Explanation of additivity over domains of integration. 
Proposition: Let $f$ be a bounded measurable function on a set of finite measure $E$.  Suppose $A$ and $B$ are disjoint measurable subsets of $E$.  Then $$\int_{A \cup B} f =\int_A f + \int_B f.$$
  Furthermore, let $E_0$ be any measurable subset of $E$.  Then $\int_{E_0} f = \int_E f\cdot \chi_{E_0}$.

The proof of the above proposition relies on the proof of $\int_{E_0} f = \int_E f\cdot \chi_{E_0}$ which was left as an exercise.  Intuitively the formula makes sense.  Proving it on the other hand is where I'm a bit stuck.  Since $f$ is bounded I thought about defining two sets of simple functions $\phi_{n,E}, \phi_{n,E_0}, \psi_{n,E}, \psi_{n,E_0}$ and then applying the definition.
Is there a more efficient way of proving this assertion: $$\int_{E_0} f = \int_E f\cdot \chi_{E_0}$$
The above proposition in Royden requires this excercised to be proven in order to fully understand (or appreciate) the proof of the proposition:

Suppose $f$ is a bounded measurable function on a set of finite measure $E$.  Let $E_0 \subseteq E$.  Then $\int_{E_0} f = \int_E f \cdot \chi_{E_0}$.


It depends on how $\int_{E_0} f$ has been defined.  I don't have Royden on hand, but if he hasn't defined it by $\int_{E_0} f = \int_E f \cdot \chi_{E_0}$, a possible alternative definition could be phrased as follows:
Let $(E, \Sigma, \mu)$ be a measure space, and let $E_0 \in \Sigma$ be some measurable subset.  We define the measure space $(E_0, \Sigma_0, \mu_0)$ by $A_0 \in \Sigma_0 \iff A_0 = A\cap E_0$, for some $A \in \Sigma$.  Note that $\Sigma_0 \subset \Sigma$, so we can simply take $\mu_0$ to be the restriction of $\mu$ to $\Sigma_0$.
Now we have a definition for $\int_{E_0} f d\mu_0$, for nonnegative measurable f as $\sup\{\int_{E_0} s(x) d\mu_0\}$, where the supremum is taken over the integral of all nonnegative simple functions on $E_0$ satisfying $s \le f$.
We'd like to see that, using this definition, $\int_{E_0} f d\mu_0 = \int_E f \cdot \chi_{E_0} d\mu$.  To prove this, first prove it under the assumption that $f$ is simple and show that the result follows in general from this.
