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Exercise 7.3.2
Recall that Thomae’s function
has a countable set of discontinuities occurring at precisely every rational number. Follow these steps to prove is integrable on with .
- (a)
- First argue that for any partition of .
- (b)
- Let , and consider the set of points . How big is ?
- (c)
- To complete the argument, explain how to construct a partition of so that .
Answers
- (a)
- Because the irrationals are dense, for any interval there must be some point in that interval where , which is also the minimum possible value of ; hence .
- (b)
- This is the size of the set of rational numbers which, when expressed in lowest terms as , have . This must be finite; let the size of be denoted as .
- (c)
-
Informally, we’ll set up the partition to “isolate” each point in
. Define the partition as
where . (If some of these isolating intervals overlap, then decrease until they don’t; this can be formalized by requring be the minimum of what it is currently and the smallest distance between two points in .) Then
where the first term comes from all partition intervals that aren’t isolating an element in , and the second term comes from the elements in .