Your counterexample is correct. You have in addition to assume that $\nu$ is also $\sigma$-finite (and apply Radon-Nidokym-Lebesgue, as you say correctly). Folland refers to the example of conditional expectation from probability theory, where $\mu$ is a (probability and hence) finite[!] measure. In this case of course, $\nu$ is also finite (and hence, $\sigma$-finite). Perhaps, when generalizing the statement, the additional assumption on $\nu$ (which is no longer automatic) was just forgotten.
Let $\mathcal A$ be an algebra on a set $X$, $\mu_0$ a premeasure on $\mathcal A$,
and $\mathcal M$ the $\sigma$-algebra generated by $\mathcal A$.
Let $\mu^*$ be the Caratheodory outer measure induced by $\mu_0$, $\mu$ the restriction of $\mu^*$ on $\mathcal M$.
Let $\mathcal C$ be the set of all $\mu^*$-measurable sets, $\bar{\mu}$ the restriction of $\mu^*$ on $\mathcal C$.
Suppose $\mu_0$ is $\sigma$-finite, i.e. there exists a sequence $E_n, n = 1, 2, \cdots$ of members of $\mathcal A$ such that $X = \cup_{n=1}^{\infty} E_n$, $\mu_0(E_n) \lt \infty$ for all $n$. I claim that $(X, \mathcal C, \bar{\mu})$ is the completion of $(X, \mathcal M, \mu)$.
Let $\bar{\mathcal M}$ be the completion of $\mathcal M$ with respect to $\mu$.
Since $\mathcal M\subset \mathcal C$, $\bar{\mathcal M}\subset \mathcal C$.
Hence it suffices to prove that $\mathcal C \subset \bar{\mathcal M}$.
Since $\mu_0$ is $\sigma$-finite, it suffices to prove that if $E\in \mathcal C$ and $\bar{\mu}(E) \lt \infty$, then $E\in \bar{\mathcal M}$.
By the definition of $\bar{\mu}$, for each integer $n \ge 1$, there is a sequence $A_j, j= 1, 2, \cdots$ such that $\sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$, where $A_j \in \mathcal{A}$, $E \subset \bigcup_{j=1}^\infty A_j.$ Let $F_n = \bigcup_{j=1}^{\infty} A_j$. Then $E \subset F_n$ and $\mu(F_n) \le \sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$.
Let $F = \cap_{n=1}^{\infty} F_n$. Then $F\in \mathcal M$, $E \subset F$, and $\bar{\mu}(E) \le \mu(F) \le \mu(F_n) \lt \bar{\mu}(E) + 1/n$ for all $n\ge 1$. Hence $\bar{\mu}(E) = \mu(F)$.
Similarly there exists $G\in \mathcal M$ such that $F - E \subset G$ and $\mu(G) = \bar{\mu}(F - E)$ = 0.
Then $E = (F - G) \cup (E\cap G)$, $F - G \in \mathcal M$, and $E\cap G$ is a subset of the $\mu$-null set $G$. Hence $E \in \bar{\mathcal M}$. This completes the proof.
Now consider the family $\mathcal A$ of finite disjoint unions of intervals of the form $[a, b)$ or $(-\infty, c)$, where $-\infty \lt a\lt b\le \infty$ and $c\in \mathbb R$.
It is elementary and well known that $\mathcal A$ is an algebra on $\mathbb R$ and there is a unique premeasure $\mu_0$ on $\mathcal A$ such that $\mu_0([a, b)) = b - a$ whenever $a$ and $b$ are finite.
Then $\mathcal M$ and $\mathcal C$ defined above are the families of Borel sets and Lebesgue measurable sets in $\mathbb R$ respectively and you get the picture.
Best Answer
No, you have equality.
This is because you can find a nondecreasing sequence of nonnegative, $\mathcal M_0$-measurable functions $(\phi_n)$ such that $\phi_n(x)\to f(x)$ for all $x\in X$. By the monotone convergence theorem applied to $\mu$ and $\mu_0$ it follows that $\int fd\mu=\int fd\mu_0$.