Let $(X, \mathcal{M})$ be a measurable space. Let $\mu$, $\nu$ be measures defined on $\mathcal{M}$.
(a) For $A \in \mathcal{M}$ define $\lambda(A)=\mu(A)+ \nu(A)$. Prove that $\lambda$ is a measure.
(b) If $\mu \geq \nu$ there is a measure $\lambda$ such that $\mu = \nu + \lambda$.
(c) If $\nu$ is $\sigma$-finite then $\lambda$ (as in b) is unique.
(d) Give an example where $\lambda$ is not unique.
I already proved (a), (b) and (c), taking $$\lambda(E)=\sup\left\{ \mu(F)-\nu(F) : F \subseteq E, F \in \mathcal{M}, \nu(F)<\infty \right\},$$ in (b). Now I'm trying to find an example where $\lambda$ is not unique, obviously $\nu$ must not be $\sigma$-finite. I found the following hint, but I haven't figured out what's the example.
We know that we are looking for a non $\sigma$-finite example. And it's pretty obvious that the non-uniqueness will follow from the non-uniqueness of the solution to the equation $\infty+x=\infty$ (since $\lambda$ must be defined pointwise as $\lambda(M)=\mu(M)-\nu(M)$, which whenever this difference can be defined - which is whenever we don't have $\infty-\infty$).
So, let's start with the simplest non-$\sigma$-finite measures: Let $\mathcal{X}=\langle X,\mathcal{M}\rangle$ be defined by $X=\{\ast\}$ and $\mathcal{M}=\{\emptyset,\{\ast\}\}$. There's only one non-$\sigma$-finite measure $\nu$ on $\mathcal{X}$, so I'll let you take it from here.