Translation Invariance of Lebesgue Measure
[11/64%]Let denote a measure that assigns finite measure to bounded -sets. Furthermore, suppose that is translation invariant: for all . Then there is such that for all .
Let be such that , and let . There exists an open interval such that
Let contain an open interval. Then there exists such that is contained in .
Let be such that . Then there exists such that is contained in .
Let be a dense subset of . Then is dense for any .
Let be an irrational number.
Let . Then is a dense subset of .
Let B=\left\{ n+m \xi : n, m \in \mathbb {Z}, n even \right\}. Then is a sense subset of .
Let C=\left\{ n+m \xi : n, m \in \mathbb {Z}, n odd \right\}. Then is a sense subset of .
For write iff , where is a fixed irrational number as in the previous exercise. Then is an equivalence relation, and hence may be partitioned into disjoint equivalence classes.
We now invoke ( ) to form a set consisting of exactly one element from each of the equivalence classes in the previous exercise. We will now show that .
There exist Borel subsets of .
Let be a Borel set. Then cannot contain any nonzero elements of , where is the set in Exercise 4.3.6.
By (b), there cannot exist an open interval containing the origin that is contained in , hence .
From (c), we have .
If and are distinct elements of , then and are disjoint.
, the countable union being disjoint.
If , then for each , hence . Therefore, since the assumption that leads to an absurdity, it must be the case that , and hence there exists a subset of that fails to be a Lebesgue set.
In conjunction with the previous exercises, it is not possible to have Lebesgue measure defined for every subset of . That is, a measure extension of [where is the identity function] from the semiring consisting of and the bounded rsc intervals to fails to exist.
Argue by contradiction for a quick answer: if such a measure extension to exists then the previous work would force , contradicting what we expect of length.
Let be such that . We will show that there is with by following the outline below. The set appearing below is the set from Exercise 4.3.7, consisting of the numbers , where and is a fixed irrational number.
is monotone and translation invariant.
, the union being disjoint.
There exists such that .
Let . Then and .
We may actually generalize further and state that any subset with [ not necessarily in ] contains a subset outside of .
By (b), we have ; and by (a) and part (d) of Exercise 4.3.8, , while at the same time ; invoke the relevant part of the Approximation Theorem.
Let , let be a dense subset of , and suppose that for every . Then either or . This exercise is independent of the previous exercises.