# Lebesgue integration definition

In building up the theory of Lebesgue integration, at some point in the exposition we get something like:

For a non-negative measurable function $f$, the integral is defined as:

$$\int_E f \, d\mu = \sup\left\{\,\int_E s\, d\mu : 0 \le s \le f,\ s\ \text{simple}\,\right\}.$$

Now the fact that $f$ is measurable is usually made as an assumption, from the expositions that I have seen. However, I don't see why $f$ needs to be measurable. The set is well-defined and non-empty for any non-negative function, why is $f$ assumed to be measurable? Seems extraneous.

The definition makes sense whether $f$ is measurable or not. In fact for nonmeasurable $f$ it defines the lower integral of $f$. The lower integral does not enjoy many of the properties of the integral, though.
• Yes that's my point, it makes sense regardless of whether or not $f$ is measurable. Yet it seems that all the expositions of this definition I have seen make measurability of $f$ part of the definition. I am just wondering why this is the case. – iMath Sep 24 '13 at 16:09