Lemma 1

— Let $\theta$ be a complex measure, and let $f$ be an element of $\overline{\mathscr{L}}^1_C(T, \theta)$. Then

*where $\mathscr{K}_1$ (resp. $\mathscr{B}_1$) denotes the set of complex functions $c$, continuous with compact support (resp. Borel), such that $|c| \leq 1$.

Let us first treat the case that $f \in \mathscr{K}(T; \mathbf{C})$. Obviously

(Ch. IV, §4, No. 2, Prop. 2). On the other hand, let $g$ be an element of $\mathscr{K}(T; \mathbf{C})$ such that $|g| \leq |f|$; $g$ is the uniform limit of a sequence $(g_n)$ of elements of $\mathscr{K}(T; \mathbf{C})$ whose supports are contained in the open set $U$ formed by the $t$ such that $f(t) \neq 0$, and one may clearly suppose that $|g_n| \leq |f|$ for every $n$. Set $c_n(t) = g_n(t)/f(t)$ for $t \in U$, $c_n(t) = 0$ for $t \notin U$; then $c_n \in \mathscr{K}_1$, $g = \lim_{n \to \infty} c_n f$, therefore $|\langle\theta, g\rangle| = \lim_{n \to \infty} |\langle\theta, c_n f\rangle|$, and finally

One concludes by observing that the first member of this inequality is equal to $\langle |\theta|, |f| \rangle$ (Ch. III, §1, No. 6, formula (12)).

Next, denote by $f$ an element of $\overline{\mathscr{L}}_C^1(\theta)$, and let us show that (4) is again true: it suffices to verify that the three members of this relation depend continuously on $f$ for the topology of $\overline{\mathscr{L}}_C^1(\theta)$, since they coincide on the dense subspace $\mathscr{K}(T; \mathbf{C})$. This results at once from the following inequalities, where $f$ and $f'$ denote elements of $\overline{\mathscr{L}}_C^1(\theta)$:

for all $c \in \mathscr{B}_1$. The lemma is thus established.

Passing to the proof of Proposition 2, let us apply the lemma to the function $uh$, where $h$ belongs to $\mathscr{K}_+(T)$. This yields:

However, the first member is also equal to

The two measures $|u| \cdot |\theta|$ and $|u \cdot \theta|$ are therefore equal.