Proposition 10

Let $\mu$ and $\nu$ be two positive measures on T. The following conditions are equivalent: a) The locally negligible sets are the same for $\mu$ and $\nu$. b) The bands generated by $\mu$ and $\nu$ in $\mathscr{M}(T)$ are identical. c) One has $\nu = g \cdot \mu$, where $g$ is locally $\mu$-integrable and $g(t) > 0$ locally almost everywhere for $\mu$.

The conditions a) and b) are equivalent by Cor. 2 of Th. 2 of No. 5. If they are satisfied, then $\nu = g \cdot \mu$ and $\mu = h \cdot \nu$, where $g$ (resp. $h$) is positive and locally integrable for $\mu$ (resp. $\nu$). Therefore (No. 4, Prop. 8) $hg$ is locally $\mu$-integrable and $\mu = (hg) \cdot \mu$. It follows (No. 3, Cor. 2 of Prop. 3) that $hg$ is equal to 1 locally almost everywhere for $\mu$, so that $g(t) > 0$ and $h(t) = 1/g(t)$ locally almost everywhere for $\mu$. Conversely, suppose that $\nu = g \cdot \mu$ with $g(t) > 0$ locally almost everywhere for $\mu$;

since $(1/g)g$ is defined locally almost everywhere and is locally $\mu$-integrable, $1/g$ is locally $\nu$-integrable and $(1/g) \cdot \nu = \mu$ (No. 4, Prop. 8).