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?

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.
