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6 let f n n 1 be a sequence of real valued functions

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6. Let { f n } n =1 be a sequence of real valued functions on a subset E of R . Define the following sets: For each triple ( n, m, k ) N 3 define F n,m,k = { x E : | f n ( x ) - f m ( x ) | < 1 k } . For each pair ( N, k ) N 2 define G N,k = \ n N,m N F m,n,k . For each k N , define H k = [ N =1 G N,k . Finally, define Z = \ k =1 H k . Prove: x Z if and only if the sequence of real numbers { f n ( x ) } converges. Solution. Assume first x Z . We prove { f n ( x ) } is a Cauchy sequence of real numbers. For this let ² > 0 be given. Let k N be such that 1 /k < ² . Since x Z , x H k by the definition of Z . By the definition of H k , there is N N such that x G N,k . By the definition of G N,k , x F n,m,k for all n, m N , thus | f n ( x ) - f m ( x ) | < 1 /k < ² if n, m N . It follows the sequence is Cauchy, hence converges. Conversely, let x R be such that { f n ( x ) } converges. Then it is Cauchy and for every k N there is N N such that | f n ( x ) - f m ( x ) | < 1 /k if n, m N , thus there is N N such that x F n,m,k for n, m N , hence x G N,k . This proves that for every k N , there is N N such that x G N,k , thus for every k N , x S N =1 G N,k = H k . Thus x H k for all k , hence x Z . 2
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7. Let E be a measurable subset of R and let f : E [0 , ] be measurable. Prove that the function 1 /f defined by (1 /f )( x ) = 1 /f ( x ) if 0 < f ( x ) < if f ( x ) = 0 0 if f ( x ) = for x E , is measurable. Solution. I’ll present two solutions. First solution, the obvious one. Let α R . It suffices to prove that { x E : 1 /f ( x ) > α } is measurable. We have to distinguish three cases: α > 0, α = 0, α < 0. If α > 0, then 1 /f ( x ) > α if and only if either f ( x ) = 0 (then 1 /f ( x ) = ) or if f ( x ) < 1 . But since 1 /α > 0, the set { f = 0 } ⊂ { f < 1 } . Thus { x E : 1 /f ( x ) > α } = { x E : f ( x ) < 1 } , the latter set being measurable because f is measurable, it follows that { x E : 1 /f ( x ) > α } is measurable if α > 0. 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 (
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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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