Corollary

— For a subset A of X to be essentially $\nu$-integrable, it is necessary and sufficient that $\pi^{-1}(A)$ be essentially $\mu$-integrable, in which case $\nu(A) = \mu(\pi^{-1}(A))$.

In particular, for every compact set $K \subset X$, $\nu(K) = \mu(\pi^{-1}(K))$. It follows from this and Cor. 3 of Prop. 2 that if $\mu$ is atomic (§ 5, No. 10) then so is $\pi(\mu) = \nu$. For, let M be the set of $t \in T$ such that $\mu(\{t\}) \neq 0$; since $\mu$ is carried by M, $\nu$ is carried by $\pi(M)$; moreover, for every $x \in \pi(M)$ we have $\nu(\{x\}) = \mu(\pi^{-1}(x)) > 0$, since $\pi^{-1}(x)$ contains at least one point of M. Therefore $\nu$ is atomic (§ 5, No. 10, Prop. 15).