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(MATH3033)[2012](f)final~cs_zxxab^_64220.pdf
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Math 3033 (Real Analysis)
Fall 2012
Final Examination (Duration: 2 Hours)
Directions: This is a closed book exam. Work with complete details must be shown legibly to receive credits. Answers alone are worth very little.
Notations: . denotes the set of all real numbers. .. denotes the Lebesgue measure on ..
1. (20 marks) Consider the system of equations
..=..2.....cos(....)+..+1 and ..=2........2
Show that near ..=(..,..,..,..)=(0,0,0,0), (..,..) can be expressed as a differentiable function of (..,..) and find the values of ........ and ........ at (..,..)=(0,0).
Solution Define ..:.2.2 by ..(..,..)=(..,..) where ..=..2.....cos(....)+..+1, ..=2........2. We have ........=2......+..sin(....),........=..2....+..sin(....)+1,........=2....,........=.1
are continuous near (..,..)=(0,0). So .. is ..1 near (..,..)=(0,0).
Also det..(0,0)=|012.1|=.20.
By the inverse function theorem, near ..=(..,..,..,..)=(0,0,0,0), (..,..) can be expressed as a differentiable function of (..,..), i.e. ...1(..,..)=(..,..) exist and differentiable. Also, ( ........(0,0)........(0,0)........(0,0)........(0,0)) =(...1)(..(0,0))=(..(0,0)).1 =|012.1|.1=1.2|.1.1.20|=|121210|
So we can get ........(0,0)=12 and ........(0,0)=1
2. (15 marks) Determine the value of lim.........5...cos(..+..)....[1...,(..+1)...] with proof.
Solution Define ....:(0,2]. by ....(..)=.......5...cos(..+..)..[1...,(..+1)...](..)
Then ....(..) is measurable. ...(0,2], |....(..)|15...10 as ... So lim......(..)=0 Also, |....(..)|15.1=14=..(..) and 14....(0,2]=12<
By LDCT, lim.........5...cos(..+..)....[1...,(..+1)...] =lim..........(0,2] =lim..........(0,2] =0....(0,2] =0
3. (15 marks) Let .. be a measurable subset of [0,1] with ..(..)>0. Prove that there exist measurable subset ..1,..2,..3 of .. such that we have ..1..2..3=.. and ..1..2=..2..3=..3..1=. and ..(..1)=2..(..2)=..(..3).
Solution Define ..:[0,1]. by ..(..)=..(..[0,..]).
If 0..<..1, then 0..(..)...(..) =..(..[0,..])...(..[0,..]) =..(..(..,..])..((..,..]) =.....
This implies ...,..[0,1], |..(..)...(..)||.....|. By sandwich theorem, lim......(..)=..(..) i.e., .. is continuous. By the intermediate value theorem, since 0=..(0)<25..(..)<35..(..)<..(..)=..(1) ...1,..2[0,1] such that ..1<..2 and 25..(..)=..(..1),35..(..)=..(..2).
Let ..1=..[0,..1], then ..(..1)=..(..1)=25..(..)
Let ..2=..(..1,..2), then ..(..2)=..(..[0,..2])...(..[0,..1]) =..(..2)...(..1)=15..(..)
Let ..3=..(..2,1), then ..(..3)=..(..[0,1])...(..[0,..2]) =..(1)...(..2)=25..(..) So ..(..1)=2..(..2)=..(..3).
4. (15 marks) Let ..:[0,1][0,) be an increasing function. Let .. be a measurable subset of [0,1] with ..(..)=... Prove that ......[0,..]........
Solution Since we have ..=..(..)=..(..[0,..])+..(...[0,..])..=..([0,..])=..(..[0,..])+..([0,..]...)} ...(.