This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Because { I n } was an arbitrary countable collection of open intervals such that A ⊂ S ∞ n =1 I n , we proved that m * ( A + x ) is a lower bound of the set of (extended) numbers of which m * ( A ) is the infimum, thus m * ( A + x ) ≤ m * ( A ). Since A ⊂ R ,x ∈ R were arbitrary, we also have m * ( A ) = m * (( A + x ) + ( x )) ≤ m * ( A + x ) for all A ⊂ R ,x ∈ R . Equality follows. (d) Chapter 3, # 8 (p. 58). Solution. Since A ⊂ A ∪ B , we have m * ( A ) ≤ m * ( A ∪ B ), by monotonicity. By σsubadditivity (which implies finite subadditivity in conjunction with m * ( ∅ ) = 0), m * ( A ∪ B ) ≤ m * ( A ) + m * ( B ) = m * ( A ). (e) Chapter 3, # 9 (p. 64). This exercise is an immediate consequence of Royden’s Exercise 7. Solution. Assume E ⊂ R is measurable and let x ∈ R . It is easy (basically trivial) to verify that the map A 7→ A + x from subsets of R to subsets of R is a Boolean isomorphism; i.e., it preserves all set operations so that for all A ⊂ R A ∩ ( E + x ) = [( A x ) ∩ E ] + x, A ∩ ( E + x ) c = A ∩ ( E c + x ) = [( A x ) ∩ E c ] + x, thus, because m * is translation invariant, m * ( A ) = m * ( A x ) = m * (( A x ) ∩ E ) + m * (( A x ) ∩ E c ) = m * ([( A x ) ∩ E ] + x ) + m * ([( A x ) ∩ E c ] + x ) = m * ( A ∩ ( E + x ) + m * ( A ∩ ( E + x ) c ) proving E + x measurable. 3 (f) Chapter 3, # 10 (p. 64). (Not graded) Solution. Since E = E \ F ∪ ( E ∩ F ) and ( E \ F ) ∩ ( E ∩ F = ∅ , m ( E ) = m ( E \ F ) + m ( E ∩ F ) . Similarly m ( F ) = m ( F \ E ) + m ( E ∩ F ) . Since the sets E \ F,F \ E,E ∩ F are pairwise disjoint, m ( E )+ m ( F ) = m ( F \ E )+ m ( E \ F )+2 m ( E ∩ F ) = m ( F \ E ∪ E \ F ∪ ( E ∩ F ))+ m ( E ∩ F ) = m ( E ∪ F )+ m ( E ∩ F ) . (g) Chapter 3, # 11 (p. 64). Solution. Let (for example) E n = ( n, ∞ ) for n = 1 , 2 ,... . Then E 1 ⊃ E 2 ⊃ ··· , T ∞ n =1 E n = ∅ , but lim n →∞ m ( E n ) = ∞ 6 = 0 = m ( ∅ ). (h) Chapter 3, # 14 a (p. 64). There are several ways of doing this. If you look at the construction I gave in the notes on the Cantor set, it should be easy to determine m ( C n ) for all n . Use all properties of Lebesgue measurable sets that have been proved so far (in Royden). That is, use that intervals, open, closed, semiopen, semiclosed, are measurable and that a measure is a measure; that is, it is σadditive, hence also additive. Then the result can be obtained as a consequence of Proposition 14 in Royden. Solution. This is essentially done in my notes on the Cantor set. (i) Chapter 3, # 14 b (p. 64). This is a bit harder. Let’s call it optional, for now. 4. ?? Let A be the family of all subsets of R that can be obtained as a finite union of sets of one of the following forms: I. ( a,b ] for some a,b,∞ ≤ a < b < ∞ , II. ( a, ∞ ) for some a,∞ ≤ a....
View
Full Document
 Spring '11
 Speinklo
 Sets, Empty set, measure, Basic concepts in set theory, Lebesgue measure

Click to edit the document details