Davidson K.R., Donsig A.P. Real Analysis with Real Applications

Подождите немного. Документ загружается.

9.5 Metric Completion 263

in X with lim

n→∞

= c. Then (x

, x

, . . . ) is another sequence converg-

ing to c. By the previous paragraph, (f(x

), f(x

), . . . ) is a Cauchy

sequence in Y . Therefore lim

n→∞

f(x

) = lim

n→∞

f(x

); and g is well deﬁned.

To verify that g is continuous, let ε > 0. Let δ > 0 be chosen as before. Sup-

pose that c, d are points in C with ρ(c, d) < δ and that (x

) and (x

) are sequences

in X converging to c and d, respectively. Then lim

n→∞

ρ(x

, x

) = ρ(c, d) < δ.

Thus there is an N so that ρ(x

, x

) < δ for all n ≥ N. It follows that

σ(g(c), g(d)) = lim

n→∞

σ(f(x

), f(x

)) ≤ ε.

Consequently, g is uniformly continuous. ¥

9.5.5. COROLLARY. The metric completion (C, ρ) of (X, ρ) is unique in the

sense that if S is an isometry of X into another completion (Y, σ), then there is a

unique isometry S

of C onto Y extending S.

PROOF. Asin Theorem 9.5.4, we consider X as a subset of C with the same metric.

The map S is an isometry and thus is uniformly continuous. Let S

be the unique

continuous extension of S from C into Y . If c, d ∈ C with c = lim

n→∞

and

d = lim

n→∞

, then

σ(S

c, S

d) = lim

n→∞

σ(Sx

, Sy

)

= lim

n→∞

ρ(x

, y

) = ρ(c, d).

So S

is also an isometry.

Conversely, there is an isometry T

of Y into C extending the map S

−1

from

SX onto X. Notice that T

is therefore a continuous extension of the identity

map on X and hence equals the identity map on C. Likewise, S

is the identity

on Y . In particular, for each y in Y , S

y) = y and so S maps C onto Y . ¥

9.5.6. EXAMPLE. Uniqueness of the Real Number System. We apply this

theory to the real number system. The real numbers have several important proper-

ties, which were discussed at length in Chapter 2. To establish uniqueness in some

sense, we need to be explicit about the required structure. First, R should be a ﬁeld

that contains the rational numbers Q. In addition to the algebraic operations, it is

important to recognize that R has an order extending the order on Q. Second, R

should be complete in some sense. The sense that ﬁts well with analysis is com-

pleteness as a metric space and containing Q as a dense subset. A subtlety that

immediately presents itself is, Where does this metric take its values? Fortunately,

the metric on Q takes values in Q, but any completions of Q will have to take met-

ric values in R! So by our construction, this would still appear to depend on the

original construction of R.

264 Metric Spaces

Therefore, it seems prudent to introduce a different and more algebraic notion

of completeness, using order. A ﬁeld is an ordered ﬁeld if it contains a subset P

of positive elements with the properties that

(i) every element belongs to exactly one of P, {0} or −P ; and

(ii) whenever x, y ∈ P , then x + y and xy belong to P .

Then we deﬁne x < y if y − x ∈ P . An ordered ﬁeld is order complete if every

nonempty set that is bounded above has a least upper bound. This notion doesn’t

mention metric and thus is free from the difﬁculty mentioned previously about

where the metric takes its values. The density of the rationals is also addressed by

order. The rationals are order dense if whenever x < y, there is a rational r so that

x < r < y.

9.5.7. THEOREM. (1) The rational numbers Q have a unique metric com-

pletion with a unique ﬁeld structure extending the ﬁeld operations on Q making

addition and multiplication continuous. The order structure is also unique.

(2) The rational numbers are contained as an order dense subﬁeld of a unique

order complete ﬁeld.

