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On a measure space $(X, \Sigma, \mu)$, the usual way to build a general Lebesgue Integral is to use the set of simple functions define as finite sum of linear combinations of characteristic functions of measurable sets (i.e. elements of $\Sigma$) and then use the non-negative ones to define the integral functional on them, and then use them to approximate from below the measurable positive functions to get their Lebesgue Integral.

I was wondering what happens if we just use the simple functions define as finite sum of linear combinations of characteristic functions of measurable sets with finite measure. The set of such simple functions has the nice properties of being a Riesz space, verifying the Stone axiom, and allowing a natural definition on it of the Integral as a linear functional, on the contrary to the previous set.

I am aware that if we do not suppose the measure $\mu$ to be $\sigma$-finite, we do not get all the integrable measurable functions (see this question Definition of the Lebesgue integral in terms of simple functions with finite measure support) but do we get the usual space $L^1(X,\mu)$ of functions with finite integral of their absolute value?

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Yes, because for $f\in L^1$, we can approximate the positive part from below only by simple functions that have a smaller and hence also finite integral. Hence, these simple functions must have finite support, so explicitly requiring them to have finite support does not matter.

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  • $\begingroup$ Thank you. That was just under my nose and I could not see it during one day! $\endgroup$
    – brunoh
    Apr 29, 2020 at 23:13

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