Download TOPOLOGICAL GROUPS - PART 1/3 Contents 1. Locally compact

TOPOLOGICAL GROUPS - PART 1/3
T.K.SUBRAHMONIAN MOOTHATHU
Contents
1.
Locally compact second countable spaces
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2.
Measure Theory on LCSC spaces
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3.
Measure Theory and Functional Analysis
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1. Locally compact second countable spaces
All topological spaces are assumed to be Hausdorff, unless stated otherwise.
Locally Compact (LC) spaces: The best after compact spaces. A topological space is LC if every
x ∈ X has a neighborhood whose closure is compact; equivalently if X has a base B such that B is
compact for every B ∈ B.
One-point compactification of a LC (non-compact) space (X, T ) is a compact Hausdorff space
(X ∪ {∞}, T 0 ) where T 0 is generated by T ∪ {{∞} ∪ (X \ K) : K ⊂ X compact}. Thus (X, T ) is
a dense open subspace of (X ∪ {∞}, T 0 ).
[101] Let X be LC Hausdorff. Then X is completely regular.
Proof. X ∪ {∞} is compact Hausdorff, hence normal, hence completely regular. And complete
regularity is passed onto all subspaces.
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[102] Let X be LC Hausdorff. If x ∈ X and if U is a neighborhood of x, then there exists open
V ⊂ X such that V is compact and x ∈ V ⊂ V ⊂ U .
Proof. X is completely regular, hence regular. Let V1 ⊂ X be open such that x ∈ V1 ⊂ V1 ⊂ U .
Since X is LC, there is open V2 ⊂ X such that x ∈ V2 and V2 is compact. Let V = V1 ∩ V2 .
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[103] Closed subspaces and open subspaces of LC spaces are LC.
Proof. If Y ⊂ X is closed and if y ∈ Y , let U ⊂ X be open with y ∈ U and U compact. If
V = Y ∩ U , then V is open in Y , y ∈ V and clY [V ] = Y ∩ U is compact since Y is closed. If Z ⊂ X
is open, then by [102], Z is LC.
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Examples: Rn , Cn are LC; discrete spaces are LC; GL(n, R) = Rn \ det−1 ({0}) is open in Rn
and hence LC; (a, b) is LC; infinite dimensional Banach spaces are not LC [∵ closed balls are not
compact]; Q is not LC [∵ Q ∩ [a, b] is not sequentially compact for reals a < b].
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T.K.SUBRAHMONIAN MOOTHATHU
[104] LC Hausdorff spaces have the Baire property. That is, countable intersection of dense open
subsets is nonempty.
Proof. Let Un ⊂ X be open and dense for n ∈ N. Choose nonempty open B1 ⊂ X such that B1
is compact B1 ⊂ U1 . At the (n + 1)th step, choose nonempty open Bn+1 ⊂ X such that Bn+1 is
T
T∞
compact and Bn+1 ⊂ Bn ∩ Un+1 . Then ∅ 6= ∞
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n=1 Bn ⊂
n=1 Un .
[105] Let X be a homogeneous LC Hausdorff space. If X is countable, then the topology on X is
discrete.
Proof. If X is not discrete, then by homogeneity {x} is not open for every x ∈ X. Let X =
T
{x1 , x2 , . . .} and Un = X \ {xn }. Then Un is open and dense in X but n Un = ∅ contradicting the
Baire property.
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Remarks: As a corollary we get that Q is not LC since Q is homogeneous. By considering {0}∪{1/n :
n ∈ N}, we see that homogeneity of the space is necessary in [105].
[106] Let X be Hausdorff (need not be LC) and Y ⊂ X be dense and LC. Then Y is open in X.
Proof. If y ∈ Y , then there is a relatively open V ⊂ Y such that y ∈ Y and clY [V ] is compact.
Then V = clY [V ]. Let U ⊂ X be open such that V = Y ∩ U . Since Y ∩ U is dense in U , we have
U = V = clY [V ] ⊂ Y . Thus U ⊂ Y .
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Corollary: R \ Q is not LC.
[107] Let X be LCSC (locally compact, second countable) Hausdorff. Then X is metrizable.
Proof. By [101], X is completely regular and hence regular. Uryzohn’s metrization Theorem says
that regular SC spaces are metrizable.
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[108] Let X be LCSC Hausdorff. Then,
(i) X has a countable base {Bn : n ∈ N} such that Bn is compact for every n.
(ii) X is σ-compact. In fact, there is a sequence (Kj ) of compact subsets such that Kj ⊂ int[Kj+1 ]
S
and X = ∞
j=1 Kj .
Proof. Let B0 be a countable base of open balls for X and let B = {B(x, ²) ∈ B0 : B(x, 2²) compact}.
(The extra factor 2 will be used to prove (ii)). Check that B is also a base for X. To prove (ii),
write B = {B(xn , ²n ) : n ∈ N}. Let (αj ) be a strictly increasing sequence in (1, 2) and put
S
Kj = jn=1 B(xn , αj ²n ).
