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# Solution a let u be open in r 2 claim there exists a

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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 ) ) 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 ) ) 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. With the claim established, the rest is trivial. Let U be open in R 2 . then U = [ n N ( I n × J n ) for intervals I n ,J n , hence h - 1 ( U ) = [ n N h - 1 ( I n × J n ) is measurable being a countable union of measurable sets. (b) The easiest way to prove h is measurable is to use the characterization by open sets: h is measurable if and only h - 1 ( U ) is measurable for every open subset U of R . Now h - 1 ( U ) = { x R : φ ( f ( x ) ,g ( x )) U } = { x R : ( f ( x ) ,g ( x )) φ - 1 ( U ) } . Here we may have to think how to apply part (a) and I admit that the fact that the h of this part has nothing to do with the h of part (a) can be confusing. Let us deﬁne H ( x ) = ( f ( x ) ,g ( x )) so H : R R 2 and h = φ H . If I,J are open intervals, then H - 1 ( I × J ) = { x R : ( f ( x ) ,g ( x )) I × J } = { x R : f ( x ) I } ∩ { x R : g ( x ) J } = f - 1 ( I ) g - 1 ( J ) so that H - 1 ( I × J ) is measurable for all open rectangles I × J . By part (a), H - 1 ( W ) is measurable for all open sets W R 2 . Returning now to our open subset U of R , since φ is continuous, φ - 1 ( U ) is open in R 2 , hence h - 1 ( 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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Solution a Let U be open in R 2 Claim There exists a...

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