Math 395 (Fall 2010). Honors Analysis I
1. Measure spaces
1.1. Denitions.
(a) A -algebra on a set X (or a -algebra of subsets of X ) is a collection of subsets of
X satisfying the following properties:
(1) If A1 , A2 , A3 , . . . , then Aj .
j =1
(2) If A
Math 395 (Fall 2010). Solutions for Problem Set 9
9.1. It is clear that if we prove all the required statements for the functions Re(F ) and
Im(F ), the statements for F itself will follow. Thus we assume from now on that F is
real-valued. First we prove
Math 395 (Fall 2010). Solutions for Problem Set 8
8.1. To show that (i) implies (ii), suppose that S is an orthonormal basis of H and S H
is an orthonormal subset such that S S . If S = S , choose v S with v S . By
the denition of an orthonormal basis, th
Math 395 (Fall 2010). Solutions for Problem Set 6
6.1. First we check that | is a norm. If f BC (X ), then 0 |f | < (the
second inequality holds because f is bounded). Moreover, if |f | = 0, then |f (x)| = 0
for each x X , whence f = 0. Next, if C, then |
Math 395 (Fall 2010). Solutions for Problem Set 5
5.1. Let (X, d) be a metric space and x, y X with x = y . By the denition of a metric,
d(x, y ) > 0. Write = d(x, y )/2. Then B (x) B (y ) = : otherwise there exists z X
such that d(x, z ) < and d(y, z ) <
Math 395 (Fall 2010). Solutions for Problem Set 4
4.4. There are three assertions in the problem. Lets label them (a), (b), (c).
(a) Let us check that L is an algebra. First of all X L because for any A A we
have X A = A and X c A = A = , and because we a
Math 395 (Fall 2010). Solutions for Problem Set 3
3.4. We proceed exactly as indicated in the hint. Let us assume that the assertion of the
problem fails. The rst part of the hint proves that we have AN |g | d for all N 1,
where AN = x X |g | > N . By hom
Math 395 (Fall 2010). Solutions for Problem Set 2
2.6. Note rst that lim fn (x) exists if and only if the two limits lim Re(fn (x) and
n
n
lim Im(fn (x) both exist. Since the functions Re(f ) and Im(f ) are measurable whenever
n
f is, and since the inters
Math 395 (Fall 2010). Solutions for Problem Set 1
1.8. Fix y Br (x) and let = |y x|. By assumption, < r, i.e., r > 0. The
triangle inequality implies that Br (y ) Br (x). This means that Br (x) is open.
Next, showing that B r (x) is closed is the same as
Math 395 (Fall 2010). Midterm exam solutions
M.1. To show that FX is right-continuous it suces to check that if cfw_an is a den=1
creasing sequence of real numbers converging to some a R, then FX (an ) FX (a) as
n . Let us write An = cfw_ X ( ) an for e
Math 395 (Fall 2010). Final exam solutions
F.1. (a) Let > 0 be given. Applying homework problem 2.11 to the measure space
(a, b], B , m), where B is the -algebra of Borel subsets of (a, b] and m is the Lebesgue
measure on B , we see that there exists > 0
Math 395 (Fall 2010). Honors Analysis I
9. Topological and metric spaces
9.1. Topological spaces. Like a measurable space, a topological space is a set together
with a collection of subsets satisfying certain axioms.
Denitions 9.1. (1) Let X be a set. A c
Math 395 (Fall 2010). Solutions for Problem Set 10
10.1. First we show that h1 := f +g and h2 := f g are holomorphic on U . Fix an arbitrary
z0 U . We will show that h1 (z0 ) = f (z0 )+ g (z0 ) and h2 (z0 ) = f (z0 ) g (z0 )+ g (z0 ) f (z0 ).
(As usual, t