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详细说明：Stein & Rami Real Analysis solution manual 普林斯顿分析学第三册解答3
since [0, 1\ Ck is a union of disjoint segments with this total length
TH
(Ck)=1
Now Ck C, so by Corollary 3.3
C)=limm(C)=1∑
b) For k-1, 2, .., let Jk be the interval of Ck which
Let I
be the interval in Cc which is concentric with Jn-1.(Thus, at the nth
step of thc iteration, the interval In is uscd to bisect thc interval n-1)
Let n be the center of In. Then in∈C°. Moreover,xn-x|≤Jn-1
since n-1 contains both an and a. Since the maximum length of the
intervals in Cn tends to 0, this implies n Finally, In EIn C cc
dInC,Jn-1→In→0
(c) Clearly C is closed since it is the intersection of the closed sets Cn. To
prove it contains no isolated points, we use the same construction from
the previous part. Let c EC. This time, let n be an endpoint of In
rather than the center. (We can actually take either endpoint, but for
specificity, we 'll take the one nearer to ) Because In is constructed
as an open interval, its endpoints lie in Cn. Moreover, successive
iterations will not delete these endpoints because the kth iteration
only deletes points from the interior of Ck-1. So n E C. We also
have m-s Jn as before, so that n r. Hence a is not an
isolated point. This proves that C is perf
(d)We will construct an injection from the set of infinite 0-1 sequences
Into
C. To do this. we number the sub-intervals of Cl. in order from
left to right. For example, C2 contains four intervals, which we denote
1oo, Io1, 110, and 111. Now, givell a sequence u=(1, a2,... of0's and
1s, let 1g denote the interval in Cn whose subscript matches the first
terms of a.(For instance, if a=0,1,0,0.. then 14=10100 C C4.
Finally, let
∈∩m.
This intersection is nonempty because 1m+1 C1n, and the intersection
of nested closcd intervals is nonempty. On thc othcr hand, it contains
only one point, since the length of the intervals tends to 0. Thus, we
have constructed a unique point in C corresponding to the sequence a
Since there is an injcction from the uncountable sct of 0-1 scqucnccs
iIlto c.c is also ullcoulltable
Problem 5: Suppose E is a given set, and and On is the open set
fr: d(, F)
Show that
4
(a)If E is compact, then m(e)=lim m(On)
(b) However, the conclusion in (a) may be false for E closed and un
boundcd, or for e open and boundcd
Proof.(a) First note that for any set E
E
since the closure of E consists of precisely those points whose distance
to E is 0. Now if E is compact, it equals its closure, so
E
Note also that O
n2+1
COn, so that On e. now since E is bounded
it is a subset of the sphere bn(o) for some N. Then 01 C BN+1(0)
so that m(o1)< o. Thus, by part (ii) of Corollary 3.3
m(e)=lim m(On)
(b) Supposc E= Z C R. Thon m(On)=oo for all m, sinco
collection of infinitely manly intervals of length 4. However, T (Z)=0
This shows that a closed unbounded set may not work To construct a
bounded open counterexample, we need an open set whose boundary
has positive measure. To accomplish this, we use one of the Canto
like sets C from Problem 4, with the e, chosen such that. m(c)>0
Let E=[0, 1C. Then E is clearly open and bounded. The boundary
of e is precisely c, since C contains no interva.l and hence has empt
intcrior.(This shows that tho boundary of E contains C; conversely,
it cannot contain any points of e because E is open, so it is exactl
equal to C. Hence F= FUOF=0, 1. Now
1
On-{x∈R:d(x,E)
∈R:d(x,E)
,1+
Clearly m(On) 1, but m(e)=l-m(c)<1
Problem G: Using translations and dilations, prove the following: Let b be
a balI in R of radius r. Then m( B)=vdr, where vd= m(B1) and Bi is
the unit ball xer: a<11
Solution. Let e> 0. Choose a covering Qi of B1 with total volume
less than m(Bi)+ e: such a covering must exist bccausc the m( bi)is
the infimum of the volumes of such cubical coverings. When we apply the
homothety hyra to R, each Qi is mapped to a cube Q,i whose side
longth is r times the sidc longth of Qj. Now (Q, is a cubical covcrin
B, with total volume less than r dm(bi)+E. This is true for any E>0, so we
must have m(Br)0,
and Cira, we define d by
6E={(61m1,,,6dd):(x
1)∈E}
Prove that se is measurable whenever e is measurable, and
(6)-61….5m(E).
