Math 515
Professor Lieberman
September 1, 2004
HOMEWORK #1 SOLUTIONS
Chapter 1
20. First we show that A0 is a σ-algebra. So suppose A ∈ A0 . Then there is a countable
subcollection C1 of C such that A is in the σ-algebra A1 generated by C1 . It follows that
à ∈ A1 , so à ∈ A0 . Next, suppose that hAn i is a sequence of elements of A0 . Then, for each
n, there is a countable subcollection Cn such that An ∈ An , the σ-algebra generated by Cn .
Now write A∞ for the σ-algebra generated by ∪n Cn . Since A∞ is a σ-algebra containing
Cn , it follows from Proposition 4 that An ⊂ A∞ for all n and hence each An is in A∞ , so
∪n An ∈ A∞ and therefore ∪n An ∈ A0 because ∪n Cn is countable by Proposition 1.7 (from
the text).
Because A0 is a σ-algebra, it follows from Proposition 4 that A ⊂ A0 . Hence, any E ∈ A
is an element of A0 , which means that E is a element of a σ-algebra generated by some
countable subcollection C0 of C.
Chapter 2
21. (a) We prove the contrapositive: If E is uncountable, then
To prove this implication, we write
En = {x ∈ E : x >
P
x∈E
x is infinite.
1
}
n
for n ∈ N. Since ∪n En = E and E is uncountable, some En must be uncountable (by
Proposition 1.7) and hence infinite. For a given M > 0 (and an n such that En is infinite),
let F be a subset of En with M n elements. Then
X
1
x > Mn
= M,
n
x∈F
so
X
x≥
x∈E
X
x > M,
x∈F
and hence the sum is infinite.P
Here’s a direct proof. Let x∈E x = S < ∞ and define En as before. Then
X
S≥
x,
x∈En
so En has at most S/n elements, which means En is finite for each n and therefore E is
countable.
(b) We consider two cases. First, if
∞
X
xn = ∞,
n=1
1
2
then, for any M > 0, there is a positive integer N such that
N
X
xn > M,
n=1
P
P
and hence x∈E x > M , which implies that x∈E x = ∞.
On the other hand, if
∞
X
xn = S < ∞,
n=1
then, for any ε > 0, there is a positive integer N such that
m
X
x
−
S
<ε
n
n=1
as long as m > N . It follows that
N
X
xn > S − ε,
n=1
and hence
X
x ≥ S − ε.
x∈E
Since ε is arbitrary, it follows that
X
x ≥ S.
x∈E
Now, if F is finite, then there is an integer k such that
F ⊂ {x1 , . . . , xk }.
(Just let k = sup{n : xn ∈ F }.) Since the xn s are all positive, it follows that
X
x∈F
≤
k
X
xn ≤ S
n=1
and hence
X
x ≤ S.
x∈E
Combining these last two inequalities yields
X
= S.
x∈E
37. First, we show the “middle thirds” characterization of the Cantor set via an induction
argument. So let x ∈
/ C, and set
∞
X
an
.
x=
3n
n=1
(If x has two ternary expansions, we choose one of them.) Also, we define An inductively
by A0 = [0, 1], and An+1 is obtained from An by deleting the (open) middle third of each
interval of An . In each An , there are 2n intervals, which we label as follows: number the
intervals from left to right in binary notation with enough leading zeros to have n bits and
3
then subtract 1 (so, for example, the fifth interval for n = 4 corresponds to 0100). If we
now replace each 1 by a 2, we get the ternary expansion of the left endpoint of the interval
and if we add infinite many twos to the end of this ternary expansion, we obtain the right
endpoint of the interval. Therefore, a number has the first n − 1 ternary terms equal to 0
or 2 if and only if that number is in An , so the Cantor set is the intersection of the An s.
Each An is the complement of a union of open intervals and hence it is closed. Since the
Cantor set is the intersection of these An s, it is also closed.
50. (a) =⇒. If f is lower semicontinuous at y, let ε > 0 be given. Since
f (y) ≤ sup
inf
f (x),
θ>0 0<|x−y|<θ
it follows that there is a δ > 0 such that
f (x) ≤ sup
inf
0<|x−y|<δ
inf
f (x) − ε
θ>0 0<|x−y|<θ
and therefore
f (y) ≤
inf
f (x) + ε
0<|x−y|<δ
and hence (because f (y) ≤ f (y)), we have f (y) ≤ f (x) + ε for all x with |x − y| < δ.
⇐=. Set L = supθ>0 inf 0<|x−y|<θ f (x) and let ε > 0 be given. With δ so that f (y) ≤ f (x)+ε
for all x such that |x − y| < δ, we see that
f (y) ≤
inf
f (x) + ε.
0<|x−y|<δ
Therefore f (y) ≤ L + ε. Since ε is arbitrary, it follows that f (y) ≤ L.
(b) This follows easily from (a) and the ε-δ definition of continuity.
(c) =⇒. Given λ ∈ R and x such that f (x) > λ, set ε = (f (x) − λ)/2. From part (a), there
is a number δ such that f (x) ≤ f (y) + ε for all y with |x − y| < δ. But this says that
f (y) ≥ f (x) − ε =
f (x) + λ
>λ
2
so {x : f (x) > λ} is open.
