Real Analysis Qualifying Exam; August, 2009.
Work as many of these ten problems as you can in four hours. Start each problem on a new sheet of paper. #1. Evaluate the iterated integral
0 0
x exp(-x2 (1 + y 2) dx dy. (Justify your answer.) #2. Let f C[0,
Problemsheet 9 in Real Analysis Math607, Fall 2013
Solutions
Problem 1. (Old Qualier). If f L1 (0, ), dene
g (s) =
est f (t)dt,
0 < s < .
0
Prove that g (s) is dierentiable on (0, ) and that
g (s) =
test f (t)dt,
0 < s < .
0
Proof. The problem follows di
Problemsheet 10 in Real Analysis Math607, Fall 13
Solutions
Problem 1. Problem 11/page 92.
Let be a positive measure on (X, M). A collection of functions (f )A
is called uniformly integrable if for every > 0 there is a > 0 such that
| E f d| < for all A w
Problemsheet 11 in Real Analysis Math607, Fall 13
Solutions
Problem 1 (Old Qualier) Let f be increasing on [0, 1] and
g (x) = lim sup
h0
f (x + h) f (x h)
,
2h
for 0 < x < 1.
Prove that if A = cfw_x (0, 1) : g (x) > 1 then
f (1) f (0) m (A).
Hint: Vitalis
Problemsheet 7 in Real Analysis Math607, Fall 2013
Solutions
Problem 1 Assume that (fn ) and fn f uniformly.
a) If (X ) < , then f L1 and limn fn d = f d.
b) If (X ) = , then conclusion of (a) may fail.
Proof. (a) Choose n0 N so that |fn (x) f (x)| 1 for
Problemsheet 6 in Real Analysis Math607, Fall 2013
Solutions
Notation: (X, M, ) measure space.
L = cfw_f : X [, ] : f M-measurable
F = cfw_f : X [, ] : f M-measurable and simple
L+ and F + are the positive function in L and F respectively.
Problem 1. Assu
Problemsheet 2 in Real Analysis Math607, Fall 2013
Solutions
Problem 1. Using the fact that BR is generated by the open intervals show
that:
BR = M(cfw_[a, ) : a rational).
Proof. Recall: Let E1 , E2 P (X ), for some set X . In order to show that
M(E1 ) =
Problemsheet 3 in Real Analysis Math607, Fall 2013
Solutions
Problem 1. Problem 8/Page 27
Proof. To show for (Ej ) M
(a)
lim inf Ej lim inf (Ej )
and provided that (
j =1 (Ej )
(b)
lim sup Ej lim sup (Ej )
j
j
<
j
j
To proof (a), note that for all n
Problemsheet 4 in Real Analysis Math607, Fall 2013
Solutions
Problem 1. Dene
n
(Q)
A
For A =
=
[ai , bi ) Q :
i=1
n
i=1 [ai , bi )
n N, cfw_ai , bi : 1 i n Q cfw_
.
and a1 < b1 < a2 < . . . bn
Q with a1 < b1 < a2 < . . . bn put
n
bi ai .
0 (A) =
i=1
a) A
Problemsheet 5 in Real Analysis Math607, Fall 2013
Solutions
Problem 1. Let (X, M) be a measurable space and fn : X R be
measurable for n N. Show that
a) lim inf n fn is measurable.
b) cfw_x X : limn fn exists M.
Proof. (a) Note that for a R and x X
lim
Problemsheet 8 in Real Analysis Math607, Fall 2013
Solutions
Problem 1 (Old Qualier) Let f : [0, 1] R be integrable (with respect to
Lebegues measure) and non negative. Dene
G = cfw_(x, y ) : 0 x 1, 0 y f (x).
Show that G is measurable in R R and that
1
f
Real Analysis Qualifying Exam, May, 2008
1. Prove that if E is a closed linear subspace of L2 (0, 1) and each element in E is bounded
(i.e. f L (0, 1) for all f E ), then E is nite dimensional.