Solution. First we note that for an open sct U, dU is also open. Wc could
see this from the fact that x -Sx is an invertible linear transformation. and
therefore a homeomorphism. More directly, if p E U, let Br(p) be a neigh
borhood of p which is contained in U; then if we define o= min(01,., dd),
we will have BSp)c&t
Next, we note that for any set S, m(SS)=51.Sam, (S). The proof of
this is almost exactly the same as problem 6 the dilation r h dr and
its inverse map rectangular coverings of s to rectangular coverings of &s
and vice versa; but since the exterior measure of a rectangle is just its area
(Page 12, Example 4), the infimum of the volume of rectangular coverings
is the same as the infimum over cubical coverings. Hence a rectangular
covering within e of the infimum for one set is mapped to a rectangular
covering within
for the other
As a morc dctailcd version of the preceding argument, supposc [Q,) is a
cubical covering of S with 2IQ, m*(S)+E. Then SQ,) is a rectangular
covering of &S with 20Q<51.. dm(S)+51.Sde. Now for each rec
tangle 8Q] we can find a cubical covering (Qik with 2kQjk <18Qj1+2
Then nj k.* is a cubical covering of S with 2, k1Q,kI< S1.5am*(S)+
(1-61∴5a)∈. This implies that m*(65)≤61…6um*(S). To get the re
verse inequality we note that another d-type transformation goes the other
direction, i.e. S=5(6S)where 8=(1751, .. 1/d)
Now let U e be an open set with m*(U\E)<8.r. Then dU dE is
an open set. Moreover, S (E\U)=SESU, so m* SU\SE=d.Sam*(U\
E)0. We will construct an open set whose
closure has boundary C. let us number the intervals involved in the Cantor
iteration as follows: If Cn is the set remaining after n iterations (with
Co=[0, 1), we number the 2n intervals in Cn in binary order, but with 2s
instead of 1s. For example, C2=100n 102 n 120 n 122. The intervals in the
complement of c, denoted by subscripted ' s, are named according to the
intervals they bisected, by changing the last digit to a 1. For instance. in
1, the interval 1 is taken away to create the two intervals Io and 12
the next iteration, lo is bisected by Joi to create loo and 102, while 12 is
bisected by J21 to create 120 and 121, ctc
Having named the intervals,letG=J1∩Jon1∩Jo21∩J201∩J221∩
be the union of the intervals in C which are removed during odd steps of
the iteration, and G=[0. 1\(Gnc)be the union of the other intervals,
i.e. the ones removed during evenl steps of the iteration. I clain that the
closure of G is GnC. Clearly this is a closed set (its complement in 0,1
the open sct G) containing G, so we nccd only show that cvcry point in
C is a limit of points in G. To do this, we first note that with the intervals
numbered as above, an interval lobe. whose subscript is k digits long ha
length less than Sk. This is so because each iteration bisects all the existing
Ts. Ill addition, an interval Jab
ith a k-digit subscript has length less
than ki because it is a subinterval of an I-interval with a(k-1)-digit
subscript. Now let∈C. Then∈∩ Cn so for each n we call finld all
interval I(n) containing a which has an n-digit subscript. Let (n)be the
J-interval with an n-digit subscript whose first n-1 digits match those of
I(m). Then I(m)and (n) are consecutive intervals
the
have length at most onI, the distance between a point in one and a
in thc othcr is at most 2n-2. Thus, if wc lot yn bc a scqucncc such that
Vn e.(n), then ym.-x. Now let gn, be the subsequence taken for odd
m,so that yn, CG. Then we have constructed a sequence of points in G
which converge to a∈C
We have shown that G= GnC. It only remains to show that a(GnC)
C. Clearly a Gnc) CC since G is open and is therefore contained in the
interior of Gn C. Now let I E C. By the same construction as above, we
subsequence yn orca. c In) which converges to c. If we now take tho
can choose a scqucncc
11. thell yi∈G′andy→x. This proves that
Ea(GnC). Hence we have shown that G is an open set whose closure
has boundary c. which has positive measure
Problem 11: Let A be the subset of 0, 1] which consists of all numbers which
do not have the digit 4 appearing in their decimal expansion. Find m(a)
Proof. A has Ineasure 0, for the saIlle reasoll as the Cantor set. We call
construct a as an intersection of cantor -like iterates. The first iterate
is the unit interval; the second has a subinterval of length 1 /10 deleted,
with segments of lengths 3/10 and 6/10 remaining.(The deleted interval
corresponds to all numbers with a 4 in the first decimal place. )The next
has 9 subintervals of length 1/100 deleted, corresponding to numbers with
a non-4 in the first decimal place and a 4 in the second. Continuing, we get
closed sets Cn of length(9/10 m, with A= nCn. Clearly A is measurable
since each Cn is; since m(Cn)0, m(A)=0
Problem 13
(a) show that a closed set is Gs and an open set Fa
(b) give an example of an Fo which is not g
(c) Give an cxamplc of a Borel sct which is neither Gs nor Fo
8
Proof
(a) lct U bc opcn. As is wcll known, U is thc union of thc opcn rational
balls that it contains. however it is also the union of the closed
rational balls t, hat it contains. To prove this, let EU and r>0 such
that Br(a)CU. Choose a rational lattice point q with a-q<33
and a rational d with a d <3. Then Ba(g)c Br(a)Cu and
C Ba(a), so any c C U is contained in a closed rational ball within
U. Thus. U is a union of closed rational balls. of which there are only
countably many. For a closed set F, write the complement R\ Fas
e open sets
(b) The
Per, 0An; then F=nB is a countable intersection
a union of rational balls b
e rational numbers are Fo since they are countable and single points
are closed. However, the Baire category theorem implies that they
are not Go.(Suppose they are, and let Un be open dense sets with
Q=nUn. Defiue Vn=Un\Tm, where r'n is the Teth rational in soine
enumeration. Note that the Vn are also open and dense, but their
intersection is the empty set, a contradiction.
