Proposition 6

— Let $(\lambda_\alpha)_{\alpha \in A}$ be an increasing directed family of measures on $T$, and let $p = \sup_\alpha \lambda_\alpha^\bullet$. For the family $(\lambda_\alpha)$ to be bounded above in $\mathscr{M}(T)$, it is necessary and sufficient that the encumbrance $p$ be locally bounded. The family $(\lambda_\alpha)$ then admits a supremum $\lambda$ in $\mathscr{M}(T)$, and $\lambda^\bullet = p$. For every compact set $K$, the measure $\lambda_K$ is the supremum of the measures $(\lambda_\alpha)_K$ in $\mathscr{M}(K)$.

If the family $(\lambda_\alpha)$ is bounded above in $\mathscr{M}(T)$, then $p$ is obviously locally bounded. Conversely, let us assume $p$ to be locally bounded, and let us show that it satisfies the conditions 1) and 2) of Prop. 2 b) of No. 2. For 2), this results from the following equalities:

On the other hand, let K be a compact set; the encumbrance $p_K$ is equal to the upper envelope of the encumbrances $(\lambda_\alpha^\bullet)_K$ and it is bounded since $p$ is locally bounded. The measures $(\lambda_\alpha)_K$ therefore admit a supremum $\lambda_K$ in $\mathscr{M}(K)$, and $\lambda_K^\bullet = p_K$ (Ch. V, §1, No. 4, Prop. 11). The condition 1) of Prop. 2 b) of No. 2 is thus satisfied, therefore there exists a measure $\lambda$ on T such that $\lambda^\bullet = p$; it is clear that $\lambda$ is the supremum of the measures $\lambda_\alpha$.