2. Let fn :=
2n
k
k=1 (1) ( k2n1 , 2k ] .
n
Prove that fn 0 w
Problems in Real Analysis (Math607) - Solutions
Prof.: Thomas Schlumprecht
Problem 1 (Old Qualier) Let f : [a, b] R be absolutely continuous.
Show that m (f (cfw_x : |f (x)| 1) b a.
Proof. First consider an interval I , and denote its closure by I . Now
f
Problemsheet 1 in Real Analysis Math607, Fall 09, Solution
Problem 1. Let f : X Y Prove that (A, B Y ) a) f -1 (A B) = f -1 (A) f -1 (B) b) f -1 (A B) = f -1 (A) f -1 (B) and give examples for the following situations c) f -1 (f (A) = A, for some A X, d)
Problems in Real Analysis Math607, Fall 09
Solutions Problem 1. Let M1 be a -algebra on X and M2 a -algebra on Y . Show that E = cfw_A B : A M1 , B M2 is an elementary system. Proof: i) = E ii) If A1 A2 and B1 B2 are in E then (A1 A2 ) (B1 B2 ) = (A1 B1
Problems in Real Analysis Math607, Fall 09
Solutions Problem 1. Define
n
A(Q) = For A =
[ai , bi )Q : n N, cfw_ai , bi i=1 n i=1 [ai , bi ) Q with - 0 (A) =
: 1 i n Qcfw_ and a1 < b1 < a2 < . . . bn . a1 < b1 < a2 < . . . bn put
n
bi - ai .
i=1
a) A(Q) is
Problems in Real Analysis Math607, Fall 09
Solutions Problem 1. Show that the Lebesgue measure is translation invariant. I.e. show that for x R and A L it follows that x + A L and m(x + A) = m(A). Proof. Let x R. It is clear that for all intervals I, m(I)
Problems in Real Analysis Math607, Fall 09
Solutions Problem 1 Problem 19/Page 59. Assume that (fn ) and fn f uniformly. a) If (X) < , then f L1 and limn fn d = b) If (X) = , then conclusion of (a) may fail. f d.
Proof. (a) Choose n0 N so that |fn (x) - f
Problems in Real Analysis Math607, Fall 09
Solutions Problem 1 (Old Qualifier) Let f : [0, 1] R be integrable (with respect to Lebegues measure) and non negative. Define G- = cfw_(x, y) : 0 x 1, 0 y f (x). Show that G- is measurable in R R and that
1
m(G-
Problems in Real Analysis Math607, Fall 09
Solutions
Problem 1 (Old Qualier). If f L1 (0, ), dene
g (s) =
est f (t)dt,
0 < s < .
0
Prove that g (s) is dierentiable on (0, ) and that
g (s) =
test f (t)dt,
0 < s < .
0
Proof. The problem follows directly fr
Problems in Real Analysis Math607, Fall 09
Solutions
Problem 1 Problem 1/88. Let be a signed measure on a measurable
space (X, M). If (Ej ) is increasing in M then
(
Ej ) = lim (Ej ).
j
j =1
If (Ej ) is decreasing in M and | (E1 )| < , then
(
Ej ) = lim
Problems in Real Analysis (Math607) - Solutions
Prof.: Thomas Schlumprecht
Problem 1 (Old Qualier) Let f be increasing on [0, 1] and
g (x) = lim sup
h0
f (x + h) f (x h)
,
2h
for 0 < x < 1.
Prove that if A = cfw_x (0, 1) : g (x) > 1 then
f (1) f (0) m (A)
Problemsheet 1 in Real Analysis Math607, Fall 2013
Solutions
Problem 1. Let f : X Y
Prove that (A, B Y )
a) f 1 (A B ) = f 1 (A) f 1 (B ) and f 1 (A B ) = f 1 (A) f 1 (B )
b) For a family (A ) P (X ), show that
f 1
f 1 A and f 1
A =
f 1 A
A =
and give e