A complex L^1 density defines a complex measure whose total variation is |h| dmu

The set function $\nu$ is finite-valued because $\mathrm{\mid }\nu (E)\mathrm{\mid }=\mid {\int }_{E}hd\mu \mid \le {\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu \le \int \mathrm{\mid }h\mathrm{\mid }d\mu <+\mathrm{\infty }$ by [L1] and [L2]. If $(E_n)$ is a pairwise disjoint measurable sequence, then ${\sum }_{n=0}^N{\mathbf{1}}_{E_n}\uparrow {\mathbf{1}}_{{\cup }_n{E}_n}$, so monotone convergence applied to the positive and negative parts of the real and imaginary parts of $h$ gives $\nu \left({\bigcup }_n{E}_n\right)={\sum }_n\nu (E_n)\mathrm{.}$ Thus $\nu$ is a complex measure.

Define $u(x):=\{\begin{array}{ll}\overline{h(x)}\mathrm{/}\mathrm{\mid }h(x)\mathrm{\mid },& h(x)\mathrm{\ne }0,\\ 0,& h(x)=0.\end{array}$ Then $\mathrm{\mid }u\mathrm{\mid }\le 1$ and $uh=\mathrm{\mid }h\mathrm{\mid }$. By [L5], the function $u$ is measurable. Apply [L6] to $\rho (F):={\int }_F\mathrm{\mid }h\mathrm{\mid }d\mu$ and obtain a finite measure $\rho$ on $(X,\mathcal{A})$. Applying [L4] to $u$ on $(E,\mathcal{A}\text{\hspace{0.17em}⁣}{\upharpoonright }_E,\rho )$ gives complex simple functions $s_n$ with $\mathrm{\mid }{s}_n\mathrm{\mid }\le 2$ and ${\int }_E\mathrm{\mid }u-s_n\mathrm{\mid }d\rho \to 0.$ For each $n$, define the clipped simple function $t_n:=\frac{s_n}{\max \{1,\mathrm{\mid }{s}_n\mathrm{\mid }\}}\mathrm{.}$ Then $\mathrm{\mid }{t}_n\mathrm{\mid }\le 1$, and because $\mathrm{\mid }u\mathrm{\mid }\le 1$ one has $\mathrm{\mid }{t}_n-u\mathrm{\mid }\le \mathrm{\mid }{t}_n-s_n\mathrm{\mid }+\mathrm{\mid }{s}_n-u\mathrm{\mid }\le 2\mathrm{\mid }{s}_n-u\mathrm{\mid }\mathrm{.}$ Hence ${\int }_E\mathrm{\mid }u-t_n\mathrm{\mid }d\rho \to 0$.

For any measurable $E$ and any countable measurable partition $E={⨆}_n{E}_n$, the inequality in step 1.1 applied on each piece gives ${\sum }_n\mathrm{\mid }\nu (E_n)\mathrm{\mid }\le {\sum }_n{\int }_{E_n}\mathrm{\mid }h\mathrm{\mid }d\mu ={\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu \mathrm{.}$ Taking the supremum over partitions shows $\mathrm{\mid }\nu \mathrm{\mid }(E)\le {\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu$.

Write the canonical representation of $t_n$ on $E$ as $t_n={\sum }_{j=1}^{m_n}{c}_{n,j}{\mathbf{1}}_{A_{n,j}}\mathrm{.}$ Because $\mathrm{\mid }{t}_{n}h\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }$, each $t_{n}h$ lies in $L^1(\mu )$. By the definition of $\nu$ and the linearity of the Lebesgue integral, ${\int }_E{t}_{n}d\nu ={\sum }_{j=1}^{m_n}{c}_{n,j}\nu (A_{n,j})={\sum }_{j=1}^{m_n}{c}_{n,j}{\int }_{A_{n,j}}hd\mu ={\int }_E{t}_{n}hd\mu \mathrm{.}$ Therefore $\mid {\int }_E{t}_{n}d\nu -{\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu \mid =\mid {\int }_E(t_n-u)hd\mu \mid \le {\int }_E\mathrm{\mid }{t}_n-u\mathrm{\mid }\mathrm{\mid }h\mathrm{\mid }d\mu ={\int }_E\mathrm{\mid }{t}_n-u\mathrm{\mid }d\rho \to 0.$ So ${\int }_E{t}_{n}d\nu \to {\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu$.

Since each $t_n$ is a unit-bounded complex simple function on $E$, [L3] gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge \mid {\int }_E{t}_{n}d\nu \mid$ for every $n$. Letting $n\to \mathrm{\infty }$ and using step 2.2 shows $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge {\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu$. Together with step 2.1, this proves $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu$.

Steps 1.1, 2.1, and 3.1 prove that $\nu$ is a complex measure and that its total variation is ${\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu$ on every measurable set $E$.