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x ) : α A } is measurable? Justify your answer. Solution. The answer is NO . For example, let A be a non-measurable subset of R and for α A define f α : R R by f α ( x ) = 1 if x = α , 0 otherwise. Then f α = 0 a.e., hence measurable for each α A . However, sup α f α = χ A is not measurable. 9. (a) Let h : R R 2 having the following property: If I, J are open intervals in R , then h - 1 ( I × J ) = { x R : h ( x ) I × J } is measurable. Prove that the inverse image under h of every open subset of R 2 is measurable. That is, prove that if W is open in R 2 , then h - 1 ( W ) is measurable in R . Three hints: Not every open subset of R 2 is a rectangle. This is sort of obvious, but people under the pressure of exams sometimes lose their minds and begin to assume nonsense just because it seems to help. The second hint is that one might consider proving that every open set in the plane is a countable union of rectangles. A third hint is, do not try to show it is a union of pairwise disjoint rectangles, that’s clearly impossible. * For measurability it suffices to have { f < α } measurable for all α R ; one gets { f < ∞} measurable as an easy consequence; for example because { f < ∞} = S n { f < n } . 3

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(b) Let E be a measurable subset of R and let f, g : E R be measurable. Let φ : R 2 R be continuous. Prove: the function h : E R defined by h ( x ) = φ ( f ( x ) , g ( x )) is measurable. Hint: Use Part (a). Solution. (a) Let U be open in R 2 . Claim: There exists a countable family of open rectangles I n × J n such that U = [ n =1 I n × J n . I consider proving this claim the essential part of this exercise , the rest is trivial. To prove the claim we use the fact that the set D = { ( x, y ) Q × Q : ( x, y ) U } is countable, thus can be written in the form of a sequence, namely D = { ( x n , y n ) : n N } . Since U is open, for each n N we can find r > 0 such that B (( x n , y n ) , r ) U . Let r n be the largest possible such r ; that is r n = sup { r > 0 : B (( x n , y n ) , r ) U . We can assume that U 6 = R 2 , otherwise U = R × R and h - 1 ( U ) is measurable. We can also assume U 6 = , so D is actually infinite countable. The fact that ∅ 6 = U 6 = R 2 implies 0 < r n < . One sees that B (( x n , y n ) , r n ) U (trivial but requiring an argument). Let I n = ( x n - ( r n / 2 , x n + ( r n / 2)), J n = ( y n - ( r n / 2 , y n + ( r n / 2)), then ( x n , y n ) I n × J n U , hence [ n N ( I n × J n ) U. Conversely, let ( x, y ) U . Then there is δ > 0 such that B (( x, y ) , δ ) U . Because D is dense in U , there is n N such that d (( x, y ) , ( x n , y n )) = p ( x - x n ) 2 + ( y - y n ) 2 < δ/ 4 . Consider B (( x n , y n ) , 3 δ/ 4). If ( u, v ) B (( x n , y n ) , 3 δ/ 4), then d (( u, v ) , ( x, y )) d (( u, v ) , ( x n , y n )) + d (( x n , y n ) , ( x, y )) < 3 δ 4 + d 4 = δ proving ( u, v ) B (( x, y ) , δ ) U . By the definition of r n , r n 3 δ/ 4. Thus | x - x n | ≤ d (( x n , y n ) , ( x, y )) < δ 4 < 3 δ 8 r n 2 , proving that x n I n . Similarly one sees that y n J n . thus ( x, y ) I n × J n . This proves [ n N ( I n × J n ) U, thus [ n N ( I n × J n ) = U.
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• Spring '11
• Speinklo
• Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

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