Corollary 3

— a) Let $(\lambda_\iota)_{\iota \in I}$ be a family of real measures on $X/H$. For the family $(\lambda_\iota)$ to be bounded above in $\mathscr{M}(X/H)$, it is necessary and sufficient that the family $(\lambda_\iota^\#)$ be bounded above in $\mathscr{M}(X)$, in which case

b) Let $\lambda$ be a real measure on $X/H$. Then $(\lambda^+)^\# = (\lambda^\#)^+$ and $(\lambda^-)^\# = (\lambda^\#)^-$.

c) Let $\lambda$ be a complex measure on $X/H$. Then $|\lambda|^\# = |\lambda^\#|$.

Assume the family $(\lambda_\iota)$ to be bounded above and let $\mu = \sup \lambda_\iota$. Since $\lambda \geq 0$ implies $\lambda^\# \geq 0$, we have $\mu^\# \geq \lambda_\iota^\#$ for all $\iota$, which shows that the family $(\lambda_\iota^\#)$ is bounded above and that

Conversely, assume the family $(\lambda_\iota^\#)$ to be bounded above and let $\nu = \sup(\lambda_\iota^\#)$. Since $\delta(\xi)\lambda_\iota^\# = \Delta_H(\xi)\lambda_\iota^\#$ for all $\xi \in H$, obviously $\delta(\xi)\nu = \Delta_H(\xi)\nu$, therefore there exists a measure $\mu' \in \mathscr{M}(X/H)$ such that $\nu = {\mu'}^\#$. Since $\lambda^\# \geq 0$ implies $\lambda \geq 0$, we have $\mu' \geq \lambda_\iota$ for all $\iota$, which shows that the family $(\lambda_\iota)$ is bounded above and that $\nu = {\mu'}^\# \geq (\sup \lambda_\iota)^\#$, whence

which completes the proof of a). The assertion b) then follows at once since, for example, $\lambda^+$ is none other than $\sup(\lambda, 0)$. To prove c), it suffices to note that $|\lambda| = \sup \mathscr{R}(\alpha \lambda)$ over the complex numbers $\alpha$ of absolute value 1, and on the other hand that $\mathscr{R}(\mu^\#) = (\mathscr{R}\mu)^\#$ for every $\mu \in \mathscr{M}(X/H)$.