⇐=. Let ε > 0 and y be given and set λ = f (y) − ε. Since {x : f (x) > λ} is open, there
is a number δ such that f (x) > λ for all x with |x − y| < δ. But this inequality says that
f (x) > f (y) − ε which implies that f (y) ≤ f (x) + ε.
(d) For f ∨ g, given ε > 0 and y, there are positive numbers δ1 and δ2 such that f (y) ≤
f (x) + ε for |x − y| < δ1 and g(y) ≤ g(x) + ε for |x − y| < δ2 . With δ = min{δ1 , δ2 }, we see
that
f (y) ≤ f (x) + ε, g(y) ≤ g(x) + ε
for |x − y| < δ. Hence,
f (y) ≤ f ∨ g(x) + ε,
g(y) ≤ f ∨ g(x) + ε
for |x − y| < δ, and therefore f ∨ g is lower semicontinuous by part (a).
For f +g, we take δ3 and δ4 so that f (y) ≤ f (x)+ε/2 for |x−y| < δ3 and g(y) ≤ g(x)+ε/2
for |x−y| < δ4 . Then, adding the inequalities gives f (y)+g(y) ≤ f (x)+g(x)+ε for |x−y| < δ
and δ = min{δ3 , δ4 }.
(e) Let ε > 0 and y be given. Then there is an index n such that fn (y) ≥ f (y) − ε/2. With
δ > 0 so that fn (y) ≤ fn (x) + ε/2 for |x − y| < δ, we see that
ε
f (y) ≤ fn (y) + ≤ fn (x) + ε ≤ f (x) + ε
2
4
for |x − y| < ε. Hence f is lower semicontinuous by part (a).
(f) =⇒. Let ε > 0 be given and let δ be given from part (a). Then there are points
y1 ∈ (xi−1 , xi ) and y2 ∈ (xi , xi+1 ) such that |x − y1 |, |x − y2 | < δ. Hence f (xi ) ≤
min{ϕ(y1 ), ϕ(y2 )} + ε by part (a). Since ε is arbitrary, we have ϕ(xi ) ≤ min{ϕ(y1 ), ϕ(y2 )}
as required.
⇐=. From part (b), ϕ is lower semicontinuous on each interval (xi , xi+1 ). At xi , we take
δ = min{xi − xi−1 , xi+1 − xi } and note that ϕ(x) is equal to the smaller of the two values
assumed in (xi−1 , xi ) and (xi , xi+1 ) if 0 < |x − xi | < δ and ϕ(x) = ϕ(xi ) if |x − xi | = 0.
Hence ϕ(xi ) ≤ ϕ(x) if |x − xi | < δ, so ϕ is also lower semicontinuous at each xi . (Note that
the choice of δ is independent of ε.)
(g) =⇒. First, for each n, we define the partition
Pn = {a = xn,0 < xn,1 < · · · < xn,2n = b}
by xn,i = a +
i2−n (b
− a). We then define
ϕn (x) = inf{f (y) : y ∈ (xn,i , xn,i+1 )}
if x ∈ (xn,i , xn,i+1 ) and ϕn (xi ) is the minimum of the values assumed in (xn,i−1 , xn,i ) and
(xn,i , xn,i+1 ). Also we take ϕn (a) to be the value assumed in (a, xn,1 ) and we take ϕn (b) to
be the value assumed in (xn,2n −1 , b). By parts (i) and (f), ϕn is lower semicontinuous. It’s
easy to see that hϕn (x)i is an increasing sequence for each fixed x since the infimum for ϕn
is taken over a smaller set than the infimum for ϕn−1 . To see that f = lim ϕn , we fix an x
and an ε > 0. By part (b), there is a δ > 0 such that
f (x) ≤ f (y) + ε
if |x − y| < δ. By taking n so large that 2−n (b − a) < δ, we see that ϕn (x) ≥ f (x) − ε and
hence ϕn (x) → f (x).
⇐=. The limit of the ϕn s is equal to the supremum, so part (e) implies that this limit is
lower semicontinuous.
(h) =⇒. With ϕn from part (g), we define ψn separately on each interval. So, on
[xn,i , xn,i+1 ], we fix points yn,i < zn,i in this interval so that the intervals |xn,i − yn,i | =
|yn,i − zn,i | = |zn,i − xn,i+1 |. Then ψn is linear on each of these intervals and agrees with ϕn
at xn,i , yn,i , zn,i , and xn,i+1 . Again, it’s easy to see that the ψn s are increasing and they
converge to f .
⇐=. The argument in part (g) shows that f is lower semicontinuous.
(i) Let hxn i be a sequence such that f (xn ) converges to the infimum I (note that the case
of the infimum being −∞ has not yet been excluded). From the Bolzano-Weierstrass theorem, there is a convergent subsequence hxnk i with limit x0 ∈ [a, b]. It follows that f (xnk )
also converges to I and part (a) implies that f (x0 ) ≤ I. Since f (x0 ) ≤ I, it follows that
f (x0 ) = I, so the infimum is achieved and hence finite.