For, let y be a point of $\bigcap_{\alpha \in A} f(K_\alpha)$; for every $\alpha \in A$, the set $L_\alpha = K_\alpha \cap f^{-1}(y)$ is compact and nonempty. The family $(L_\alpha)_{\alpha \in A}$ is directed downward, therefore its intersection L is nonempty. Now, $L = K \cap f^{-1}(y)$, whence $y \in f(K)$. We have thus proved the inclusion $f(K) \supset \bigcap_{\alpha \in A} f(K_\alpha)$, and the reverse inclusion is obvious.

a) For every compact subset K of T, one has $K = \bigcap_{i \in I} p_i^{-1}(p_i(K))$.

b) Let K and L be two disjoint compact subsets of T. There exists an $i \in I$ such that $p_j(K)$ and $p_j(L)$ are disjoint for $j \geq i$.

a) Let x be a point of $\bigcap_{i \in I} p_i^{-1}(p_i(K))$; for every $i \in I$, the set $K_i$ of points y in K such that $p_i(y) = p_i(x)$ is a nonempty closed subset of K. For $i \leq j$ we have $K_i \supset K_j$, and, since K is compact, the set $\bigcap_{i \in I} K_i$ is therefore nonempty. Let y be a point of $\bigcap_{i \in I} K_i$; we have $y \in K$ and $p_i(y) = p_i(x)$ for all $i \in I$, whence $y = x$; finally, $x \in K$, which proves the inclusion $K \supset \bigcap_{i \in I} p_i^{-1}(p_i(K))$; the reverse inclusion is obvious.

b) For every $i \in I$, set $M_i = p_i^{-1}(p_i(K)) \cap L$; this is a closed subset of the compact space L, we have $M_i \supset M_j$ for $i \leq j$, and, by a),

Consequently, there exists an index i such that $M_i = \varnothing$. For $j \geq i$, we have $p_j^{-1}(p_j(K)) \cap L = M_j \subset M_i = \varnothing$, whence $p_j(K) \cap p_j(L) = \varnothing$.

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.

For there to exist a bounded measure $\mu$ on $T$ such that $p_i(\mu) = \mu_i$ for all $i \in I$, it is necessary and sufficient that the following condition be satisfied:

(P) *for every $\varepsilon > 0$, there exists a compact subset $K$ of $T$ such that $\mu_i^\bullet(T_i - p_i(K)) \leq \varepsilon$ for all $i \in I$.

When this is so, the measure $\mu$ is uniquely determined and*

for every compact set $K$ in $T$.

Let us first prove the uniqueness of $\mu$. Let $\mu$ be a bounded measure on $T$ such that $p_i(\mu) = \mu_i$ for all $i \in I$. Let $K$ be a compact subset of $T$; by Prop. 2 of No. 1, the set $K$ is the intersection of the decreasing directed family $(\overline{p_i^{-1}(p_i(K))})_{i \in I}$ of closed subsets of $T$. By the Cor. of Prop. 5 of § 1, No. 6, we therefore have

which establishes the formula (2). Since two measures that coincide on the set of compact sets are equal (§ 1, No. 2, Cor. of Prop. 2), it follows that $\mu$ is unique.

By Prop. 3, there exists a bounded measure $\pi$ on $T$ such that $\pi^\bullet(K) = \inf_{i \in I} \mu_i^\bullet(p_i(K))$ for every compact subset $K$ of $T$. By formula (2), the existence of a bounded measure $\mu$ on $T$ such that $p_i(\mu) = \mu_i$ for all $i \in I$ is therefore equivalent to the relation:

(P') $p_i(\pi) = \mu_i$ for all $i \in I$.

For $i \leq j$, we have $\mu_i = p_{ij}(\mu_j)$, whence $\mu_i^\bullet(T_i) = \mu_j^\bullet(T_j)$; since $I$ is directed, there exists a finite number $c \geq 0$ such that $\mu_i^\bullet(T_i) = c$ for all $i \in I$. By Prop. 3, the measure $\mu_i - p_i(\pi)$ is positive, hence is zero if and only if its total mass is zero, that is, if $\mu_i(T_i) = p_i(\pi)^\bullet(T_i)$. Since $p_i(\pi)^\bullet(T_i) = \pi^\bullet(T)$, the condition (P') is thus equivalent to $\pi^\bullet(T) = c$, that is (§ 1, No. 2, Remark 3) to the property:

(P'') $\sup_{K \in \mathscr{K}} \pi^\bullet(K) = c$, where $\mathscr{K}$ is the set of compact subsets of $T$.

Now, for $K \in \mathscr{K}$, we have

and this formula immediately implies the equivalence of (P) and (P'').

Q.E.D.

