Proposition 11

Let $\mu$ be a positive measure on $T$. If $T$ is countable at infinity, then there exists a continuous function $h$ such that $h(t) > 0$ for all $t \in T$ and such that the measure $\nu = h \cdot \mu$ (equivalent to $\mu$) is bounded.

Let $(K_n)$ be a sequence of compact sets forming a covering of $T$ and, for every $n$, let $f_n$ be a function in $\mathscr{H}(T)$ such that $0 \leq f_n \leq 1$ and $f_n(t) = 1$ on $K_n$ (Ch. III, §1, No. 2, Lemma 1). Let $(a_n)$ be a sequence of numbers $> 0$ such that $\sum_n a_n < +\infty$; the series $h = \sum_n a_n f_n$ is then normally convergent in $T$, consequently $h$ is a continuous function on $T$, such that $h(t) > 0$ for all $t \in T$, by construction. Setting $\nu = h \cdot \mu$, we then have (Prop. 3 and Ch. IV, §1, No. 3, Prop. 13)

On taking for example $a_n = 2^{-n} (\int f_n \, d\mu)^{-1}$ when $\int f_n \, d\mu > 1$, and $a_n = 2^{-n}$ otherwise, we have $\sum_n a_n < +\infty$ and $\nu^*(1) < +\infty$, which proves the proposition.