A sigma-finite positive measure has a nonnegative Radon–Nikodym density version

Choose increasing countable finite-measure covers $(A_i)$ for $\mu$ and $(B_j)$ for $\nu$, using [F1]. Set $X_n={\bigcup }_{i,j\le n}(A_i\cap B_j)$. Then $X_n\uparrow X$ and both $\mu (X_n)$ and $\nu (X_n)$ are finite, since each finite union is contained in finite unions of sets from the corresponding cover. By [F2], $\mathrm{\mid }\nu \mathrm{\mid }(X_n)=\nu (X_n)<\mathrm{\infty }$. Thus [F3] applies. Fix any finite-valued representative $f$ of the nonempty class; it satisfies $\nu (E)={\int }_{E}fd\mu$ for every measurable $E$ and ${\int }_{X_n}\mathrm{\mid }f\mathrm{\mid }d\mu <\mathrm{\infty }$.

For $n,m\ge 1$, let $E_{n,m}=X_n\cap \{f\le -1\mathrm{/}m\}$, which is measurable by [F5]. On $E_{n,m}$, $-f\ge 1\mathrm{/}m$; because $f$ is integrable on $X_n$, [F6] and [F7] give ${\int }_{E_{n,m}}fd\mu \le -\mu (E_{n,m})\mathrm{/}m$. If $\mu (E_{n,m})>0$, the representation in step 1.1 would give $\nu (E_{n,m})<0$, contradicting positivity of $\nu$. Hence every $E_{n,m}$ is $\mu$-null.

The countable union $N={\bigcup }_{n,m\ge 1}{E}_{n,m}$ is measurable and $\mu$-null by [F5]. Since the sets $X_n$ cover $X$, $N=\{f<0\}$.

Define $\tilde{f}=\max \{f,0\}$. By [F7], $\tilde{f}$ is measurable; it is nonnegative, equals $f$ off $N$, and is zero on $N$. For every measurable $E$, absolute continuity gives $\nu (E)=\nu (E\setminus N)$ by [F4], while step 1.1 gives $\nu (E\setminus N)={\int }_{E\setminus N}fd\mu ={\int }_E\tilde{f}d\mu$. Thus $\tilde{f}$ represents the Radon–Nikodym class. The argument also shows that any finite-valued representative is nonnegative outside a null set.