MATH 525a ASSIGNMENT 7 SOLUTIONS
FALL 2008
Prof. Alexander
Chapter 2
(46) For xed y , D (x, y ) = 1 only for one x, so D (x, y ) (dx) = 0. Hence
0.
For xed x, D (x, y ) = 1 for exactly one y , so D (x, y ) (dy ) = 1. Hence
1.
Now
D d d =
D d d =
()
( )(D)
MATH 525a ASSIGNMENT 1 SOLUTIONS
FALL 2013
Prof. Alexander
(1)(a) Let
A = cfw_all maps N cfw_0, 1 (sequences of 0s and 1s)
B = cfw_all maps N A (sequences of sequences)
A map f B corresponds to an innite array with the sequence f (n) as nth row. Then
card
MATH 525a ASSIGNMENT 2 SOLUTIONS
FALL 2013
Prof. Alexander
(5) M = (E ). Let G = cfw_E M : E (F ) for some countable F E. If E G then
E (F ) for some countable F E , so E c (F ), so E c G . Thus G is closed under
complements.
If E1 , E2 , . G then each En
MATH 525a SAMPLE MIDTERM SOLUTIONS
FALL 2013
Prof. Alexander
(1)(a) In the text.
1
(b) Consider En = cfw_x : f (x) > M + n . We have
f d
M (En )
M+
En
1
n
(En ),
so (En ) = 0. Therefore (cfw_x : f (x) > M ) = (n En ) = 0, that is, f M a.e.
(2)(a) Let A
MATH 525a ASSIGNMENT 8 SOLUTIONS
FALL 2013
Prof. Alexander
Chapter 3
(11)(a) For a single f L1 () we have f d
d so by Theorem 3.5, given > 0 there
exists > 0 such that (E ) < = | E f d| < . Thus a single integrable function is
uniformly integrable.
For cf
MATH 525a ASSIGNMENT 9 SOLUTIONS
FALL 2008
Prof. Alexander
Chapter 3
(33) Let G(x) = F (x+), so G F . Let a < b and
increasing. Therefore
> 0. Then G(b ) F (b), since F is
F (b) F (a) G(b ) G(a)
= G (a, b ])
(G )ac (a, b ]) (G )ac is the absolutely cont.
MATH 525a ASSIGNMENT 10 SOLUTIONS
FALL 2008
Prof. Alexander
Chapter 5
(12)(recommended problem) Let X be a normed vector space and M a proper closed subspace.
(a) Let K, = 0. Then y M if and only if y M, so for all x,
x + M = inf cfw_ (x + y ) : y M = | i
MATH 525a
SAMPLE FINAL EXAM SOLUTIONS
FALL 2013
Prof. Alexander
(1)(a) Clearly , X M. Now E M implies either (i) A E or (ii) E A = which
implies either (i) A E c = or (ii) A E c . Either way this shows E c M. Suppose
E1 , E2 , . M. If all Ei A = , then (i
MATH 525a MIDTERM 1 SOLUTIONS
FALL 2013
Prof. Alexander
(1) For x < y in A, we have s(x) y (by denition of successor), so Ix and Iy are disjoint.
Hence cfw_Ix : x A is a collection of disjoint open intervals so is at most countable (because
each interval
MATH 525a ASSIGNMENT 5 SOLUTIONS
FALL 2008
Prof. Alexander
Chapter 2
(20) We may assume all the functions are real-valued, otherwise consider real and imaginary
parts separately. Then gn + fn 0 and gn fn 0, so by Fatous Lemma,
g + lim inf
fn = lim
gn + li
MATH 525a FINAL EXAM SOLUTIONS
FALL 2013
Prof. Alexander
(1) Neither implies the other. To show (i) = (ii), consider f with a jump, for example
f ( x) =
0, x [0, 1 ],
2
1, x ( 1 , 1].
2
1
If f = g a.e., then for every > 0 there are points x ( 2 , 1 + ) wh
MATH 525a ASSIGNMENT 6 SOLUTIONS
FALL 2008
Prof. Alexander
Chapter 2
(33) By denition of lim inf, there is a subsequence cfw_fnk for which fnk lim inf fn .
Since fnk f in measure, there exists a further subsequence cfw_fmj of cfw_fnk for which
fmj f a.
MATH 525a ASSIGNMENT 4 SOLUTIONS
FALL 2008
Prof. Alexander
Chapter 2
(3) Since fn is measurable for all n, so is g = lim sup fn lim inf fn , by 2.6 and 2.7. Hence
cfw_x : limn fn (x) exists = g 1 (0) is a measurable set.
(4) The sets (r, ] generate BR so
MATH 525a ASSIGNMENT 3 SOLUTIONS
FALL 2008
Prof. Alexander
Chapter 1
(18)(a) By the denition of , there exist cfw_Aj , j 1 with Aj A, E Aj , and
1
(Aj ) (E ) + . Let A = Aj . Then A A , E A and (A) (Aj ) =
1
j =1
1
1 (Aj ) (E ) + .
(b) Suppose rst that E