Is the Converse of the Radon-Nikodym Theorem true

Question

I'm curious as to whether the converse of the Radon-Nikodym theorem holds:

Converse Theorem: let $\mu.\nu$ be measures on $(\Omega,\Sigma)$ such that
$$\nu : A\mapsto \int_A f \ d\mu$$
for some measurable function $f$. Then

1. $\nu\ll\mu$, and
2. $\mu$ and $\nu$ are $\sigma$-finite.

Proof: the first part is easy, as
$$\mu(A) = 0\implies \int_A f \ d\mu = 0.$$

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Does the second part, regarding $\sigma$-finiteness, hold?

Answer 1

It does not. Take any measure $\mu$. Then

$$\mu(A)=\int_A~\mathrm{d}\mu.$$