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 $\int_{E_0} f = \int_E f \cdot \chi_{E_0}$.