Proposition 1

*Let $X$ be a $\mu$-measurable subset of $T$. The encumbrance $(\mu_X)^\bullet$ is equal to the encumbrance $(\mu^\bullet)_X$ induced by $\mu^\bullet$ on $X$ (§1, No. 1). In other words, $(\mu_X)^\bullet(g) = \mu^\bullet(g^0)$ for every function $g \in \mathscr{F}_+(X)$.

Let $f \in \mathscr{F}_+(X)$ and let $f^0$ be the extension by 0 of $f$ to $T$. One has $(\mu^\bullet)_X(f) = \mu^\bullet(f^0) = \sup_L \mu^\bullet(f^0 \varphi_L)$, where $L$ runs over the set of compact subsets of $T$ (§1, No. 2, Prop. 2); similarly $(\mu_X)^\bullet(f) = \sup_K \mu_K^\bullet(f_K) = \sup_K \mu^\bullet(f^0 \varphi_K)$, where $K$ runs over the set of compact subsets of $X$. Thus it all comes down to showing that $\mu^\bullet(f^0 \varphi_L) = \sup_K \mu^\bullet(f^0 \varphi_K)$ for every compact subset $L$ of $T$, where $K$ runs over the set of compact subsets of $L \cap X$. Now, let $(K_n)$ be an increasing sequence of compact sets contained in $L \cap X$, such that $(L \cap X) - \bigcup_n K_n$ is locally $\mu$-negligible (§1, No. 8, Prop. 11); $f^0$ being zero outside $X$, $f^0 \varphi_L$ is zero outside $L \cap X$ hence is equal locally almost everywhere to the upper envelope of the sequence $(f^0 \varphi_{K_n})$. This implies that $\mu^\bullet(f^0 \varphi_L) = \sup_n \mu^\bullet(f^0 \varphi_{K_n})$, whence the desired result.*