I've run into an issue with an exercise and I think that more assumptions are needed in order for the result to be true. Here is the exercise:
Let $\nu:\mathcal{M}\to\mathbb{R}$ be a signed measure and $(X,\mathcal{M},\mu)$ be $\sigma$-finite. Show that there is a function $f\in L^{1}(X,\mu)$ for which $$\int_{X}g\,d\nu=\int_{X}g\cdot f\,d\mu \quad \forall g\in L^{\infty}(X,\mu)$$
It is assumed that $X$ is complete with respect to $\mu$ (an assumption that seems irrelevant for this problem). Do we not need to assume that $X$ is finite with respect to $\nu$ and that $\nu \ll \mu$? I ask because it seems that ultimately this boils down to whether or not I can express $\nu$ as $$\nu(E)=\int_{E}f\,d\mu,$$ which by (a corollary to) the Radon-Nikodym Theorem I must have these properties. If this is the case, then I can just use the RNT, the Kantorovitch Representation Theorem, and a previous exercise which says that if $\nu \ll \mu$ and $f\geq0$ is measurable then $$\int_{X}f\,d\nu=\int_{X}f\,\frac{d\nu}{d\mu}d\mu.$$ I figure either this or I am possibly misunderstanding something about the problem statement. Any hints or tips are greatly appreciated.
You are right. Note that $$\tag1\int_{X}g\,d\nu=\int_{X}g\cdot f\,d\mu \quad \forall g\in L^{\infty}(X,\mu)$$ implies that $\nu\ll \mu$, so that hypothesis cannot be avoided.
The typical counterexample would be to take $X=\mathbb R$, with $\mu$ the Lebesgue measure, and $\nu=\delta_0$, the Dirac measure. Then you cannot have $(1)$, since $$ g(0)=\int_X g\,d\nu $$ is not even well-defined on $L^\infty(\mathbb R)$.