Corollary 4

— Let S be the support of $\mu$. If $\pi$ is continuous, then the support of $\pi(\mu)$ is $\pi(S)$.

For, it follows from Cor. 3 that $\pi(\mu)$ is concentrated on $\pi(S)$, therefore if $S'$ is the support of $\pi(\mu)$ then $S' \subset \pi(S)$. On the other hand, $\pi^{-1}(X - S')$ is a locally $\mu$-negligible open set (Cor. 2), hence is $\mu$-negligible (Ch. IV, §5, No. 2, Cor. 2 of Prop. 5). Therefore $\pi^{-1}(X - S') \subset T - S$, consequently $\pi(S) \subset S'$, which proves the corollary.