In my current lecture we have the following proposition:
Let $(\Omega, \mathcal{A})$ be a measurable space and let $\nu, \mu$ be measures on $(\Omega, \mathcal{A})$ such that $\nu \ll \mu$ with $\mu$-density $f$. Then for every measurable positive $h: \Omega \rightarrow \mathbb{R}$ holds
$$\begin{equation*} \tag{$*$} \int h \,d\nu = \int f h \,d\mu \end{equation*}$$
In particular we have for every measurable function $h: \Omega \rightarrow \mathbb{R}$
$$\int |h| \,d\nu < \infty \Longleftrightarrow \int |hf| \,d\mu < \infty$$
In the case of a finite result $(*)$ holds as well.
Now naturally I'm wondering why we need the assumption of non-negativity. But I can't come up with a counterexample. Where can this fail?
I'm assuming that $\nu$ and $\mu$ are positive measures, things are different for signed or complex measures.
The positivity (non-negativity) assumption is one of two simple (to state) ways to ensure that both integrals appearing in $(\ast)$ make sense. The other simple way is assuming integrability of $h$ (with respect to $\nu$).
But the non-negativity assumption is much simpler to prove. It is immediate from the definitions (and the fact that a product of measurable functions is measurable) that both sides of $(\ast)$ make sense for a measurable $h \geqslant 0$. And then it is straightforward to show that both sides are equal for such $h$.
Furthermore, it is easy to derive the integrable case from it (decompose the real and imaginary parts of an integrable function into differences of their respective positive and negative parts and use linearity). And by the same method, that $(\ast)$ holds whenever $\int h\,d\nu$ makes sense.
On the other hand, proving the integrable case (that both sides of $(\ast)$ make sense for $\nu$-integrable $h$, and then that they are equal) without first proving the non-negative case seems to be at least difficult. I think it is possible, but I don't see a promising strategy. The natural way passes through the non-negative case.