Proposition 6

— Let $\pi$ be a $\mu$-proper mapping of T into a topological space X, and let B be a $\pi(\mu)$-measurable subset of X. Set $A = \overline{\pi^{-1}(B)}$, and denote by $\pi'$ the mapping of A into B that coincides with $\pi$ on A. The set A is then $\mu$-measurable, $\pi_A$ and $\pi'$ are $\mu_A$-proper, and

The set A is $\mu$-measurable by Prop. 5 applied to $\varphi_B$; the mapping $\pi_A$ is clearly $\mu_A$-measurable by the definition of induced measures (No. 1), and it follows that $\pi'$ is measurable. Let $f$ be an element of $\mathscr{F}_+(B)$; denoting by zero exponents the extensions by 0 in X and in T, we have

whence $(\pi(\mu)_B)^\bullet = \pi'(\mu_A^\bullet)$. Since the encumbrance $(\pi(\mu)_B)^\bullet$ is locally bounded, the same is true of $\pi'(\mu_A^\bullet)$ and therefore $\pi'$ is $\mu_A$-proper. The measures $\pi'(\mu_A)$ and $(\pi(\mu))_B$ have the same essential upper integral, hence are equal. The other relation may be established in an analogous manner.