Corollary 5

If $\mu$ is a moderated measure on T, there exists a sequence $(\mu_n)_{n \in \mathbf{N}}$ of measures with compact support such that $\mu = \sum_{n \in \mathbf{N}} \mu_n$.

By hypothesis, the constant function 1 is $\mu$-moderated. Let us apply Cor. 3 to the case $f = 1$; thus, there exists a sequence $(K_n)_{n \in \mathbf{N}}$ of pairwise disjoint compact subsets of T such that $1 = \sum_{n \in \mathbf{N}} \varphi_{K_n}$ $\mu$-almost everywhere. Let $\mu_n$ be the measure on T defined by the measure $\mu_{K_n}$ on $K_n$ (No. 3, Example 2). One knows (No. 6, Remark 2) that $\mu_n$ has compact support, and that $\mu_n^\bullet(f) = \mu^\bullet(f \varphi_{K_n})$ for $f \in \mathscr{F}_+(T)$. Now, $f$ is equal to $\sum_{n \in \mathbf{N}} f \varphi_{K_n}$ $\mu$-almost everywhere, whence

It follows that $\mu = \sum_{n \in \mathbf{N}} \mu_n$ (No. 7, Prop. 7).