Let $(T_i, p_{ij})$ be an inverse system of topological spaces. Set $T = \lim_{\leftarrow} T_i$ and denote by $p_i$ the canonical mapping of $T$ into $T_i$. Generalizing Def. 2 of Ch. III, § 4, No. 5, we shall say that a bounded measure $\mu$ on $T$ is the inverse limit of an inverse system $(\mu_i)_{i \in I}$ of measures if $\mu_i = p_i(\mu)$ for all $i \in I$. Th. 1 provides a criterion for the existence of inverse limits of measures. When the spaces $T_i$ are compact, and the mappings $p_{ij}$ surjective, $T$ is compact and $p_i(T) = T_i$ for every $i \in I$; the condition (P) is therefore fulfilled, and in this case we recover Prop. 8, (iv) of Ch. III, § 4, No. 5.

We shall first treat the case that $I = \mathbf{N}$ and set $q_n = p_{n,n+1}$. Let $\varepsilon > 0$. Define recursively a sequence of compact sets $L_n \subset T_n$ as follows: $L_0$ is a compact subset of $T_0$ such that $\mu_0^\bullet(T_0 - L_0) \leq \varepsilon/2$, and for $n \geq 0$ the compact set $L_{n+1}$ is contained in $\overline{q_n^{-1}(L_n)}$ and satisfies

This construction is possible by virtue of Remark 3 of §1, No. 2. We have because $\mu_n = q_n(\mu_{n+1})$; by induction on $p$, one deduces that Since $T$ is a closed subspace of $\prod_{n \in \mathbf{N}} T_n$ and the product space $\prod_{n \in \mathbf{N}} L_n$ is compact, the subset $L = T \cap \prod_{n \in \mathbf{N}} L_n = \bigcap_{n \in \mathbf{N}} \overline{p_n^{-1}(L_n)}$ of $T$ is compact. Let $n \in \mathbf{N}$; we have $p_n(L) = \bigcap_{m \geq n} p_{nm}(L_m)$ (GT, I, §9, No. 6, Prop. 8) and $p_{nm}(L_m) \supset p_{nm'}(L_{m'})$ for $m' \geq m \geq n$, whence

But, for $m \geq n$, the measure $\mu_n$ is the image of $\mu_m$ under $p_{nm}$, whence

passing to the limit with respect to $m$, we obtain $\mu_n^\bullet(T_n - p_n(L)) \leq \varepsilon$. In other words, Prokhorov’s condition (P) is satisfied, and there exists a bounded measure $\mu$ on $T$ such that $\mu_n = p_n(\mu)$ for all $n \in \mathbf{N}$ (No. 2, Th. 1).

Let us pass to the general case: there exists in $I$ an increasing cofinal sequence $(i_n)_{n \in \mathbf{N}}$. The mapping $t \mapsto (p_{i_n}(t))_{n \in \mathbf{N}}$ is a homeomorphism of $T$ onto the inverse limit of the inverse system $(T_{i_n}, p_{i_n i_m})$ (GT, I, §4, No. 4). By the first part of the proof, there exists therefore a bounded measure $\mu$ on $T$ such that $\mu_{i_n} = p_{i_n}(\mu)$ for all $n \in \mathbf{N}$. Let $i \in I$; there exists an $n \in \mathbf{N}$ with $i \leq i_n$, whence

Q.E.D.

Theorem 2 is often used in the following situation: let $D$ be a countable set and $(X_t)_{t \in D}$ a family of topological spaces. Let $\mathfrak{F}$ be the set of finite subsets of $D$, ordered by inclusion. For $J$ in $\mathfrak{F}$, set $X_J = \prod_{t \in J} X_t$, and for $J \subset J'$ let $p_{JJ'}$ be the canonical projection of $X_{J'}$ onto the partial product $X_J$. Also set $X = \prod_{t \in D} X_t$ and denote by $p_J$ the canonical projection of $X$ onto the partial product $X_J$. One shows easily (cf. S, III, §7, No. 2, Remark 3) that the family $(p_J)_{J \in \mathfrak{F}}$ defines a homeomorphism of $X$ onto $\lim_{\leftarrow} X_J$. An inverse system of measures is then a family of bounded measures $\mu_J$ on $X_J$ such that $\mu_J = p_{JJ'}(\mu_{J'})$ for $J \subset J'$. There exists one and only one bounded measure $\mu$ on $X$ such that $\mu_J = p_J(\mu)$ for every finite subset $J$ of $D$ (Kolmogoroff’s theorem). One sometimes says that $\mu$ is the measure on $\prod_{t \in D} X_t$ having margins $\mu_J$.

In particular, suppose given, for every $t \in D$, a measure $\nu_t$ on $X_t$ of total mass 1. Set $\mu_J = \bigotimes_{t \in J} \nu_t$ for every finite subset $J$ of $D$. Let $J \subset J'$ be two finite subsets of $D$ and let $K = J' - J$; identifying $X_{J'}$ with $X_J \times X_K$, one has $\mu_{J'} = \mu_J \otimes \mu_K$, and since the measure $\mu_K$ has total mass 1, the projection of $\mu_J \otimes \mu_K$ on $X_J$ is equal to $\mu_J$. The measure on $X$ admitting the margins $\mu_J$ is denoted $\bigotimes_{t \in D} \nu_t$ and is called the product of the family $(\nu_t)_{t \in D}$. When the spaces $X_t$ are compact, we recover the construction of Ch. III, §4, No. 6.