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[109] Let X be LCSC Hausdorff. Then X ∪ {∞} is SC and metrizable.
Proof. If D ⊂ X is compact, then D ⊂ X =
S∞
j=1 Kj
⊂
S∞
j=1 int[Kj+1 ],
so D ⊂ Kj+1 for some j.
This means that {{∞} ∪ (X \ Kj ) : j ∈ N} is a countable base at ∞. Combining with a countable
TOPOLOGICAL GROUPS - PART 1/3
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base for X (since X is open in X ∪ {∞}), we get a countable base for X ∪ {∞}. Being compact
Hausdorff, X ∪ {∞} is normal and hence regular. Now apply Urysohn’s metrization theorem.
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Thus if X is LCSC (Hausdorff), its one-point compactification is a compact metric space. Since
compact metric spaces are ‘nice’, we may expect LCSC spaces also to possess many ‘nice’ features.
[110] Let (Y, d) be a complete metric space and let X ⊂ Y be a nonempty Gδ set. Then X admits
a complete metric. In particular, open subsets of complete metric spaces are completely metrizable.
Proof. Suppose X =
T∞
n=1 Un
where each Un is open in Y . We may assume Un 6= Y . We embed
X as a closed subspace of Y × RN to say X is completely metrizable. Let f : X → Y × RN
be f (x) = (x, z), where z = (zn ), zn = 1/d(x, Y \ Un ). Easy to see that f is a 1-1 continuous
function with f −1 : f (X) → X continuous. To show f (X) is closed, consider (y, w) ∈ Y × RN
and let (xm ) be a sequence in X such that (f (xm )) → (y, w). Then (xm ) → y and therefore
f (y) = limm→∞ f (xm ) = (y, w) so that it now suffices to show y ∈ X. If y ∈
/ X, then y ∈ ∂Un for
some n, and then wn = limm→∞ [d(xm , Y \ Un )]−1 = ∞, a contradiction.
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Example: If Y = R and X = R \ {0}, then a complete metric on X is d(a, b) = |a − b| + | a1 − 1b |.
Try to write down a complete metric on R \ Q.
[111] Let X be LCSC. Then X admits a complete metric.
Proof. X ∪ {∞} is a compact metric space and X is open in X ∪ {∞}.
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Exercise 1 : Let X, Y be non-compact LC Hausdorff spaces and let f : X → Y be a continuous
proper map (i.e., f −1 (K) is compact for every compact K ⊂ Y ). Show that (i) f has a continuous
extension fe : X ∪ {∞} → Y ∪ {∞0 }. (ii) f is a closed map. [Hint: (i) Put fe(∞) = ∞0 . (ii) If
A ⊂ X is closed in X and B is the closure of A in X ∪ {∞}, then B is compact and B = A or
A ∪ {∞}. Consider Y ∩ fe(B).]
Exercise 2 : Let X be LC Hausdorff and consider K ⊂ U ⊂ X, where K is compact and U is open.
Then there exists open V ⊂ X such that V is compact and K ⊂ V ⊂ V ⊂ U . [Hint: X is regular.
So for each x ∈ K, there is open Vx such that Vx is compact and x ∈ Vx ⊂ Vx ⊂ U .]
Exercise 3 : [Difference between compact and LC spaces] (i) Continuous image of a LC space need
Q
not be LC. (ii) Infinite product of LC spaces need not be LC; in fact, α Xα is LC iff every Xα
is LC and Xα is compact except for finitely may α. (iii) A continuous real valued map on a LC
space need not be uniformly continuous. [Hint: (i) Identity map from Q with discrete topology to
Q with the usual topology. (ii) RN . (iii) f (x) = x2 on R.]
Exercise 4 : Let X be LC Hausdorff. (i) U ⊂ X is open if U ∩ K is open in K for every compact
K ⊂ X. (ii) If Y is Hausdorff and f : X → Y is a function such that f |K : K → Y is continuous
for every compact K ⊂ X, then f is continuous. [Hint: (i) If x ∈ U , choose open V ⊂ X such that
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T.K.SUBRAHMONIAN MOOTHATHU
x ∈ V and V is compact. Since U ∩ V is open in V , U ∩ V is open in V and hence open in X. (ii)
follows from (i).]
Space of continuous functions C(X, Y ): We know that pointwise limit of a sequence of continuous
functions need not be continuous. But the limit function is continuous if the convergence is locally
uniform (i.e., every point has a neighborhood where the convergence is uniform). If the domain X
is LC, then it may be seen that locally uniform convergence is equivalent to uniform convergence
on compact subsets of X. Thus:
[112] Let X be LC and Y be a metric space. If a sequence (fn ) in C(X, Y ) converges to a function
f : X → Y uniformly of compact subsets of X, then f ∈ C(X, Y ).
