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# Assume next α 0 we have 1 f x 0 if and only if f x

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Assume next α = 0. We have 1 /f ( x ) > 0 if and only if { f ( x ) < ∞} and this last set is measurable because f is measurable * . Finally, assume α < 0. Then { x E : 1 /f ( x ) > α } = E is measurable. Second solution. Deﬁne f n ( x ) = max( n,f ( x ) + 1 n ) for x E , n = 1 , 2 , 3 ... . Because f is measurable, all functions f n are measurable. Moreover, f n takes values in the interval [1 /n,n ], in which y 7→ 1 /y is continuous. It follows that 1 /f n is measurable for every n = 1 , 2 ,... , being the composition of a measurable function with a continuous one. It is clear (or trivial to prove) that { f n } converges pointwise to f and that { 1 /f n } converges pointwise to 1 .f , hence 1 /f is measurable. Comment For some credit I needed to see at least that if α > 0 then 1 /f > α is equivalent to f < 1 . 8. For this exercise one must, of course, assume (as one usually does) the existence of some non-measurable set. It was proved in class that if f n : R R is measurable for all n N , then sup n N f , the function deﬁned by (sup n N f )( x ) = sup { f n ( x ) : n N } is measurable. Assume now that A is an arbitrary set and for each α A , f α : R R is measurable. Is it always true that sup α A f α , the function deﬁned by (sup α A f α )( x ) = sup { f α ( 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.
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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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Assume next α 0 We have 1 f x 0 if and only if f x and...

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