PROOF. (1) The metric space of real numbers R is a metric completion of Q, by

the completeness of R (Theorem 2.7.4). By Corollary 9.5.5, it follows that if S is

any metric space completion of Q, then there is an isometry T of R onto S that

extends the identity map on Q. The Extension Theorem guarantees that there is

only one way to extend addition and multiplication on Q to a continuous function

on S. One such way is to transfer the operations on R by T x + T y = T (x + y)

and T x · T y = T (xy). Uniqueness says that this is the only way, and thus the

ﬁeld operations on S are equivalent to those on R. The positive elements of R are

exactly those nonzero elements that have a square root. Since this is determined by

the algebra structure, we see that order is also preserved.

(2) The real numbers are order complete by the Least Upper Bound Princi-

ple (2.5.3), and the rationals are order dense by construction (see Exercise 2.2.F).

Suppose that S is another order complete ﬁeld containing Q as an order dense sub-

ﬁeld. Deﬁne a map T from R to S by sending each real number x to the least upper

bound in S of {r ∈ Q : r < x}. Notice that T q = q for q ∈ Q. Indeed, q is an

upper bound for {r ∈ Q : r < q}. If T q < q, the order density of Q would imply

that there is an r ∈ Q such that T q < r < q, contradicting the fact that Tq is the

least upper bound for this set.

Let us show that T is a bijection. Clearly, if x ≤ y, then T x ≤ T y. If x < y

in R, then there is a rational r with x < r < y. Consequently, T x < r < T y

and so T is one-to-one and preserves the order. Now suppose that s ∈ S and let

S = {r ∈ Q : r < s}. This set has a least upper bound x in R. Then T x is the least

upper bound of S in S. Again s is an upper bound by deﬁnition and if T x < s, the

order density would yield a rational r with T x < r < s. Hence T x = s and the

map T is onto.

9.5 Metric Completion 265

Finally, we verify that T preserves addition and multiplication. The fact that

P + P ⊂ P means that addition preserves order: if x

< x

amd y

< y

, then

+ y

< x

+ y

. Thus in both R and S, we have that

x + y = sup{r + s : r, s ∈ Q, r < x, s < y}.

Since T preserves order and therefore sups, it follows that T x + T y = T (x + y).

Similarly, if x, y are positive,

xy = sup{rs : r, s ∈ Q ∩ P, r < x, s < y}.

So (T x)(T y) = T (xy) for x, y ∈ P . The rest follows since

(−x)y = x(−y) = −(xy) and (−x)(−y) = xy

allow the extension of multiplication to the whole ﬁeld. This establishes that the

map T preserves all of the ﬁeld operations and the order. ¥

So part (2) establishes, without any reference to any properties based on our

construction of R, that there is only one ﬁeld that has the properties we want. This

we call the real numbers. We could, for example, deﬁne the real numbers using

expansions in base 2 or base 3. We have implicitly assumed up to now that these

constructions yield the same object. We now know this to be the case.

There are other constructions of the real numbers that do not depend on a base.

This has a certain esthetic appeal but does not in itself address the uniqueness ques-

tion. One approach brieﬂy mentioned in Section 2.2 is Dedekind’s construction

based on an abstract deﬁnition of the sets {r ∈ Q : r < x} used in the proof here.

Another method alluded to at the beginning of this section is to consider Cauchy

sequences of rational numbers as representing points. We have to decide when two

Cauchy sequences should represent the same point. A little thought shows that

if you take (r

) and (s

) and combine them as (r

, s

, r

, s

, . . . ), then this new

sequence is Cauchy if and only if they have the same limit. Thus we are led to

dealing with equivalence classes of Cauchy sequences. Both of these constructions

yield a complete structure; Dedekind’s is order complete and Cauchy’s is metri-

cally complete. But there remains in both cases the tedious, but basically easy, task

of deﬁning addition and multiplication and verifying all of the ordered ﬁeld prop-

erties. Indeed, we did not verify all of these details for our decimal construction

either.

Another practical approach is to take any construction of the real numbers and

show that every number has a decimal expansion. This is an alternate route to

the uniqueness theorem. The reader interested in the details of these foundational

issues should consult [3].

Exercises for Section 9.5

A. Show that the metric completion of a normed vector space is a complete normed vector

space. HINT: Use the Extension Theorem to extend the vector space operations.

