Let $\mu$ and $\nu$ be finite measures on a measure space $(X,\mathcal{A})$. Show that there is a nonnegative measurable function $f$ on $X$ such that for all $E \in \mathcal{A}$, $$\int_E(1-f)d\mu=\int_Efd\nu$$
I have no clue how to show the above:
I started using below but I stuck
$$\mu(E)=\int_Efd\nu + \int_Efd\mu = \int_Efd(\nu+\mu)$$
defining $\psi=\nu+\mu$ , I have to show that $\mu\ll\psi$?
You want to show $\mu = f \cdot (\mu + \nu)$ for some $f.$
If the result were true for positive $\mu$ and $\nu,$ then $\mu^+ = f \cdot (\mu^+ + \nu^+)$ and $\mu^- = g \cdot (\mu^- + \nu^-),$ then $\mu = \mu^+ - \mu^- = f \cdot \mu^+ - g \cdot \mu^- + f \cdot \nu^+ - g \cdot \nu^-.$ Since $\mu^+ \perp \mu^-,$ it turn outs $fg = 0$ (one of them is always zero). Thus, $h = f - g$ satisfies $f \cdot \mu^+ = h \cdot \mu^+$ and similarly for $\mu^-, \nu^\pm.$ Hence, $\mu = h \cdot (\mu + \nu).$ Thus, suffices to show the result for positive measures.
The result is trivial for positive measures for you can do what you did already and the relation $\mu(N) \leq \psi(N)$ shows that $\mu$ is absolutely continuous relative to $\psi.$ Q.E.D