This is Exercise 15c Chapter 1 from Folland. I know that if there is an $F\subseteq E$ with $0<\mu(F)<\infty$, then $\mu_0(E)=\mu(E)$. If $\mu(E)=\infty$, and $F\subseteq E$,$\mu(F)<\infty$ imply $\mu(F)=0$, then $\mu_0(E)=0$.
I tried
$$\nu(E)=
\begin{cases}
0, & \text{if }\mu(E)=\mu_0(E)\\
\infty, & \text{otherwise}.
\end{cases}$$
This satisfies $\mu=\mu_0+\nu$, but it is not a measure. If $\{E_n\}\subseteq\mathcal M$ is a collection of disjoint sets, and $\nu(E_n)=\infty$ for some $n$, and $\nu(E_m)=0$ for $m\ne n$ then
$$\sum_{n=1}^\infty \nu(E)=\infty,$$
but there are $F\subset\bigcup E_n$ such that $0<\mu(F)<\infty$, since such sets are contained in any $E_m$ distinct from $E_n$; so,
$$\nu\left(\bigcup_{n=1}^\infty E_n\right)=0.$$
Letting $\nu(E)=\infty$ if and only if $\mu(E)=\infty$ doesn't work either. We only need a countable collection of sets of finite measure, such that the union has infinite measure to see that $\nu$ is not a measure in this case. So, I'm stuck.
Best Answer
I'm not sure if this proof is correct, but it seems to be. Can someone verify??
Define a set $E \in \mathcal{M}$ to be $\sigma$-finite for $\mu$ if there exists a collection $\{F_n\}_{n \in \mathbb{N}} \subset \mathcal{M}$ such that $ \mu(F_n) < \infty$ (for all $n$) and $E=\bigcup_n F_n$.
Define the measure $\nu: \mathcal{M} \to \{0,\infty\}$ as
$\nu(E)= \left\{ \begin{array}{} 0 & \text{if } E \text{ is } \sigma\text{-finite for } \mu\\ \infty & \text{if } E \text{ is not } \sigma \text{-finite for } \mu\\ \end{array} \right.$
It's easy enough to see that $\mu = \mu_0+\nu$. This is because if $E$ is $\sigma$-finite for $\mu$ then $\mu(E)=\mu_0(E)=\mu_0(E)+0=\mu_0(E)+\nu(E)$; and if $E$ is not $\sigma$-finite for $\mu$ then it necessarily holds that $\mu(E)=\infty= \nu(E)=\mu_0(E)+\nu(E)$.
To prove that $\nu$ is a measure, consider a collection of disjoint sets $\{E_n\}_{n \in \mathbb{N}} \subset \mathcal{M}$. If all of the $E_n$ are $\sigma$-finite for $\mu$, then $\bigcup_n E_n$ is $\sigma$-finite for $\mu$ (easy proof), and thus $\nu(\bigcup_nE_n)=0=\sum_n \nu(E_n)$. On the other hand, if any one of the sets $E_n$ is not $\sigma$-finite for $\mu$, then $\bigcup_n E_n$ cannot possibly be $\sigma$-finite for $\mu$ (easy proof), and thus $\nu(\bigcup_nE_n)=\infty=\sum_n \nu(E_n)$.