Remark 2

Def. 1 may be extended to the case that $\mu$ is a complex measure on $T$. To show in this case that the premeasure $\mu_X$ is a measure, it suffices to observe that $|\mu_K| = |\mu|_K$ for every compact subset $K$ of $X$ (§1, No. 2).

By Prop. 1, a subset $Y$ of $X$ is $\mu_X$-measurable (resp. locally $\mu_X$-negligible) if and only if it is $\mu$-measurable (resp. locally $\mu$-negligible). If $Y$ is $\mu_X$-measurable, hence $\mu$-measurable, the induced measures $(\mu_X)_Y$ and $\mu_Y$ are obviously equal by virtue of Prop. 1 (transitivity of induced measures).

Remark 3). — Let X be a $\mu$-measurable subset of T. By Prop. 10 of §1, No. 8, applied to $g = \varphi_X$, there exists a crushing $(K_\alpha)_{\alpha \in A}$ of T such that for each $\alpha \in A$, either $K_\alpha \subset X$ or $K_\alpha \subset \mathbf{C}X$. If one modifies the topology of T by the procedure of the Scholium of §1, No. 8, the space $X'$ obtained by equipping X with the topology induced by $T'$ is locally compact, and one knows how to associate to $\mu$ (resp. to $\mu_X$) a measure $\mu'$ (resp. $\nu$) on $T'$ (resp. on $X'$) that admits the same essential upper integral as $\mu$ (resp. $\mu_X$): this implies that $\mu'_{X'} = \nu$. Since the locally negligible sets, measurable mappings, and essentially integrable functions with values in a Banach space, are the same for $\mu$ and $\mu'$, and for $\mu_X$ and $(\mu_X)' = \nu = \mu'_{X'}$, the theory of integration with respect to an induced measure reduces to that which was treated in Ch. V, §7 in the special case of locally compact spaces. We leave to the reader the task of transcribing the results.