Math 131A/2 Winter 2002 Handout #1a Instructor: E. Eros, MS 6931. Lecture Meeting Time: MWF 2:00PM-2:50PM Location: MS 5117 Recitation TA: to be announced, MS 5117 T 2:00P-2:50P Oce hours (tentative): TF 4-5. Text Fundamental Ideas of Analysis by Michael
Elementary Analysis
Kenneth A. Ross
Selected Solutions
Angelo Christopher Limnios
EXERCISE 1.2
Claim:
P (n) = 3 + 11 + + (8n
Proof :
5) = 4n2
n 8n 2 N
By induction. Let n = 1. Then 3 = 4(1)2 (1) = 3, which will serve as the
induction basis. Now for the in
4.1. Simple Functions
115
B(i) of finite measure and ai for which (4.1.1) holds, not that all such B(i)
must have finite measure.) Some examples of simple functions on R are the
step functions, where each B(i) is a finite interval.
Any finite collection o
5
p
L Spaces; Introduction to Functional Analysis
The key idea of functional analysis is to consider functions as points in a
space of functions. To begin with, consider bounded, measurable functions
on a finite measure space (X, S , ) such as the unit in
62
General Topology
2.5.7. Theorem Let (Sn , dn ) for n = 1, 2, . . . , be a sequence of complete
metric spaces. Then the Cartesian product n Sn , with product topology, is
complete with the metric d of Proposition 2.4.4.
Proof. A Cauchy sequence cfw_xm m
2.3. Complete and Compact Metric Spaces
45
determined, and
0.d1 d2 d3 . . . dm1 dm 9999 . . . = 0.d1 d2 d3 . . . dm1 (dm + 1)0000. . . .
In all other cases, the digits d j are unique given x.
In practice, we work with only the first few digits of decimal
88
Measures
Proof. Let us first show that is finitely additive. Suppose (a, b] =
1in (ai , bi ] where the (ai , bi ] are disjoint and we may assume the intervals
are non-empty. Without changing the sum of the (ai , bi ]), the intervals can
be relabeled so
162
L p Spaces; Introduction to Functional Analysis
5.3.3. Cauchy-Bunyakovsky-Schwarz Inequality for Inner Products. For
any semi-inner product (,) and f, g H ,
|( f, g)|2 ( f, f )(g, g).
Proof. By Proposition 5.3.2, f can be multiplied by any complex c w
300
Convergence of Laws and Central Limit Theorems
9.4.2. Proposition (a) The characteristic functions of normal laws on R are
given by
(b)
(c)
ei xu N (m, 2 )(d x) = exp(imu 2 u 2 /2).
e xu N (0, 1)(d x) = exp(u 2 /2)
x 2n N (0, 1)(d x) = (2n)!/(2n n!)
f
10
Foundations; Set Theory
of relation. Its inverse f 1 is not necessarily a function. In fact, a function
f is called 11 or one-to-one if and only if f 1 is also a function. Given a
relation E, one often writes x E y instead of x, y E (this notation is u
4.5. Daniell-Stone Integrals
143
Remark. For any vector space L of functions, the constant function 0 belongs
to L. Thus if L is a vector lattice, then for any f L, the functions f + :=
max( f, 0) and f := min( f, 0) belong to L, and are nonnegative. So t
5.4. Orthonormal Sets and Bases
169
space, a Hamel basis is just a basis, but in an infinite-dimensional Banach
space, such as a Hilbert space, there are no obvious Hamel bases, although
they can be shown to exist with the axiom of choice. In analysis, on
8.3. Laws of Large Numbers
261
happen almost surely (a.s.). Thus a sequence Yn of (real) random variables
is said to converge a.s. to a random variable Y iff P(Yn Y ) = 1. (The
set on which a sequence of random variables converges is measurable, as
shown
256
Introduction to Probability Theory
a -finite measure space, although in this example it was a countable product
of finite measure spaces. By contrast, the countable product of probability
spaces will again be a probability space. Here are some definit
References
219
Polish journal Studia Mathematica beginning in 1936. His career was cut short by the
1939 invasion (in 1940, papers of his appeared in Annals of Math. in the United States
and in Revista Ci. Lima, Peru). When Studia was able to resume publi
232
Measure, Topology, and Differentiation
Recall that f is continuous from the right at x iff f (x + ) = f (x) (as in
3.1 and 3.2). Thus 1[0,1) is continuous from the right, but 1[0,1] is not, for
example. For any finite, nonnegative measure , and a R, f
9.5. Uniqueness of Characteristic Functions
309
theorem, the characteristic function of P is
j
2
dr tr 2 = exp(Dt, t)/2).
