ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

30 ANALYSIS QUALS
Problem 6: Define for each n = 1, 2, 3, . . . , the Cantor-like set Cn as [0, 1]
with its central open interval of length 1
2n · 1
3 removed, then with the two
central open intervals of length 1
2n · 1
32 removed from the remaining two
closed intervals and so on (at the jth stage, 2j−1 intervals of length 1
2n · 1
3j are removed), continuing with j = 1, 2, 3, . . ..
(a) With µ=Lebesgue measure, show that µ([0, 1]\ ∞
n=1 Cn) = 0.
(b) Show that if E is a subset of [0, 1] which is not Lebesgue measurable
(you may assume such an E exists without proof), then for some n ≥ 1,
E ∪ Cn fails to be Lebesgue measurable.
(c) Use part (b) to show that there is a continuous, strictly increasing
function f : R → R with f(R) = R and a Lebesgue measurable set
A ⊂ R such that f(A) is not Lebesgue measurable.
Solution. (a) Cn is constructed by removing intervals of total length 2−n 2
3
at step j. But
1
3
∞
j=0
j 2
=
3
1
1 − 2
3
= 3,
so Cn has a total of 2 −n removed. Thus, Cn has total measure 1−2 −n ,
so
n Cn has measure 1, and hence, µ([0, 1]\ ∞ n=1 Cn) = 0.
(b) E not measurable implies that there exists ε > 0 and a set A such that
µ(E) < µ(E∩A)+µ(E∩A c )−ε. Choose n such that µ(Cn) > 1−ε/10.
Then
µ(E ∩ Cn) < µ(E) < µ(E ∩ Cn ∩ A) + µ(E ∩ Cn ∩ A c ) − 4ε/5,
so E ∩ Cn is not measurable.
(c) Let f be a strictly increasing homeomorphism from C0 to Cn, where
n as in part (b). Such a function exists: both sets are homeomorphic
to {0, 1} N by considering at each point in a Cantor set, whether it is
in the left or right side at each stage of removing central intervals.
Thus, just as the strictly monotonic Cantor-Lebesgue function can be
constructed for C0 onto [0, 1], so too can such a function be created
for Cn, and taking an appropriate composition yields f.
Extend f linearly to all of R. Then f −1 (E∩Cn) is a subset of C0, which
has measure zero. Therefore f −1 (E ∩ Cn) is measurable. However, by
part (b), E ∩ Cn is not measurable.
Problem 8: Suppose f : U → C is a holomorphic function with intU |f| 2 < ∞
where U = {z ∈ C : 0 < |z| < 1} and the integral is the usual R 2 -area
integral. Prove that f has a removable singularity at z = 0.
Solution. Let w be a point in the annulus around z = 0 of radii ε and 1/2,
for 0 < ε < 1/2. Then by Cauchy integral formula, we have that
f(ξ)
f(ξ)
f(w) =
dξ −
ξ − w ξ − w dξ.
|ξ|=1/2
|ξ|=ε
j−1·