Lemma 1

Let U and V be two compact spaces, h a continuous mapping of U onto V. The mapping $\lambda \mapsto h(\lambda)$ of $\mathscr{M}_+(U)$ into $\mathscr{M}_+(V)$ is then surjective.

For, let a be the linear mapping $f \mapsto f \circ h$ of $\mathscr{C}(V)$ into $\mathscr{C}(U)$; since h is surjective, a is an isometry of $\mathscr{C}(V)$ onto a subspace H of $\mathscr{C}(U)$. Let $\theta$ be a positive measure on V; then $\theta \circ a^{-1}$ is a continuous linear form on H, which is extendible to a linear form $\eta$ on $\mathscr{C}(U)$ with the same norm, by virtue of the Hahn–Banach theorem (TVS, II, §3, No. 2, Cor. 3 of Th. 1); $\eta$ is then a measure on U, and $\theta(f) = \eta(f \circ h)$ for all $f \in \mathscr{C}(V)$, so that $\theta = h(\eta)$. Finally, $\theta(1) = \| \theta \| = \| \eta \|$, and $\theta(1) = \eta(1)$, so that $\eta$ is positive (Ch. V, §5, No. 5, Prop. 9).

Let us now prove the sufficiency of the condition stated in a). The condition implies the existence of a sequence $(K_n)_{n \geq 1}$ of compact subsets of T, such that the restriction of $\pi$ to each $K_n$ is continuous, and such that, for every $n$, $\nu^\bullet(X - \pi(K_n)) < 1/n$. The sequence $(K_n)$ can be assumed to be increasing. Set $L_n = \pi(K_n)$ and denote by $\nu'_n$ the measure $\varphi_{L_n - L_{n-1}} \cdot \nu_{L_n}$ on $L_n$, with the convention $L_0 = \varnothing$.

The restriction $\pi_{K_n}$ being continuous, there exists a measure $\mu'_n$ on $K_n$ such that $\pi_{K_n}(\mu'_n) = \nu'_n$ (Lemma 1). Let $\mu_n$ be the image of $\mu'_n$ under the canonical injection of $K_n$ into T, and let g be an element of $\mathscr{F}_+(X)$. Using successively the fact that $\nu$ is concentrated on $\bigcup L_n$, Prop. 4 of §1, No. 5; Prop. 2 of §1, No. 2; Prop. 4 of No. 3, and finally Prop. 7 of No. 3, we have

Taking $g = 1$ in this formula, one sees that the family $(\mu_n)$ is summable and that its sum is a bounded measure $\mu$ (§1, No. 7, Prop. 7). By Prop. 5 of No. 3, the mapping $\pi$ is $\mu_n$-measurable for all $n$, because $\pi_{K_n}$ is continuous, hence $\mu'_n$-measurable; it follows that $\pi$ is $\mu$-measurable (§1, No. 7, Prop. 8), hence $\mu$-proper since $\mu$ is bounded. The above relations then prove that the measures $\pi(\mu)$ and $\nu$ have the same essential upper integral, hence are equal (§1, No. 2, Cor. of Prop. 2).

Finally, let us assume that $\pi$ is injective, and let us prove b). Let f be an element of $\mathscr{F}_+(T)$; since $\pi$ is injective, there exists a function $g \in \mathscr{F}_+(X)$ such that $f = g \circ \pi$ and, setting $\nu = \pi(\mu) = \pi(\mu')$, by Prop. 4 of No. 3 we have

The two measures $\mu$ and $\mu'$ thus have the same essential upper integral, which implies their equality (§ 1, No. 2, Cor. of Prop. 2).