Does $f \in L(E) \rightarrow$ that $f$ is measurable? So this arose because a proof I was going over stated that since $f\in L(E)$ then $f$ is measurable. This wasn't immediately clear to me. I guess that Lebesgue integration, at least in Royden, is defined explicitly for a measurable function. So is this true only by default?
If this is the case then I am still curious about what the attempted Lebesgue integral would represent, that is:
$\sup _{\phi(x) \le f} \{\int \phi\}$
I know the conclusion that a function can be represented  as the pointwise limit of a sequence of simple functions relies on that function being measurable. But still, I wonder if a Lebesgue type integration of a non-measurable function wouldn't have some meaning? Thanks for your time!
 A: For simplicity let's suppose we're working with a bounded function $f$ on a set $M$ of finite measure.  Then you can define "upper" and "lower" integrals of $f$ over $M$ as the inf of $\int \phi$ over measurable $\phi \ge f$, and the sup of $\int \phi$ over measurable $\phi \le f$.  If $f$ is measurable these coincide, but for nonmeasurable $f$ they don't have to.  In the case where $f$ is an indicator function, these are the outer and inner measures of the corresponding set.
A: To add to Robert's answer, I would say that the integral is defined only for a measurable function. As you know, we have a sequence of simple functions increasing pointwise to a given measurable function. And the integral is defined to be the limit of the integrals of these simple functions.
If the function is non-measurable, then this approximation will not exist. (e.g. characteristic function of a non-measurable set...there is no obvious way to construct a sequence for this function).
If you work with the equivalent definition that $\int f = \sup_{\phi \leq f}\int \phi$ then as pointed out above, it will not be equal to the "upper" integral for a non-measurable function. The interpretations are given above.