(c) Let A-(Qn(0, 1)U((R\Q)n2, 3) consist of the rationals in(0, 1)
together with the irrationals in 2, 3. Suppose A is Fo, say A= UFr
where Fn is closed. Then
(R\Q)∩[2,3=A∩2,3=(UF7)∩[23]=U(Fn∩[2,3)
is also Fo since the intersection of the two closed sets Fn and 2, 3is
closed. But th
Q∩(2,3)=n(F2n(2,3)
is Gs because Fi n(2, 3) is the intersection of two open sets, and
therefore opell, for each T. But then if T'n is all enumeration of the
rationals in(2, 3),(FCn(2, 3)\Tn is also open, and is dense in(2, 3)
Hence n(Fnn(2,3))\tri is dense in(2, 3 )by the Baire Category
Theorem. But this set is empty, a contradiction. Hence A cannot be
Similarly, suppose A is Gs, say A=nGn where Gn is open. Then
Q∩(0,1)=A(0,1)=(∩Gn)∩(0,1)=∩(Gn∩(0,1)
is also Gs since Gn n(0, 1) is the intersection of two open sets and
therefore open. But then if an is an enumeration of the rationals in
0,1),(Gn. n(0, 1))\n) is open and is dense in(0, 1),so
∩((Gn∩(0,1){qn})
Illust be dense in(0, 1. But this set is empty, a contradictiON. Hence
A is not gs
Problem 16: Borel-Cantelli Lemma: Suppose Ek is a countable family
of measurable subsets of rd and that
m(er)<
9
e=aER: 2 E Ek for infinitely many k)= lim sup Ek
Show that e is measurable
· Prove n7(E)=0
Let
n=∪F
be the set of c which are in some Ek with k n. Then c is in infinitely
Ek iff u∈ B for all
E
B
E
1=
This is a countable intersection of a countable union of measurable
sets, and hence is measurable
Lete>0. Since∑m(Ek) converges,彐 N such that
k=N
Then
m(BN)=m(∪E)s∑m(Ex)<∈
y subadditivity. But m(nBn)s m(Bn) by monotonicity, so m(E)<
E for all e. Hence m(E)=0
Problem 17: Let in be a sequence of measurable functions on 0, 1 with
Ifn(al< oo for a e. x. Show that there exists a sequence Cr of positive
real numbers such that
C
Solution. We are given that for each n
fn(a)
since this sct is prcciscly thc sct whcrc f(a)|=oo. Sincc thcsc scts arc
nested, this implies
li
1e-
ence
c suci
({x:1(9)<
Define
E
Ifn()
10
Then m(En)<2n,so
m∩∪E=0
m=l i=m
by the borel-Cantelli lemma. But the complement of this set consists of
precisely those points that are in finitely many En, i. e. those points for
which In(al is eventually less than 3. Hence we have found a set of measure
0 such tha.t
fn(c)
→0 on the complement
Problem 18: Prove the following assertion: Every measurable function is the
limit, a.e. of a sequence of continuous functions
Proof. Let f R-R be measurable.(The problem didn't specify whether
f can have +oo as a value, but I'm assuming not. Let Bn=[
by Lusin's Theorem, there exists a closed(hence compact) subset En C Bn
with m(Bn\en)< on and f continuous on En. Then by Tietze's Extension
Theorem, we can extend f to a continuous function fn on all of R, where
f n=f on En.(Explicitly, such an extension could work as follows: Define
fn R+R by fn(a)=f(a)for a E En; for a En, since the complement
is open, a is in some open interval(a, b)CEn or in some unbounded oper
interval (O,a)CEm or(b,o)
Let fn(a)=f(a)+b-af(b)in the
first casc and fn()=f(a) in thc other two cascs
I claim that fn -f almost everywhere. Suppose a is a point at which
fn f. Then x E(B)U(Bn En)for infinitely many n since otherwise
fn(a)is eventually equal to f(a). Now a given a can be in only finitely many
c, so it must be in infinitely many (Bn En). i.e. a E lim sup(Bn En
But lim sup(Bn En)has measure 0 by the Borel-Cantelli Lemma. Hence
the set of x at which fn(a)t f(a) is a subset of a set of measure 0, and
therefore has measure 0
Problem 20: Show that there exist closed sets A and B with m(A)=m(B)
0,butm(4+B)>0:
a) In R, let A=C, B=C/2. Note that A+B) [0,1
(b) In R2, observe that, if A=I X 0) and B=10 I(where /=[0, 1)
then 4+B=I×I
Solution
(a)As noted, let C be the Cantor set, A=C, and B=C/2. Then A
consists of all numbers which have a ternary expansion using only 0s
and 2s, as showI on d previous homework set. This implies that B
consists of all numbers which have a ternary expansion using only 0s
and 1s. Now any number a c 0, 1 can be written as a t b whore
a E A and b e b as follows: Pick any ternary expansion 0.C122... for
r. Define
0
Is e
0
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