The semifinite portion of a measure $\mu$ Let $\mu$ be a measure and define $\mu_1$ such that $\mu(E)=\mu_1(E)$ for $\mu(E)$ finite.  And for $\mu(E)$ infinite definite $\mu_1$ such that: 
(i) if $E$ contains finite subsets of arbitrarily large measure then $\mu_1(E)=\infty.$
(ii) and if not then $\mu_1(E)=0$. 
Prove $\mu_1$ is a measure.$$$$
So I'm really having trouble seeing how countable (even finite) additivity holds for sets $E$ such that $\mu(E)=\infty$ but which don't contain finite subsets of arbitrarily large measure.  For instance if $E$ can be partitioned into $A\cup B$ with $\mu(A)=1$ and $\mu(B)=\infty$, then $$0=\mu_1(E)=\mu_1(A\cup B)\neq\mu_1(A)+\mu_1(B)=1 + 0 = 1.$$

I agree with you that $\mu_1$ as defined in Royden-Fitzpatrick does not yield a measure. An example that shows this is obtained by considering the measure on $\mathbb{N}$ defined on all subsets $A \subseteq \mathbb N$ by
$$
\mu(A) = \begin{cases} 0, & \text{if } A = \emptyset, \\ 1, & \text{if } A = \{0\} \\ \infty, & \text{otherwise.} \end{cases}
$$
For this measure $\mu$ we have $\mu_1(\emptyset) = 0$, $\mu_1(\{0\}) = 1$ and $\mu_1(A) = 0$ for all other $A \subseteq \mathbb{N}$, so $\mu_1$ is obviously not a measure.
The usual definition of the semi-finite version $\mu_{\rm sf}$ of a measure $\mu$ on $(X,\Sigma)$ is given by
$$
\mu_{\rm sf}(E) = \sup\{\mu(A) \mid A \subseteq E \text{ measurable, } \mu(A) \lt \infty\} \quad \text{for }E \in \Sigma
$$
In this definition it is clear that $\mu_{\rm sf}$ is non-negative, that $\mu_{\rm sf}(\emptyset) = 0$ and that $\mu(E) =\mu_{\rm sf}(E)$ whenever $E \in \Sigma$ has finite $\mu$-measure.  Using this, it is not hard to check that $\mu_{\rm sf}$ is $\sigma$-additive on $\Sigma$, hence it is a measure on $(X,\Sigma)$. Moreover, $\mu_{\rm sf}$ is semi-finite and we have $\mu_{\rm sf} = \mu$ if and only if $\mu$ is semi-finite.
One can also show that $\mu_{\rm sf}$ and $\mu$ have the same integrable functions (up to $\mu_{\rm sf}$-null sets) and that $\mu_{\rm sf}$ is complete only if $\mu$ is complete.
