Proposition 2

— Let X be a topological space and T a subspace of X.

a) Suppose that T is a Radon space. Then, for every function I defined on $\mathscr{B}(X)$ that is positive, countably additive and bounded, one has

Moreover, T is universally measurable in X.

b) Conversely, suppose that X is a Radon space and that T is universally measurable in X; then T is a Radon space.

Let us prove a). We denote by $\alpha$ the right side of (6); for every $n \in \mathbf{N}$, there exists a set $C_n \in \mathscr{B}(X)$ containing T, such that $I(C_n) \leq \alpha + 2^{-n}$. Setting $C = \bigcap_n C_n$, we then have $T \subset C$, $I(C) = \alpha$. If $A \in \mathscr{B}(T)$, let us choose a Borel set B of X such that $A = B \cap T$ (GT, IX, §6, No. 3) and set $J(A) = I(B \cap C)$. This number does not depend on the choice of B, for if $B'$ is a second Borel set in X such that $A = B' \cap T$, then $B \cap C$ and $B' \cap C$ differ only by a Borel set M contained in $C - T$, and $I(M) = 0$ by the construction of C. Clearly $J(K) = I(K)$ for every compact set $K \subset T$. Let $(A_n)$ be a sequence of Borel sets of T, pairwise disjoint, and, for each $n$, let $B_n$ be a Borel set of X such that $B_n \cap T = A_n$. Replacing $B_n$ by $B_n - \left( \bigcup_{k < n} B_k \right)$ if necessary, we can suppose that the sets $B_n$ are pairwise disjoint. Set $A = \bigcup_n A_n$ and $B = \bigcup_n B_n$; then

J is thus a countably additive and bounded function on $\mathscr{B}(T)$. Since T is by hypothesis a Radon space, there exists a bounded measure $\mu$ on T such that $J(A) = \mu^\bullet(A)$ for every $A \in \mathscr{B}(T)$; consequently

by the definition of $\mu^\bullet$. Formula (6) is thus established.

Let us now show that T is universally measurable. Let $\lambda$ be a bounded measure on X; the preceding argument may be applied to the set function $I : A \mapsto \lambda^\bullet(A)$ on $\mathscr{B}(X)$, thus there exists a sequence $(K_n)$ of compact subsets of T such that (with the above notations)

Set $K' = \bigcup_{n \in \mathbf{N}} K_n$; $K'$ is Borel in $X$, $K' \subset T \subset C$, $\lambda^*(K') = \lambda^*(C)$, therefore these three sets differ only by $\lambda$-negligible sets, and so $T$ is $\lambda$-measurable. This completes the proof of a).

Let us pass to b). Suppose that $X$ is a Radon space, and that $T$ is universally measurable in $X$. Let $I$ be a positive function on $\mathscr{B}(T)$ that is countably additive and bounded; the function $A \mapsto I(A \cap T)$ on $\mathscr{B}(X)$ is then positive, countably additive and bounded, therefore there exists a bounded measure $\nu$ on $X$ such that $I(A \cap T) = \nu^*(A)$ for all $A \in \mathscr{B}(X)$. Now, $T$ is $\nu$-measurable; the preceding relation shows that $\nu^*(K) = 0$ for every compact subset $K$ of $X$ that is disjoint from $T$, therefore $\nu$ is concentrated on $T$. Consequently, for every Borel set $A$ of $X$, we have $I(A \cap T) = \nu^*(A \cap T) = \mu^*(A \cap T)$, where $\mu$ is the measure induced by $\nu$ on $T$. Finally, it follows that $I(B) = \mu^*(B)$ for every set $B \in \mathscr{B}(T)$ (GT, IX, §6, No. 3, Remark 2), and $I$ is indeed inner regular.