2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – iMath Sep 24 '13 at 16:09
  • $\begingroup$ My guess is that most books avoid nonmeasurable functions entirely (other than perhaps to point out that they exist) and so don't bother to point out that things like "lower integrals" make sense. $\endgroup$ – Umberto P. Sep 24 '13 at 16:13
  • $\begingroup$ I guess so, I just thought it was happening too often to be merely the writer's personal approach. Thanks for your replies. $\endgroup$ – iMath Sep 24 '13 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.