f P (t) = exp
r =1
Thus P has the characteristic function of a law N (0, C) as defined, so a law
N (0, C) exists and by the uniquene
74
General Topology
to show that f is a homeomorphism, first note that it is 11 since for any
x
= y in X , by complete regularity, there is a g G(X ) with g(x)
= g(y)
so f (x)
= f (y). Let W be any open set in X . To show that the direct image
f (W ) :
358
Conditional Expectations and Martingales
9. Give an example of a martingale cfw_X n n1 which converges in probability
but not a.s. Hints: Let cfw_u n n1 be i.i.d. uniformly distributed in [0, 1] (with
distribution = Lebesgue measure). Let X 1 0. If X
102
Measures
3.3.2. Proposition For any measure space (X, S , ), and the collection S
of sets almost equal to sets in S , we have S = S N (), the smallest
-algebra including S and N ().
Proof. If E B N () and B S , then E\B N () and B\E N (),
so E = (B (E
258
Introduction to Probability Theory
decreasing sequence of sets in A and for some > 0, P(A j ) for all j,
we must show j A j
=
.
Let P (0) := P on A. For each n 1, let (n) := m>n m . Let A(n) and P (n)
be defined on (n) just as A and P were on . For
322
Convergence of Laws and Central Limit Theorems
Now, more concrete conditions for convergence will be brought in. For
any real random variable X , the truncation of X at 1 will be defined by
X if |X | 1
X 1 :=
or
0 if |X | > 1.
Recall that for any real
92
Measures
/2. Now (B ) (B ) < , and since on S ,
(A) = (B ) (B \A) (Bk ) (B \A)
(Bk ) /2 = (Bk ) /2 (since Bk A and by Lemma 3.1.6)
(B ) (A) .
Letting 0 gives (A) (A), so (A) = (A). Or if An are disjoint
with (An ) < , then
(A An ) =
(A An ) = (A)
(A
292
Convergence of Laws and Central Limit Theorems
total variation measure, | := + + 0. We say that Pn P0 in total
variation iff as n , |Pn P0 |(S) 0, or equivalently
sup |(Pn P0 )(A)| 0.
A B
Convergence in total variation, however, is too strong for most
128
Integration
cfw_x1 , . . . , xn , and for each j n,
cfw_x: d(x, xi ) > d(x, x j )
cfw_x: d(x, xi ) d(x, x j ).
f n1 (cfw_x j ) =
i< j
jin
The latter is an intersection of an open set and a closed set, hence a Borel set,
so f n is measurable. Clearly,
126
Integration
Now let I = [0, 1] with its usual topology. Then I I with product topology
is a compact Hausdorff space by Tychonoffs Theorem (2.2.8). Such spaces
have many good properties, but Theorem 4.2.2 does not extend to them (as
range spaces), acco
146
Integration
found 0 gn L with gn 1 A . Thus M E . Clearly (A) < for each A
M. Now f 1 (1, )g 1 (1, ) = ( f g)1 (1, ), and f 1 (1, )
g 1 (1, ) = ( f g)1 (1, ), so if C M and D M then C D M
and C D M. Since 1C\D = 1C 1CD , it follows that C\D E . Then
112
Measures
following Kolmogorov (1933), as a (countably additive) measure on a -algebra S of
subsets of a set X with P(X ) = 1. Among the relatively few researchers in probability
who work with finitely additive probability measures, a notable work is t
Math 131 A, Lecture 3
Real Analysis
Final Exam
Instructions: You have three hours to complete the exam. There are ten problems, worth a
total of one hundred points. Write your solutions in the space below the questions. If a question
is in multiple parts,
Object1
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MATH 131A-2 MIDTERM 01 SOLUTION
Exercise 1. (3 + 3 + 4 = 10 pts) Write down the sup and inf for the following sets.
(1) cfw_(2)n | n Z, (2) cfw_x | x2 < 3, (3) cfw_(1)n + n1 | n N.
Proof. (1) inf
= and sup= +.
(2) inf = 3 and sup = 3.
(3) inf = 1 and sup