Remark

— Let $(\mu_i)_{i \in I}$ be an inverse system of measures on the inverse system of spaces $\mathscr{T} = (T_i, p_{ij})$. Assume given a topological space $T'$ and continuous mappings $p'_i : T' \to T_i$; assume that the family $(p'_i)_{i \in I}$ is coherent, but not necessarily separating. If Prokhorov's condition (P) is satisfied by the family $(p'_i)_{i \in I}$, there exists a measure $\mu'$ (not necessarily unique) on $T'$ such that $p'_i(\mu') = \mu_i$ for all $i \in I$.

For, set $T = \lim_{\leftarrow} T_i$ and $p' = (p'_i)_{i \in I}$, and denote by $p_i$ the canonical mapping of $T$ into $T_i$; Prokhorov's condition is satisfied by $T$ and the $p_i$, because $p_i(p'(K')) = p'_i(K')$ and $p'(K')$ is compact in $T$ for every compact subset $K'$ of $T'$. By Th. 1, there exists a bounded measure $\mu$ on $T$ such that $p_i(\mu) = \mu_i$ for all $i \in I$. Let $K'$ be a compact set in $T'$; then $\mu^\bullet(p'(K')) = \inf_{i \in I} \mu_i^\bullet(p'_i(K'))$, whence

Let $\varepsilon > 0$; since Prokhorov’s condition (P) is satisfied by the $p_i'$, one can therefore find a compact subset $K'$ of $T'$ such that $\mu^\bullet(T - p'(K')) \leq \varepsilon$. Prop. 8 of §2, No. 4 then establishes the existence of a bounded measure $\mu'$ on $T'$ with $\mu = p'(\mu')$, whence $\mu_i = p_i(\mu) = p_i(p'(\mu')) = p_i'(\mu')$ for all $i \in I$.