Let $(X, M)$ be a measurable space. If $\nu$ and $\lambda$ are measures on $(X,M)$, define measure $\mu = \nu + \lambda$ to be $\mu(E) = \nu(E) + \lambda(E)$.
Assume $\mu$ and $\nu$ are measures on $(X,M)$ such that $\mu(A) \geq \nu(A)$ for all $A \in M$.
1) If $\nu$ is $\sigma$-finite, show that if $\lambda$ is a measure on $(X,M)$ such that $\mu = \nu + \lambda$ then $\lambda$ is unique. Also give a counterexample to show that this result is false without $\nu $ being $\sigma$-finite
2) Show that there always exists a $\lambda$ such that $\mu = \nu + \lambda$.
I was thinking that to show $\lambda$ is unique, we need to show $\lambda_1(E) = \lambda_2(E)$ for all $E$ if $\nu(E) + \lambda_1(E) = \nu(E) + \lambda_2(E)$. I thought if the sum is finite, then we are done. If the sum is infinite, we somehow want to show $\nu(E)$ is finite. But then I got stuck, I didn't know how to get this from the $\sigma$-finiteness and $\mu \geq \nu$ conditions. I also had a hard time coming up with counter examples. Can someone provide a solution to this problem? Thanks.