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Hi everyone: Let $f$ be a function defined on an open set of $\mathbb{R}^{N}$ $(N\geq1)$. Is there any difference between the following two statements? 1) $f$ is locally integrable 2) $f$ admits a locally integrable majorant. Thanks for your reply.

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  • $\begingroup$ Well, (2) appears to allow $f$ to be nonmeasurable. How exactly do you define "a majorant", by the way? $\endgroup$ – user147263 Jun 30 '14 at 1:44
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Yes, there is. A function with locally integrable majorant does not have to be measurable, e.g. it can be the characteristic function of a Lebesgue non-measurable set. If it is measurable then it depends on what you mean by "majorant". If it is just $f\leq g$ then $f=-1/x^2$ is non-integrable with $g=0$ locally integrable. However, if $f$ is measurable, $|f|\leq g$ and $g$ is locally integrable then so is $f$, essentially by definition.

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