Let $(\mathbb{R}^n,\mathbb{\beta}^n)$ be a measurable space and $S$ is complete sufficient statistics. Let's consider a measure $$ \nu(B) = \int\limits_{S^{-1}(B)}h(x)\mu^n(dx), $$ where $h(x)$ is a measurable function.
The question is why $\nu^{\pm}(dt) = \phi^{\pm}(t)\nu(dt)$ is the $\sigma-$finite measure, where $\phi-$measurable function.