B. Prove that the map taking each bounded uniformly continuous function on a met-

ric space X to its continuous extension on the completion C is an isometry from

BU C(X) onto BUC(C), where BUC(X) is the normed space of bounded uniformly

continuous functions on X.

266 Metric Spaces

C. Show by example that there is a bounded continuous function on (0, 1) that does not

extend to a continuous function on the completion.

D. Let (X, ρ) and (Y, σ) be metric spaces with completions (C, ρ) and (D, σ).

(a) Show that d

, y

), (x

, y

)

= ρ(x

, x

) + σ(y

, y

) is a metric on the product

space X × Y .

(b) Show that C × D is the completion of X × Y .

E. Let (X, ρ) be the metric on all ﬁnite sequences of letters with the dictionary metric of

Example 9.1.I. Describe the metric completion.

HINT: Consider the natural extension of this metric to all inﬁnite words.

F. Show that the completion of a metric space (X, ρ) is compact if and only if X is totally

bounded.

G. The p-adic numbers. Fix a prime number p. For each n ∈ Z, let ord

(n) denote the

largest integer d such that p

divides n. Extend this function to all rational numbers by

setting ord

(a/b) = ord

(a) − ord

(b).

(a) Show that ord

(r) is independent of the representation of r as a fraction.

(b) Set |r|

= p

−ord

(r)

and ρ

(r, s) = |r − s|

. Prove that ρ

is a metric on Q.

HINT: In fact, ρ

(r, t) ≤ max{ρ

(r, s), ρ

(s, t)}.

be the metric completion of (Q, ρ

). Prove that addition extends to a con-

tinuous operation on Q

(d) Show that |xy|

= |x|

|y|

. Hence show that there is a continuous extension of

multiplication to Q

(e) Verify that every nonzero element of Q

has an inverse.

(f) Why doesn’t this complete ﬁeld containing Q contradict the uniqueness of the real

numbers?

H. We develop further properties of the p-adic numbers, although this requires a bit of

number theory.

(a) If r ∈ Q and |r|

= p

−k

, ﬁnd a

∈ {0, 1, . . . , p −1} so that |r −a

≤ p

−k−1

HINT: Write r = p

where gcd(p, b) = 1. By the Euclidean algorithm, we can

write 1 = mb + np. Let a

≡ am (mod p).

(b) Hence show that every element r ∈ Q with |r|

≤ 1 is the limit in the ρ

metric of a

sequence of integersof the form a

p+···+a

, where a

∈ {0, 1, . . . , p−1}.

with |x|

≤ 1. Hence deduce that Z

, the

closure of Z in Q

, consists of all these points.

(d) Prove that Z

is compact.

9.6. The L

Spaces and Abstract Integration

In Section 7.7, the L

norms for p ≥ 1 were introduced and shown to be norms

via the H

older and Minkowski inequalities. However, these norms were put on

C[a, b], and it is easy to show that C[a, b] is not complete in any of these norms.

The completion process of the last section may be used to remedy this situation.

One caveat to the reader is that there is a better, although more involved, method of

deﬁning the L

spaces using measure theory. Measure theory provides a powerful

way to developnot only integration on R, but also integration on many other spaces,

9.6 The L

Spaces and Abstract Integration 267

and provides a setting for probability theory. Nevertheless, we are able to achieve

some useful things by our abstract process, including recognizing L

as a space of

functions (or, more precisely, as equivalence classes of functions).

9.6.1. DEFINITION. The space L

[a, b] is the completion of C[a, b] in the L

norm kfk

|f(x)|

1/p

At this point, we know that L

[a, b] is a complete normed space by Exer-

cise 9.5.A. In particular, kf k

is the distance in L

(a, b) to the zero function.

However, the elements of this space are no longer represented by functions. We

will rectify that in this section. The ﬁrst step is to extend our integration theory to

spaces.

Keeping in mind that we expect to represent elements of L

(a, b) as functions,

we will write elements as f, g, and so on. In particular, we shall say that f ≥ 0 if

it is the limit in the L

norm of a sequence of positive continuous functions f

. So

we say f ≤ g if g − f ≥ 0.

