Proposition 1

Every Lindelöf$^{(1)}$* Radon space is strongly Radon.*

Let T be a Lindelöf space that is Radon, and let I be a set function on the tribe $\mathscr{B}(T)$ that is positive, countably additive and locally bounded. The open sets V such that $I(V) < +\infty$ form a covering of T, from which one can extract a countable covering $(V_n)_{n \in \mathbf{N}}$. Set $G_n = V_0 \cup V_1 \cup \cdots \cup V_n$ for every $n \in \mathbf{N}$; set $H_0 = G_0$ and $H_n = G_n - G_{n-1}$ for $n \geq 1$; finally, denote by $I_n$ the set function $A \mapsto I(A \cap H_n)$ on $\mathscr{B}(T)$, which is obviously countably additive and bounded. Since the sets $H_n$ form a partition of T, we have $I = \sum_n I_n$. The space T being Radon, for every $n \in \mathbf{N}$ there exists a bounded measure $\mu_n$ on T such that $\mu_n^\bullet(A) = I_n(A)$ for all $A \in \mathscr{B}(T)$; therefore also $\sum_n \mu_n^\bullet(A) = I(A)$. Since I is locally bounded, the family $(\mu_n)$ is summable (§ 1, No. 7, Prop. 7); if $\mu$ denotes $\sum_n \mu_n$, we have $\mu^\bullet(A) = I(A)$ for all $A \in \mathscr{B}(T)$, and it follows that I is inner regular. In other words, T is strongly Radon.

Recall that a subset A of a topological space T is said to be universally measurable if A is $\mu$-measurable for every measure $\mu$ on T. This is equivalent to saying that A is $\mu$-measurable for every measure $\mu$ on T with compact support (§ 1, No. 8, Prop. 9).

(1) Recall (GT, I, §9, Exer. 14; TG, IX, Appendix I) that a Lindelöf space is a topological space T such that every open covering of T contains a countable covering.