Proposition 15

For every scalar $\alpha > 0$ and every pair of positive measures $\mu, \nu$ on $X$,

Moreover, the relation $\mu \leq \nu$ implies $\mu^* \leq \nu^*$.

Let us prove the relation (15). Set $\lambda = \mu + \nu$; thus $\lambda(f) = \mu(f) + \nu(f)$ for $f \in \mathscr{K}_+$; for $f \in \mathscr{J}_+$, the value of $\lambda^*(f)$ (resp. $\mu^*(f)$, $\nu^*(f)$) is the limit of $\lambda(g)$ (resp. $\mu(g)$, $\nu(g)$) as $g$ runs over the directed set (for $\leq$) of all $g \in \mathscr{K}_+$ such that $g \leq f$; therefore $\lambda^*(f) = \mu^*(f) + \nu^*(f)$. Finally, if $f$ is any function $\geq 0$ defined on $X$, then $\lambda^*(f)$ (resp. $\mu^*(f)$, $\nu^*(f)$) is the limit of $\lambda^*(h)$ (resp. $\mu^*(h)$, $\nu^*(h)$) as $h$ runs over the directed set (for $\geq$) of all functions $h \in \mathscr{J}_+$ such that $h \geq f$; again, by passage to the limit, we therefore have $\lambda^*(f) = \mu^*(f) + \nu^*(f)$, which proves (15). The relation (14) is established similarly. Finally, if $\mu \leq \nu$ one can write $\nu = \mu + (\nu - \mu)$, where $\nu - \mu \geq 0$, therefore $\nu^* = \mu^* + (\nu - \mu)^*$, which shows that $\mu^* \leq \nu^*$.