Math630-3 - Real Analysis Math 630 Homework Set#3 Chapter 3 by Bobby Rohde Problem 18& Show that(v does not imply(iv in Propostion 18 by

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Unformatted text preview: Real Analysis - Math 630 Homework Set #3 - Chapter 3 by Bobby Rohde 9-21-00 Problem 18 & Show that (v) does not imply (iv) in Propostion 18 by constructing a function f such that { x : f ( x ) > 0} = E , a given non-measurable set, and such that f assumes each value at most once. & Counter-Example Let E = the standard non-measurable set on [0, 1] Let f ( x ) = & x , x & E ¡ x , x ¢ E defined on domain [0, £ ) Thus { x : f ( x ) > 0} = E , and f ( x ) = ¤ has at most one solution for any ¤ , namely ¤ or - ¤ . Hence { x : f ( x ) = ¤ } is measurable ¥ ¤ but { x : f ( x ) > 0} is not measurable. Problem 19 & Let D be a dense set of real numbers. Let f be an extended real-valued function on & such that { x : f ( x ) > & } is measurable ¡ & ¢ D . Show that f is measurable. & Proof We wish to show that { x : f ( x ) > & } is measurable ¡ & ¢ D is equivalent to the condition { x : f ( x ) > & } is measurable ¡ & . Take £ ¤ D then define A n = { x : f ( x ) > ¥ n , with ¥ n ¢ ( £- 1 ¦¦¦¦¦ n , £ ) & D }. We know that ( £- 1 ¦¦¦¦¦ n , £ ) & D is non-empty ¡ n since, D is dense. Consider & n § 1 ¨ A n © & n § 1 ¨ ¡ x : f ¢ x £ ª £ « 1 ¦¦¦¦¦ n ¤ = { x : f ( x ) ¬ £ }. But { x : f ( x ) ¬ £ } © & n § 1 ¨ A n since ¡ n , ¥ n < £ . Thus & n § 1 ¨ A n = { x : f ( x ) ¬ £ } which means that { x : f ( x ) ¬ £ } is measurable and hence by Proposition 18, { x : f ( x ) > £ } is measurable, so { x : f ( x ) > & } is measurable ¡ & . QED Problem 20 & Show that the sum and product of two simple functions are simple. & Proof £ ¤ A & B ¥ ¤ A ¦ ¤ B ­ A & B = ¡ 1, if x ¢ A & B 0, if x ¤ A & B ­ A ® ­ B = ¡ 1, if x ¢ A and x ¢ B 0, if x ¤ A or x ¤ B But x ¢ A and x ¢ B ¯ x ¢ A & B and, x ¤ A or x ¤ B ¯ x ¤ A & B, thus ­ A & B § ­ A ® ­ B , QED. MATH630-3.nb 2 & ¡ A & B ¢ ¡ A £ ¡ B ¤ ¡ A ¥ ¡ B & A & B = ¡ 1, if x ¡ A & B 0, if x ¢ A & B & A £ & B ¤ & A ¥ & B = & A £ & B ¤ & A ¢ B = ¡ 1, if x ¡ A or x ¡ B 0, if x ¢ A and x ¢ B But x ¡ A or x ¡ B ¦ x ¡ A & B and, x ¢ A and x ¢ B ¦ x ¢ A & B, thus & A & B § & A £ & B ¤ & A ¥ & B . QED....
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This note was uploaded on 03/23/2010 for the course MATH 515 taught by Professor Staff during the Spring '08 term at Iowa State.

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Math630-3 - Real Analysis Math 630 Homework Set#3 Chapter 3 by Bobby Rohde Problem 18& Show that(v does not imply(iv in Propostion 18 by

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