Remark

If $f$ is a mapping of $X$ into a Banach space $F$, it comes to the same to say that the restriction of $f \circ \pi$ to $S$ is $\mu$-measurable or to say that the function $(f \circ \pi)g$ (defined on $T$) is $\mu$-measurable, since $g$ is $\mu$-measurable, is not zero in $S$, and is zero on $T - S$ (Ch. IV, §5, No. 10, Prop. 15).