D chapter 3 8 p 58 solution since a a b we have m a m

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(d) Chapter 3, # 8 (p. 58). Solution. Since A A B , we have m * ( A ) m * ( A B ), by monotonicity. By σ -subadditivity (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
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(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, semi-open, semi-closed, are measurable and that a measure is a measure; that is, it is σ -additive, 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. The sets in the list above include R , which is of the second form with a = -∞ . “Finite unions” include unions with no sets at all, putting the empty set into the mix, and unions involving only one set. So A contains R , , all intervals of the form ( a, b ], all intervals of the form ( a, ), all unions of two sets of the type just mentioned, all unions of three such sets, etc. Prove A is an algebra. Hint: The only difficult part (i.e., non-trivial part) is to see that A ∈ A implies A c ∈ A . For this purpose you should prove first: If A, B ∈ A , then A B ∈ A . Once you have this, you can proceed by induction. Let us call the sets listed in I, II basic sets. Go by induction on n , proving: If A is a union of n basic sets, then A c ∈ A . But if you find a better way, by all means use it.
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