Exercise 7.3.2

Recall that Thomae’s function

t ( x ) = { 1  if  x = 0 1 n  if  x = m n Q { 0 }  is in lowest terms with  n > 0 0  if  x Q

has a countable set of discontinuities occurring at precisely every rational number. Follow these steps to prove t ( x ) is integrable on [ 0 , 1 ] with 0 1 t = 0 .

(a)
First argue that L ( t , P ) = 0 for any partition P of [ 0 , 1 ] .
(b)
Let 𝜖 > 0 , and consider the set of points D 𝜖 2 = { x [ 0 , 1 ] : t ( x ) 𝜖 2 } . How big is D 𝜖 2 ?
(c)
To complete the argument, explain how to construct a partition P 𝜖 of [ 0 , 1 ] so that U ( t , P 𝜖 ) < 𝜖 .

Answers

(a)
Because the irrationals are dense, for any interval there must be some point in that interval where t ( x ) = 0 , which is also the minimum possible value of t ; hence L ( t , P ) = 0 .
(b)
This is the size of the set of rational numbers which, when expressed in lowest terms as m n , have n 2 𝜖 . This must be finite; let the size of D 𝜖 2 be denoted as | D 𝜖 2 | .
(c)
Informally, we’ll set up the partition to “isolate” each point in D 𝜖 2 . Define the partition as
P 𝜖 = { d + k 2 : d D 𝜖 2 } { d k 2 : d D 𝜖 2 } { 0 , 1 }

where k = 𝜖 ( 2 | D 𝜖 2 | ) . (If some of these isolating intervals overlap, then decrease k until they don’t; this can be formalized by requring k be the minimum of what it is currently and the smallest distance between two points in D 𝜖 2 .) Then

U ( t , P 𝜖 ) < 𝜖 2 + k | D 𝜖 2 | = 𝜖

where the first term comes from all partition intervals that aren’t isolating an element in D 𝜖 2 , and the second term comes from the elements in D 𝜖 2 .

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2022-01-27 00:00
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