Math 676
Problem Set #8
Solutions
1. (page 93, problem 19) Observing that
Z
m(E ) =
E
(x) dx
Rd
we have
Z
1
m(E )d =
0
Z
1
0
=
Z
Rd
=
Z
Rd
=
Z
Rd
Z
Z
E
Rd
1
E
0
Z
jf (x)j
(x) dx
d
(x) d
!
dx
1d
0
dx
jf (x)j dx
where in the second and third lines we used Tonnelli’s theorem.
2. (page 93, problem 20) Let S be the -algebra of subsets E of R2 so that
each slice E y is a Borel subset of R. If E is open, then E y is also open
since, for each (x; y) 2 E, there is a ball B about (x; y) contained in E,
and B y is an open interval that contains x. Hence E y is open. Thus S
contains the open sets. Next, suppose that fEi g is a sequence of sets in
S and let E = [i Ei . Since E y = [Eiy is follows that each E y is Borel.
c
y
Finally if E 2 S, since (E y ) = (E c ) , it follows that E c 2 S . Hence S is
a -algebra containing the open sets, and hence the Borel sets.
3. (page 146, problem 4) Suppose that f 2 LR1 and f 6= 0 on a set of nonzero
measure. Then, there is a ball B0 with B0 jf j = C1 > 0. Without loss
we’ll assume that B0 is centered at 0. For any x 2 Rd , let B be a ball
centered at x and containing B0 . There is a constant C2 so that the radius
of B is no more than C2 jxj if jxj 1. For such a ball we have
Z
Z
1
C1
1
jf j
jf j =
f (x)
d
d
d
m(B) B
cd C2 jxj B0
cd C2d jxj
as was
R to be proved. Next, suppose that f has support in the unit ball
with jf j = 1. For any x, the ball Bjxj+1 (0) contains the support of f so
f (x)
1
d
cd (jxj + 1)
and hence the set E = fx : f (x) > g contains the set of x with
1
d
cd (jxj + 1)
> :
Hence, Ea contains the set of x with
jxj
If
< cd 1 2
d
!
1
1 :
1=d
(cd )
1=d
, then (cd )
< 1=2 and this set in turn contains the ball
1
jxj
which has measure at least 2
d
1=d
2 (cd )
= . Hence,
2
m (fx : f (x) > g)
d
as was to be shown.
4. (Page 146, problem 5)
(a) We can compute the improper Riemann integral (with r = log x)
2
Z
1=2
1
2
r (log r)
0
dr = 2
Z
log 2
1
2
dw
=
2
w
log 2
(b) Observe that, since B(0; jxj + ") contains x, for jxj < 1=2 we have
Z
1
1
f (x)
dy
m(B0 (jxj + "))) jyj jxj+" jyj (log jyj)2
Z jxj
1
1
=
dt
(jxj + ") 0 t (log t)2
1
1
=
(jxj + ") log (1= (jxj + "))
for any " > 0, hence
f (x)
1
jxj log (1= jxj)
for jxj < 1=2 . We may estimate
Z
f (x) dx
2
Z
0
1=2
1
dt
t log(1=t)
and compute the right-hand integral as an improper Riemann integral:
lim
b#0
Z
b
1=2
1
dt = lim
b#0
t log (t)
Z
log 2
log(b)
1
dw
w
= lim [log jlog bj
b#0
log(log 2)]
which diverges as b # 0. It follows that f is not integrable.