Migros was founded in 1954 as a jointventure between Migros Cooperatives Union
of Switzerland and Istanbul Municipality.
In first years, Migros reached their
customers with 20 vehicles and its first
store opened at 1957 in Beyolu.
_ t< 5 Y
Q1) F or the given LP in the below:
min axl + bxz
@5 7-3.4; x1 x2 2 2
21) Draw feasible region.
b) Give an example for a and b such that LP will have a unique optimal solution. For the
chosen a and b,
Sets, Logic and Categories
Solutions to Exercises: Chapter 1
1.1 Show that the empty set is a subset of every set.
Let x be any set. Then for any set z, the implication (z 0) (z x) is true, since
(z 0) is false; thus 0 x.
1.2 Which of the following
14. Measurability Criteria
Theorem 14.1. L is a -algebra of subsets of Rn , and
is a measure on L.
Proof. ; 2 L0 , so ; 2 L. If A 2 L, then Ac \ M = M \ A = M \ (A \ M ) 2 L0 for each M 2 L0 ,
so Ac 2 L. Finally, if Ai 2 L for each i 2 N an
16. Measurable Functions
Recall that the -algebra B of Borel sets in R is the smallest -algebra of subsets of R
which contains the open sets.
Denition 16.1. Let M be a -algebra of subsets of a set X . A function f : X ! R is said
to be M-measur
17. Measurable Functions, Simple Functions
Proposition 17.1. If f, g : X ! R are M-measurable, then f + g and f g are M-measurable.
Proof. For each t 2 R,
cfw_x | f ( x ) + g ( x ) < t =
cfw_x | f (x) < r \ cfw_x | g (x) < t
18. Lebesgue Integration: Non-negative Functions
Recall that S denotes the collection of all non-negative Lebesgue measurable simple functions on Rn .
Proposition 18.1. For ', 2 S :
(i) R ' d R .
(ii) R (c') d = c ' d for each c 2 [0, 1).
> 0 and let N 1 . We have that for n N
supcfw_|fn (x) f (x)| = supcfw_fn (x)
So fn 0 uniformly. However
fn d =
([0, n]) = 1
and f d = 0. This does not contradict the Monotone convergence
theorem since f1 (1) = 1 < f2 (1) = 1/2 whi
1. We know that f L and so f , f M + and max
. For > 0 we let
f + d,
f 1 d <
A+ = cfw_x : f + (x) and A = cfw_x : f (x) .
Both these sets are measurable and cfw_x : |f (x)| = A A+ / Let
= A+ and note that
(A+ ) =
f + d < .
Thus (A+ ) < and we can si
Exercise sheet 1
1. Show that if f : R2 R is a continuous function and A R is an
open set then f 1 (A) = cfw_(x, y ) R2 : f (x, y ) A is an open set in
R2 (take the usual Euclidean metric on R2 ).
2. Show that
(i) [a, b] =
(ii) (a, b) =
n=1 (a 1/n,
1. We x n N and let An = x n , x + n B. We have that for each
A . Thus
n An An+1 and that cfw_x = n=1 n
(cfw_x) = lim (An ) = lim
Now if X = cfw_x1 , . . . , xn , . . . = cfw_xi is a countable set then
(X ) = (cfw_xi ) = 0.
1. Let (x1 , x2 ) f 1 (A) and y = f (x1 , x2 ) A. We need to show that
there exists > 0 such that B (x1 , x2 ), ) f 1 (A). Since A R is
open there exists > 0 such that B (y, ) A. Furthermore by the
continuity of f there exists > 0 such that if (y1 , y2 )
Exercise sheet 4
NB: Throughout this sheet (X, X, ) refers to a measure space.
1. Let f L(X, X, ) and a > 0. Show that cfw_x : |f (x)| a has nite
measure. Show that cfw_x : f (x) = 0 has -nite measure (i.e for each
n N there exists An X such that nN An =
Real Analysis HW 5 Solutions
Problem 9: Let E have measure zero. Show that if f is a bounded function on E , then f
is measurable and E f = 0.
Solution: Clearly f is measurable, since the preimage cfw_x E | < f (x) c E is
a set of measure 0. It suces to c
Exercise sheet 3
Notation: (X, X, ) will denote a measure space and (R, B, ) will be the
specic measure space of the real numbers with the Borel sigma algebra and
1. Working on (R, B, ) let fn = (1/n)[0,n] and f = 0. Show that the
Math 318 HW #9 Solutions
1. Let A be a bounded, measurable set and let (fn ) be a sequence of measurable functions on A
converging to f . Suppose L(A) and that |fn (x)| (x) for all x A and all n = 1, 2, . . .
f g dm
fn g dm =
if g is m
Exercise sheet 2
1. Consider the measure space (R, B, ) where is Lebesgue measure.
For any x R show that (cfw_x) = 0. Thus show that any countable
set X R satises (X ) = 0.
2. Again let denote Lebesgue measure. Show the following.
a) If E is a non-empty o
Consider the exterior Lebesgue measure m introduced in Chapter 1. Prove that a set E in Rd is
Carathodory measurable if and only if E is Lebesgue measurable in the sense of Chapter 1.
Proof. () Suppose E is Cara
Real Analysis HW 4 Solutions
Problem 4: Suppose f is a real-valued function on R such that f 1 (c) is measurable for
each number c. Is f necessarily measurable?
Solution: No. Let V be a non-measurable subset of (0, 1) and consider the function f (x) =
ANALYSIS QUALIFYING EXAM PROBLEMS
I make no claims about the infallibility of these solutions. In addition to
Let cfw_fn be a sequence of measurable functions on [0, 1] with |fn (x)| < for almost every x. Show that
there exists a sequence cn of positive real numbers such that
0 for almost every x.
Proof. For a
Math 209A Homework 1
Problem 8 - If (X; M; ) is a measure space and fEj g1
(lim inf Ej )
lim inf (Ej ). Also, (lim sup Ej )
lim sup (Ej ) provided
that ([1 Ej ) < 1.
Solution Dene Fk = \1 k Ej . Then clearly F1