9.6.2. THEOREM. The Riemann integral has a unique continuous extension to

(a, b) denoted by

f. It has the properties

(1)

sf + tg = s

f + t

g for all f, g ∈ L

(a, b) and s, t ∈ R.

(2)

≤

|f| = kfk

for f ∈ L

(a, b).

(3) If f ≥ 0, then

f ≥ 0.

PROOF. The Riemann integral on C[a, b] satisﬁes the properties (1)–(3). In partic-

ular, it is uniformly continuous in the L

norm by (2) since for f, g ∈ C[a, b],

f(x) dx −

g(x) dx

≤

|f(x) − g(x)|dx = kf − gk

Therefore, by the Extension Theorem, there is a unique continuous extension to

(a, b), which we call

f. Properties (1)–(3) follow by taking limits. ¥

The integral deﬁned in this manner is called the Daniell integral. The alter-

native is the Lebesgue integral, which is constructed using measure theory. Al-

though the constructions are different, the two integrals have the same theory. They

are powerful for two reasons. The ﬁrst reason is that the L

spaces of integrable

functions are complete. For the Lebesgue integral, this is a theorem, proved us-

ing properties of the integral; whereas in our case, the complete space comes ﬁrst

and the properties of the integral are a theorem. The second reason is that there are

much better limit theorems than for the Riemann integral. Here is ﬁrst of these. The

best limit theorem, the Dominated Convergence Theorem, we leave as an exercise.

This result, like Theorem 2.5.4, is traditionally called the Monotone Conver-

gence Theorem. It is usually easy to distinguish between the two, as one result

applies to sequences of real numbers and the other to sequences of functions.

268 Metric Spaces

9.6.3. MONOTONE CONVERGENCE THEOREM.

Suppose (f

) ∈ L

(a, b) is an increasing sequence such that sup

n≥1

< ∞.

Then f

converges in the L

(a, b) norm to an element f and

f = lim

n→∞

PROOF. By property (2),

is an increasing sequence of real numbers that is

bounded above. Hence by the Monotone Convergence Theorem for real numbers,

L = lim

n→∞

exists. So if ε > 0, there is an integer N so that

L − ε <

≤

≤ L for all n ≥ N.

Thus when N ≤ m ≤ n,

− f

k =

− f

| =

− f

−

< ε.

Therefore, (f

) is Cauchy in L

(a, b). It follows that f = lim

n→∞

exists. Since

the integral is continuous,

f = lim

n→∞

. ¥

The preceding analysis would appear to deal only with L

(a, b). For L

(a, b),

we obtain much the same result by extending the H

older inequality. Minkowski’s

inequality, which is just the triangle inequality, is immediately satisﬁed because the

metric d(f, g) = kf − gk

satisﬁes the triangle inequality.

9.6.4. THEOREM. Let 1 < p, q < ∞ satisfy

= 1. If f ∈ L

(a, b) and

g ∈ L

(a, b), then fg ‘belongs’ to L

(a, b) and

≤ kfk

kgk

PROOF. We make L

(a, b) ×L

(a, b) into a complete metric space with the metric

, g

), (f

, g

)

= kf

−f

+kg

−g

. By Exercise 9.5.D, this is the com-

pletion of C[a, b] × C[a, b] with the same metric. By the H

older inequality (7.7.3),

the map Φ(f, g) = fg maps C[a, b] × C[a, b] into L

(a, b) and

kfgk

|f(x)g(x)|dx ≤ kfk

kgk

This map is not uniformly continuous on the whole product, but it is uniformly

continuous on balls. So ﬁx r ≥ 0 and consider

(f, g) ∈ C[a, b] × C[a, b] : kfk

≤ r and kgk

≤ r

Then for (f

, g

) and (f

, g

) in X

kΦ(f

, g

) − Φ(f

, g

≤ kf

− g

+ k(f

− f

≤ kf

− g

+ kf

− f

≤ r

− f

+ kg

− g

= r d

, g

), (f

, g

)

