Proposition 3

Let there be given an inverse system of topological spaces $\mathscr{T} = (T_i, p_{ij})$ indexed by $I$, a topological space $T$, a coherent and separating family of continuous mappings $p_i : T \to T_i$ (for $i \in I$) and a sub-inverse system $(\mu_i)_{i \in I}$ of measures on $\mathscr{T}$. For every compact subset K of $T$, set

Then there exists a bounded measure $\pi$ on $T$, and only one, such that $\pi^\bullet(K) = J(K)$ for every compact subset $K$ of $T$. One has $\mu_i \geq p_i(\pi)$ for all $i \in I$, and $\pi$ is the largest measure on $T$ satisfying this condition.

Let us first prove that $J(K)$ is the limit of $\mu_i^\bullet((p_i(K)))$ with respect to the section filter $\mathfrak{F}$ of the directed preordered set $I$: for this, it suffices (GT, IV, §5, No. 2, Th. 2) to show that $\mu_i^\bullet(p_i(K)) \geq \mu_j^\bullet(p_j(K))$ for $i \leq j$; now, setting $\mu_{ij}' = p_{ij}(\mu_j)$, we have $\mu_{ij}' \leq \mu_i$ and $p_j(K) \subset \overline{p_{ij}}^{-1}(p_i(K))$, whence

Let us now pass to the study of the properties of the function $J$: 1) It is clear that $J(K) \leq J(L)$ when $K \subset L$. 2) Let $K$ and $L$ be two compact subsets of $T$. For every $i \in I$, we have $p_i(K \cup L) = p_i(K) \cup p_i(L)$, whence

passing to the limit with respect to the filter $\mathfrak{F}$, we obtain $J(K \cup L) \leq J(K) + J(L)$. 3) Suppose that the compact sets $K$ and $L$ are disjoint. By Prop. 2 of No. 1, there exists an $i \in I$ such that $p_j(K) \cap p_j(L) = \varnothing$ for $j \geq i$. For $j \geq i$ we therefore have

whence $J(K \cup L) = J(K) + J(L)$ on passing to the limit with respect to the filter $\mathfrak{F}$. 4) Let $(K_\alpha)_{\alpha \in A}$ be a decreasing directed family of compact subsets of $T$, with intersection $K$. By Prop. 1 of No. 1, $p_i(K) = \bigcap_{\alpha \in A} p_i(K_\alpha)$ and therefore $\mu_i^\bullet(p_i(K)) = \inf_{\alpha \in A} \mu_i^\bullet(p_i(K_\alpha))$ for all $i \in I$ (§ 1, No. 6, Cor. of Prop. 5). From this, one deduces

5) Let us choose an $i \in I$ and set $c = \mu_i^\bullet(T_i)$. Then $c$ is finite and $J(K) \leq \mu_i^\bullet(p_i(K)) \leq \mu_i^\bullet(T_i)$, thus $J(K) \leq c$ for every compact set $K$ in $T$.

The preceding properties permit applying Th. 1 of § 3, No. 1; we conclude that there exists one and only one bounded measure $\pi$ on $T$ such that $\pi^\bullet(K) = J(K)$ for every compact subset $K$ of $T$. For every $i \in I$, let us denote by $\nu_i$ the measure on $T_i$ that is the image of $\pi$ under $p_i$. Let $i \in I$, $A$ a compact subset of $T_i$, and $\mathscr{L}$ the set of compact subsets of $\overline{p_i^{-1}(A)}$. By Remark 3 of § 1, No. 2, we have $\pi^\bullet(\overline{p_i^{-1}(A)}) = \sup_{K \in \mathscr{L}} \pi^\bullet(K)$; moreover, $\nu_i^\bullet(A) = \pi^\bullet(\overline{p_i^{-1}(A)})$ and $J(K) = \pi^\bullet(K)$ for $K \in \mathscr{L}$, whence $\nu_i^\bullet(A) = \sup_{K \in \mathscr{L}} J(K)$.

For $K \in \mathscr{L}$, we have $p_i(K) \subset A$, whence $J(K) \leq \mu_i^\bullet(p_i(K)) \leq \mu_i^\bullet(A)$ and finally $\nu_i^\bullet(A) \leq \mu_i^\bullet(A)$. Since $A$ is an arbitrary compact set in $T_i$, we conclude that $\nu_i \leq \mu_i$. The last assertion of the proposition is obvious.

Q.E.D.