Proposition 5

Let $\pi$ be a $\mu$-proper mapping of T into a topological space X, and let f be a mapping of X into a topological space F (Hausdorff or not). For f to be $\pi(\mu)$-measurable, it is necessary and sufficient that $f \circ \pi$ be $\mu$-measurable.

Let us take up again the proof of Prop. 4, and commence with the special case treated at the beginning, with the same notations; g is measurable for the measure $\pi(\mu) = \nu$ if and only if $g_L$ is $\nu'$-measurable (§1, No. 5, Example); now, this is equivalent to saying that $g_L \circ \pi' = (g \circ \pi)_K$ is $\mu_K$-measurable (Ch. V, §6, No. 2, Prop. 3), and finally that $g \circ \pi$ is $\mu$-measurable (§1, No. 5, Example). Let us now pass to the general case, with the same notations as in the proof of Prop. 4; f is $\nu$-measurable if and only if f is $\nu_\alpha$-measurable for every $\alpha \in A$ (§1, No. 7, Prop. 8), hence if and only if $f \circ \pi$ is $\mu_\alpha$-measurable for every $\alpha \in A$ (the preceding special case) and finally if and only if $f \circ \pi$ is $\mu$-measurable ($§ 1$, No. 7, Prop. 8).