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Integrability in the product space can follow from a property of the Nemytskii operator?

Nemytskii operators on Lebesgue spaces are funny objects with a lot of implicit structure. In fact, if $\mathcal N_f$ maps $L^p(\Omega)$ into $L^q(\Omega)$, then $f$ must necessarily satisfy the ...
Hannes's user avatar
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1 vote

For what sets does the Lebesgue Differentiation Theorem hold in one dimension?

In chapter 2 of Federer's book, Geometric Measure Theory, it is developed the theory of Vitali's coverings. That theory is exactly the answer to your question
Luigi De Pascale's user avatar

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