Lebesgue integration definition

Question

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.

score 5 · Answer 1 · answered Sep 24 '13 at 16:03

5

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.

answered Sep 24 '13 at 16:03

Umberto P.

52,165

1

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
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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. – Umberto P. Sep 24 '13 at 16:13
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I guess so, I just thought it was happening too often to be merely the writer's personal approach. Thanks for your replies. – iMath Sep 24 '13 at 16:16
UmbertoP. and @iMath relevant? For instance, in some sense the Lebesgue integral is "just" trying to prove the properties of integrals without using the continuity of f - so lebesgue instead of riemann removes continuity of $f$ and now you're talking about a lebesgue integral without measurability of $f$ ? Btw i posted an answer. – BCLC Feb 12 '23 at 07:26

score 0 · Answer 2 · answered Feb 12 '23 at 07:29

I could be missing something but this is based on the definition of integrals of simple functions and then the way you define integral of simple functions is like $\int (\sum a_i1_{A_i}) := \sum a_i \mu(A_i)$

If $s = \sum a_i1_{A_i}$ is not measurable then how are you going to do $\mu(A_i)$ ?

Or do you mean even if each simple $s$ is measurable, the $f$ in question actually need not be measurable?

Lebesgue integration definition

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