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Differentiation holds along families shrinking nicely
Statement
Assume the Axiom of Countable Choice (The Axiom of Countable Choice ()).
Let , let , and suppose that for each there is a family shrinking nicely to with constant . Then for almost every , hence
Facts & Assumptions
Given: The Axiom of Countable Choice, a locally integrable function , a set , and for each a family shrinking nicely to .
Shrinking nicely means that for each there is a constant such that for every . (A family shrinking nicely to a point)
Almost every point of is a Lebesgue point. (Almost every point is a Lebesgue point of a locally integrable function)
Proof
Let be a Lebesgue point of , as supplied by [L2]. For every [L1, L2, given, algebra] , [L1] gives
Because is a Lebesgue point, the right-hand side of step 1.1 tends to [step 1.1] as . Therefore the left-hand side also tends to .
Using [step 2.1, algebra] step 2.1 immediately gives the second limit as well.
Step 3.1 holds at every Lebesgue point of , hence for almost every .
Depends on
Used by
Dependency tree · two levels
17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Theorem 3.21 (standard reference, not scraped)
- Terence Tao, An Introduction to Measure Theory, Exercise 1.6.15 (standard reference, not scraped)