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Nov 20, 2011 at 20:08 answer added Florian timeline score: 4
Oct 19, 2011 at 17:05 comment added Pietro Majer To sketch the standard argument: Since countable intersections of full-measure sets is a full-measure set, a countable set of integrable functions admits a common full-measure set $S\subset\Omega$ of differentiation (that is, such that the first formula you wrote holds for every function of the family and for every $x\in S$). Consider in particular the functions $u_q(y):=|u(y)-q|$, $q$ a rational number. Then it is easy to see that for all $x\in S$ the second formula you wrote is true. Check for instance Wheeden- Zygmund, Measure and integral.
Oct 19, 2011 at 14:12 comment added Gerald Edgar Such a point is called a "Lebesgue point" of $u$. Look for that in your textbook. Here it is in Wikipedia: en.wikipedia.org/wiki/Lebesgue_point
Oct 19, 2011 at 13:30 history asked Florian CC BY-SA 3.0