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
294
Then
Convergence of Laws and Central Limit Theorems
f d Pm( j) g d Pm( j) < for all j, so
lim sup f d Pm( j) lim inf f d Pm( j) 2.
j
j
Letting 0, we see that f d Pm( j) converges for all f C(K ).
For each r = 1, 2, . . . , take compact sets K such th
228
Measure, Topology, and Differentiation
unions of closed sets Fmn , where Fmn := cfw_ f : |x y| 1/m implies
| f (x) f (y)| 1/n.
12. Let be a finite Borel measure on a separable metric space. Show that
for every atom A of , as defined in 3.5, there is a
4.1. Simple Functions
117
Figure 4.1
Proof. For n = 1, 2, . . . , and j = 1, 2, . . . , 2n n 1, let E n j := f 1 ( j/2n ,
( j + 1)/2n ]), E n := f 1 (n, ]). Let
f n := n1 En +
n
2
n1
j1 En j 2n .
j=1
In Figure 4.1, gn = f n for the case f (x) = x on [0, ]
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
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
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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
8
Introduction to Probability Theory
Probabilities are easiest to define on finite sets. For example, consider a toss
of a fair coin. Here fair means that heads and tails are equally likely. The
situation may be represented by a set with two points H and
68
General Topology
Definition. Given a set S, a uniformity on S is a filter U in S S with the
following properties:
(a) Every A U includes the diagonal D := cfw_s, s: s S.
(b) For each A U , we have A1 := cfw_y, x: x, y A U .
(c) For each A U , there is
Problems
285
and for almost all x, F (x) exists and equals f (x) (by Theorem 7.2.1). A
function f on R is a probability density function if and only if it is measurable,
nonnegative a.e. for Lebesgue measure, and
f (x) d x = 1.
If P has a density f, the
5.4. Orthonormal Sets and Bases
171
Remark. Having a countable set with dense linear span is actually equivalent
to being separable (having a countable dense set), as one can take rational
coefficients in the linear span and still have a dense set.
For th
218
Convex Sets and Duality of Normed Spaces
f > 0 is an interval J , and on this interval f is concave: f (u + (1 )v)
f (u) + (1 ) f (v) for all u, v J , and 0 1.
Proof. If x, u C and y, v C, then since C is convex, x + (1 )y,
u + (1 )v C. Thus
Cu+(1)v
174
L p Spaces; Introduction to Functional Analysis
5.5.1. Theorem (Riesz-Frechet) A function f from a Hilbert space H into
K is linear and continuous if and only if for some h H , f (x) = (x, h) for
all x H . If so, then h is unique.
Proof. If follows fr
264
Introduction to Probability Theory
X j := min(X j , 0) are independent of each other. Thus it will be enough to
prove convergence separately for the positive and negative parts. So we can
assume that X n 0 for all n.
Let Y j := X j 1cfw_X j j where 1c
4
Integration
The classical, Riemann integral of the 19th century runs into difficulties with
certain functions. For example:
(i) To integrate the function x 1/2 from 0 to 1, the Riemann integral itself
does not apply. One has to take a limit of Riemann i
Problems
159
Proof. As noted after the Minkowski-Riesz inequality (Theorem 5.1.5), ( p
is a seminorm on L p ; it follows easily that it defines a norm on L p . To prove it
is complete, let cfw_ f n be a Cauchy sequence in L p . If p = , then for almost
a
7.2. Lebesgues Differentiation Theorems
229
Figure 7.2
Hence if q is the largest i such that Vi is defined,
q
q
n
Uj
(Wi ) = 3
(Vi ).
t < (K ) <
j=1
i=1
i=1
7.2.3. Lemma If is a finite Borel measure on an interval [a, b] and
A is a Borel subset of [a, b
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Measure, Topology, and Differentiation
set A with (X \A) < such that f n f uniformly on A, that is,
lim supcfw_d( f n (x), f (x): x A = 0.
n
Proof. For m, n = 1, 2, . . . , let
Amn := cfw_x: d( f k (x), f (x) 1/m for all k n.
Each Amn is measurable by
84
General Topology
Tietze, Heinrich (1915). Uber
Funktionen, die auf einer abgeschlossenen Menge stetig
sind. J. reine angew. Math. 145: 914.
(1923). Beitrage zur allgemeinen Topologie. I. Axiome fur verschiedene
Fassungen des Umgebungsbegriffs. Math. A
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Convergence of Laws and Central Limit Theorems
so all are equal to some P by the uniqueness theorem. Then Pn P by
Proposition 9.3.1.
Example. The functions f n := exp(nt 2 /2) are the characteristic functions
of the laws N (0, n), which are not unifor
Notes
183
Holder (1889, p. 44) proved the historical Holder inequality of Problem 9 for finite
sums. F. Riesz (1910, p. 456) proved the current form for integrals. Rogers did other
work of lasting interest; see Stanley (1971). Holder is also well known fo
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Convergence of Laws and Central Limit Theorems
functions which, after some errors were corrected, led to proofs much shorter than
Lindebergs. Levy (1922c) first gave the uniqueness theorem (9.5.1) and a form of the
continuity theorem (9.5.5). Both Lev
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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
Math 131A: Preparation Questions for Quiz 2
1. For each of the following sequences (sn )
n=1 prove that (sn ) converges. You should use the
definition of convergence given in lectures (which, at the moment of typing, is definition 7.2.1;
tell me if this s