Lemma 5

Let m be a complex measure on T, f an m-integrable complex function. Then $\int |f|\, d|m| = \sup |\int fh\, dm|$, as h runs over the set of functions in $\mathscr{H}_C(T)$ such that $|h(t)| \leq 1$ for all $t \in T$.

⁶Cf. Ch. III, §1, No. 8.

If $m = g \cdot |m|$, then $\int |f| d|m| = \int |f g| d|m|$ and $\int f h \, dm = \int f g h \, d|m|$. Set $\zeta(t) = 0$ when $f(t)g(t) = 0$, and $\zeta(t) = \frac{f(t)g(t)}{|f(t)g(t)|}$ when $f(t)g(t) \neq 0$; $\zeta$ is $|m|$-measurable, thus for every $\varepsilon > 0$ there exists a compact subset K of T such that $\int_{T-K} |f| \, d|m| \leq \varepsilon$, the restriction of $\zeta$ to K is continuous, and $|\zeta(t)| = 1$ on K. Therefore, by virtue of Urysohn’s theorem, there exists a continuous function $\zeta_1$ defined on T, with complex values, such that $\zeta_1 = \zeta$ on K and such that $|\zeta_1(t)| \leq 2$ and $\zeta_1(t) \neq 0$ for every $t \in T$; setting $h(t) = \zeta_1(t)/|\zeta_1(t)|$, we see that $h$ is continuous on T, coincides with $\zeta$ on K, and is such that $|h(t)| = 1$ for all $t \in T$. Finally, let $u$ be a continuous mapping of T into $[0,1]$, equal to 1 on K and with compact support; setting $h_1 = h^{-1}u$, we have

which proves the lemma.