegative measurable representative of $d\eta \mathrm{/}d\rho$ with values in $[0,1]$.