A real L^1 density defines a finite signed measure with its canonical Hahn and Jordan data

Because $f$ is real-valued, every $\nu (E)={\int }_{E}fd\mu$ is a real number. The complex-density theorem [L1] shows that the same set function is countably additive and satisfies $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_E\mathrm{\mid }f\mathrm{\mid }d\mu \mathrm{.}$ Also $\mathrm{\mid }\nu (E)\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(E)\le \int \mathrm{\mid }f\mathrm{\mid }d\mu <+\mathrm{\infty },$ so $\nu$ takes no infinite values. Hence $\nu$ is a finite signed measure. The finiteness of $\mathrm{\mid }\nu \mathrm{\mid }(X)$ follows from $f\in L^1(\mu )$.

By [L3], the sets $P=\{f>0\}$ and $N=\{f\le 0\}$ are measurable and form a partition of $X$. If $E\subseteq P$ is measurable, then $f\ge 0$ on $E$, so $\nu (E)={\int }_{E}fd\mu \ge 0$; if $E\subseteq N$, then $f\le 0$ on $E$, so $\nu (E)\le 0$. Thus $P$ is positive and $N$ is negative, so [L4] makes them canonical Hahn sets up to null sets.

The formulas in [L2] give $\nu (E)={\int }_E{f}^{+}d\mu -{\int }_E{f}^{-}d\mu \mathrm{.}$ On subsets of $P$ one has $f^{-}=0$ and $f=f^{+}$, while on subsets of $N$ one has $f^{+}=0$ and $-f=f^{-}$. Therefore the positive measures $E\mapsto {\int }_E{f}^{+}d\mu$ and $E\mapsto {\int }_E{f}^{-}d\mu$ are mutually singular and decompose $\nu$. By uniqueness in [L4], they are exactly ${\nu }^{+}$ and ${\nu }^{-}$. The total-variation formula from step 1.1 and [L2] then becomes $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_E\mathrm{\mid }f\mathrm{\mid }d\mu$.

Steps 1.1 through 3.1 prove the signed-measure, Hahn, Jordan, and total-variation claims.