Proposition 3

— Let f be a mapping of T into a topological space F. If f is $\mu$-measurable, then the mapping $(t, t') \mapsto f(t)$ is $\nu$-measurable. Conversely, if $\mu' \neq 0$, and if this mapping is $\nu$-measurable, then the function f is $\mu$-measurable.

The first assertion follows from Cor. 1 of Prop. 10 of § 6, No. 6. Suppose that $\mu' \neq 0$, denote by $\mu'_1$ a nonzero measure with compact support that is bounded above by $\mu'$, by $\nu_1$ the measure $\mu \otimes \mu'_1$, and set $a = \| \mu'_1 \|$. The projection $\mathrm{pr}_1$ of $T \times T'$ onto T is then $\nu_1$-proper, and the image measure $\mathrm{pr}_1(\nu_1)$ is equal to $a \mu$ (§ 6, No. 6, Prop. 10). If $(t, t') \mapsto f(t)$ is $\nu$-measurable, then it is also $\nu_1$-measurable, therefore $f$ is measurable with respect to the measure $a \mu$ (§ 6, No. 2, Prop. 3), whence the result since $a \neq 0$.

The preceding statement extends immediately to complex measures (Ch. III, § 4, No. 2, Prop. 3), as do the following corollaries.