So Φ is uniformly continuous on X

. By the Extension Theorem (Theorem 9.5.4),

Φ extends to a uniformly continuous function on the completion of X

, which is

9.6 The L

Spaces and Abstract Integration 269

its closure in L

(a, b) × L

(a, b), namely

(f, g) ∈ L

(a, b) × L

(a, b) : kf k

≤ r and kgk

≤ r

In particular, if f ∈ L

(a, b) and g ∈ L

(a, b), then Φ(f, g) = fg belongs

to L

(a, b). Moreover, kΦ(f, g)k

≤ kfk

kgk

by continuity since this holds on

C[a, b]×C[a, b]. Moreover, since

is uniformly continuous on L

(a, b), we obtain

older’s inequality:

Φ(f, g)

≤ kΦ(f, g)k

≤ kf k

kgk

. ¥

We turn to the problem of representing each element of L

as a function. We

already know we must allow discontinuous functions, as C[a, b] is not complete in

the L

norm. But this allows the possibility of having a function that is zero except

at a countable set of points. Such a function will have integral zero, and then the

norm will not be positive deﬁnite. So we must identify this function with the

zero function.

In general, we must identify functions that are equal on “negligble” sets. That

is, each element of L

will be an equivalence class of functions that are equal almost

everywhere, in the sense of Section 6.6. We return to this issue after proving the

next theorem.

Measure theory has something useful to add to this picture, since there is a

notion of measurable function that provides a natural set to choose the L

functions

from. However, this approach cannot get around the essential difﬁculty that an

element of L

(a, b) is not actually a function.

9.6.5. THEOREM. If f ∈ L

(a, b), we may choose a sequence (f

) in C[a, b]

converging to f in L

(a, b) so that lim

n→∞

(x) converges almost everywhere (a.e.)

to a function f(x). Moreover, if (g

) is another sequence converging to f in

(a, b) and pointwise to g(x) a.e., then g(x) = f(x) a.e.

PROOF. Choose any sequence (f

) in C[a, b] such that kf −f

< 4

−n

for n ≥ 1.

Let

= {x ∈ [a, b] : |f

(x) − f

n+1

(x)| > 2

−n

This is an open set since |f

−f

n+1

|is continuous. Hence there are disjoint intervals

, d

) so that U

i≥1

, d

). Now

− f

n+1

≤ kf

− fk

+ kf − f

n+1

< 2 · 4

−n

On the other hand,

− f

n+1

≥

i≥1

(x) − f

n+1

(x)|

dx ≥ 2

−np

i≥1

− c

Therefore,

i≥1

− c

< (2 · 4

−n

)

= 2

(1−n)p

270 Metric Spaces

Let E =

k≥1

n≥k

. This set may be covered by the open intervals mak-

ing up

n≥k

, which have total length at most

n≥k

(1−n)p

= 2

−kp

1−2

−p

This converges to 0 as k increases and thus may be made less than any given ε > 0.

Therefore, E has measure zero.

If x ∈ [a, b]\E, then there is an integer N = N(x) so that x is not in

n≥N

Hence |f

(x) − f

n+1

(x)| ≤ 2

−n

for all n ≥ N . Thus if N ≤ n ≤ m,

(x) − f

(x)| ≤

m−1

k=n

(x) − f

k+1

(x)| ≤

m−1

k=n

−k

< 2

1−n

Therefore (f

(x)) is a Cauchy sequence, and lim

n→∞

(x) converges, say to f(x).

That is, this sequence converges almost everywhere.

Now suppose that (g

) is another sequence in C[a, b] converging to f in the L

norm and converging almost everywhere to a function g. Drop to a subsequence

) so that kg

− fk

< 4

−k

for k ≥ 1. Let

= {x ∈ [a, b] : |g

(x) − f

(x)| > 2

−k

This is an open set because |g

− f

| is continuous. Now

− f

≤ kg

− fk

+ kf − f

< 2 · 4

−k

Arguing as previously, the intervals making up V

have length at most 2

(1−n)p

Consequently, as before, the set F =

k≥1

n≥k

has measure zero.

If x ∈ [a, b] \ F , then there is an N = N(x) so that x is not in

n≥N

Hence lim

k→∞

(x) −f

(x)| = 0. Now f

(x) converges to f(x) for x ∈ [a, b] \E.

