Lebesgue Integral over set of measure of zero Is it defined for a non-measurable, non-negative function? It would make sense, as clearly $s=0$ is a simple function, and $s\leq f$ for any $f$, whether it is measurable or not. So the clasically defined $$\int_E f\,d\mu=\sup\left\{\int_E s\,d\mu\ : \ s \text{ is a measurable simple function }, s\leq f \right\} $$
would exist and be equal to $0$, if $E$ is of measure $0$.
I don't think it is defined for non-measurable functions, but what I wrote above would seem like a convenient addition - couldn't it help with the whole idea of "almost everywhere" and make it easier/simpler, at least in some cases?
 A: I think you are thinking about this the wrong way. One should adopt the definition:
$$\int_E f d \mu := \int_{\mathbb{R}} f \chi_E d \mu.$$
Because the Lebesgue measure is complete, $f \chi_E$ really is a measurable function, so the right side is well-defined, and provides a suitable definition of the left side. Specifically you do indeed get $\int_E f d \mu = 0$.
An alternative is to make $E$ a measure space in its own right by equipping it with the $\sigma$-algebra $\mathcal{A}$ consisting of elements of $\mathcal{M}$ which are subsets of $E$, and the restriction of $\mu$ to $\mathcal{A}$. Then again because the Lebesgue measure is complete, $\mathcal{A}$ is actually the power set of $E$, and so $f$ is measurable as a function from this new measure space. The end result is the same.
A: No, because by the definition of $\int_E f\,d\mu$, it is limit of $\int_E s\,d\mu$ for $s$ simple measurable function, and a simple measurable function is by definition is a linear combination of indicator functions of measurable sets, and its the measurability of this sets thats ensures the measurability of the simple functions, which make all the rest make sense. If these sets are not measurable, you cannot conclude to the measurability of the simple "wannabe" measurable function.
