If $f$ is a bounded function on an interval E, and E has measure $0$ , Is $f$ measurable? What is the value of it's $\int_E f$? If $f$ is a bounded function on an interval E, and E has measure $0$, Is $f$ measurable? What is the value of $\int_E f$?
I have the question above in Royden Analysis 4e. 
Intuition suggests that f is measurable because E is , and that $\int_E f = 0$  because the lebesgue integral ignores intervals of measure 0.
If anybody could show me a more rigorous proof, I'd appreciate it. 
 A: The original exercise from Royden is as follows:

Let $E$ have measure zero. Show that if $f$ is a bounded function on $E$, then $f$ is measurable and $\int_E f = 0.$

Let $F$ be a measurable subset of $\mathbb R$. Then $f^{-1}(F)$ is a subset of $E$, and since subsets of measure zero sets have measure zero, $f^{-1}(F)$ has measure zero. In particular $f^{-1}(F)$ is measurable, so $f$ is measurable. Now modify the proof of Theorem 4 from section 4.2 (bounded measurable functions over a set of finite measure are integrable) to conclude that $\int_E f = 0$. 
A: I'm going to remove the restriction that $E$ is an interval, because otherwise it will be trivial.
Since $f$ is bounded on $E$, there exists $y \geq 0$ such that $|f\restriction_E(x)| < y$, so $|f\chi_E| < y \chi_E$, where $\chi_E$ is the characteristic function of $E$. Then
$$ \left| \int_E f\,d\lambda \right| \leq \int \left|f\chi_E \right|\,d\lambda
\leq \int y \chi _E\,d\lambda = y\lambda(E) = 0
$$
Since $\lambda$ is a complete measure, the above integrands are all integrable.
The measurability of $f$ cannot be determined if you only know that it is bounded on $E$.