Therefore, g(x) = f(x) for every x ∈ [a, b] \ (E ∪ F ). Thus g = f a.e. ¥

One of the subtleties of L

spaces is that the elements of L

[a, b] are not really

functions. The preceding theorem shows that they may be represented by functions

but that representation is not unique. Two functions that differ on a set of measure

zero both represent the same element of L

[a, b]. The correct way to handle this

is to identify elements of L

[a, b] with an equivalence class of functions that agree

almost everywhere. (See Appendix 1.6.)

Another point that bears repeating is that in this approach, the integral of these

limit functions is determined abstractly. The Lebesgue integral develops a method

of integrating these functions in terms of their values. This is an important view-

point that we do not address here. We refer the interested reader to [17] or [15].

Exercises for Section 9.6

A. Show that every piecewise continuous function on [a, b] is simultaneously the limit in

(a, b) and the pointwise limit everywhere of a sequence of continuous functions.

B. (a) Show that the characteristic function

of any open set U ⊂ [a, b] is the increasing

limit of a sequence of continuous functions.

(b) Hence show that

belongs to L

(a, b). If U is the disjoint union of open intervals

= (c

, d

), show that

= k

9.6 The L

Spaces and Abstract Integration 271

C. Suppose that f ∈ L

[a, b] and f ≥ 0 and

f = 0. Prove that f = 0 a.e.

HINT: Apply the Monotone Convergence Theorem (MCT) with f

= nf.

D. (a) Suppose that A ⊂ [a, b] has measure zero. Show that there is a decreasing sequence

of open sets U

containing A such that lim

n→∞

= 0.

(b) If f is a positive element of L

(a, b) that is represented by a function f(x) such

that f(x) = 0a.e., prove that kfk

= 0.

HINT: Let A

= {x : 0 < f (x) < n}. Consider f

(a, b) is represented by a function f (x) such that f(x) = 0a.e.

Prove that

|f| = 0 (i.e. f = 0 in L

(a, b)).

E. Suppose that f

∈ C[a, b] is a decreasing sequence of continuous functions converging

pointwise to 0. Prove that lim

n→∞

(x) dx = 0.

HINT: Exercise 8.1.I

F. (a) Show that if f, g, h ∈ L

[a, b], then f ∨g :=

(f + g + |f −g|) satisﬁes f ≤ f ∨g

and g ≤ f ∨g.

(b) If f ≤ h and g ≤ h, show that f ∨ g ≤ h. That is, f ∨ g = max{f, g}.

G. Dominated Convergence Theorem. Suppose that f

, g ∈ L

[a, b] and |f

| ≤ g for

all n ≥ 1. Also suppose that lim

n→∞

(x) = f(x) a.e.

(a) Deﬁne u

= f

∨f

m+1

∨···∨f

and l

= f

∧f

m+1

∧···∧f

. Show that

= lim

n→∞

belongs to L

(a, b), as does l

= lim

n→∞

. HINT: MCT

(b) Show that u

is a decreasing sequence of functions converging to f almost every-

where, and l

is an increasing sequence converging to f almost everywhere.

m→∞

= lim

m→∞

f. HINT: MCT

(d) Prove that lim

n→∞

f. HINT: l

≤ f

≤ u

H. Lebesgue Measure. Let Σ consist of all subsets A of [a, b] such that

belongs to

(a, b) in the sense that there is an f ∈ L

(a, b) with f(x) =

a.e. Deﬁne a

function m : Σ → [0, +∞) by m(A) =

(a) Show that if A, B ∈ Σ, then A ∩ B, A ∪ B and A \ B belong to Σ.

(b) Show that m(A ∩ B) + m(A ∪ B) = m(A) + m(B) for all A, B ∈ Σ.

(d) Countable Additivity. If A

are disjoint sets in Σ, show that

n≥1

belongs to

Σ, and m

n≥1

m(A

(e) Show that Σ contains all open and all closed subsets of [a, b]; and verify that

m((c, d)) = m([c, d]) = d − c.