Because of this and certain other reasons, the ‘topology of uniform convergence on compact
subsets’ is the ‘right’ topology on C(X, Y ). This topology is metrizable when X is LCSC, and this
topology can be presented in three ways. Suppose that X is LCSC and let (Y, ρ) be a metric space.
Replacing ρ with min{1, ρ}, we may assume ρ ≤ 1. Three (equivalent) topologies on C(X, Y ) are
described below.
(i) TU C , topology of uniform convergence on compact subsets: For f ∈ C(X, Y ), compact K ⊂ X
and ² > 0, let B(f, K, ²) = {g ∈ C(X, Y ) : ρ(f (x), g(x)) < ² ∀x ∈ K}. If g ∈ ∩2i=1 B(fi , Ki , ²i ), note
that the continuous function x → ρ(fi (x), g(x)) attains maximum on compact Ki , and therefore
there exists αi > 0 such that ρ(fi (x), g(x)) + αi < ²i for every x ∈ Ki , for i = 1, 2. If α =
min{α1 , α2 }, then B(g, K1 ∪ K2 , α) ⊂ ∩2i=1 B(fi , Ki , ²i ). It follows that the collection {B(f, K, ²) :
f ∈ C(X, Y ), K ⊂ X compact , ² > 0} is a base for a topology on C(X, Y ). This topology is TU C .
(ii) TCO , compact-open topology: The subbase {S(K, V ) : K ⊂ X compact ,V ⊂ Y open} generates this topology, where S(K, V ) = {f ∈ C(X, Y ) : f (K) ⊂ V }.
[Remark : TP C , the topology of pointwise convergence on C(X, Y ) is the subspace topology coming
from the inclusion C(X, Y ) ⊂ Y X . Now, {S(F, V ) : F ⊂ X finite ,V ⊂ Y open} is a subbase for
TP C on C(X, Y ); and therefore, TCO = TP C on C(X, Y ) when X is discrete.]
Tσ , a topology on C(X, Y ) coming from the σ-compactness of X: Write X =
S∞
j=1 Kj ,
where Kj ’s
are compact and Kj ⊂ int[Kj+1 ]. Let dj be the supremum metric on C(Kj , Y ), and define a metric
Q
d on C(X, Y ) making use of the embedding of C(X, Y ) in ∞
j=1 C(Kj , Y ) by the natural map
P∞ dj (f, g)
f 7→ (f |K1 , f |K2 , . . .). That is, d(f, g) =
, where dj (f, g) = dj (f |Kj , g|Kj ) to be
j=1
2j
precise. The topology induced by this metric on C(X, Y ) will be denoted by Tσ . Observe that if
X is compact, then Tσ is induced by the supremum metric on C(X, Y ). In the general case, the
following result says in particular that Tσ is independent of the choice of Kj ’s.
[113] Let X be LCSC and (Y, ρ) be a metric space with ρ ≤ 1. Then, TU C = TCO = Tσ on C(X, Y ).
Proof. TU C ⊂ TCO : Consider B(f, K, ²). Choose δ > 0 such that ρX (a, b) < δ implies ρ(f (a), f (b)) <
² for every a, b ∈ K. Cover K with finitely many balls B1 , . . . , Bn having compact closures
TOPOLOGICAL GROUPS - PART 1/3
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and diameter < δ. Let Ai = K ∩ Bi , ai ∈ Ai and let Vi = Bρ (f (ai ), ²/2) ⊂ Y . Check that
T
f ∈ ni=1 S(Ai , Vi ) ⊂ B(f, K, ²).
S
TCO ⊂ Tσ : Consider S(K, V ) and f ∈ S(K, V ). Since K ⊂ X = ∞
j=1 int[Kj+1 ], there is j with
K ⊂ Kj . Also since f (K) is compact and is contained in the open set V , there is ² > 0 such that
N² (f (K)) ⊂ V . If g ∈ Bd (f, ²/2j ), then dj (f, g) ≤ 2j d(f, g) < ², and therefore g(K) ⊂ N² (f (K)) ⊂
V . Thus f ∈ Bd (f, ²/2j ) ⊂ S(K, V ).
P
Tσ ⊂ TU C : Consider Bd (f, ²) and choose k ∈ N so that j>k 2−j < ²/2. If g ∈ B(f, Kk , ²/2),
P
P
then dj (f, g) ≤ ²/2 for j ≤ k so that kj=1 2−j dj (f, g) ≤ ²/2( kj=1 2−j ) < ²/2. Hence d(f, g) ≤
Pk
P
−j
−j < ². Thus, f ∈ B(f, K , ²) ⊂ B (f, ²).
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k
d
j=1 2 dj (f, g) +
j>k 2
Exercise-5 : Let X be LCSC and (Y, ρ) be a metric space with ρ ≤ 1. (i) If we consider the topology
of pointwise convergence on C(X, Y ), then the evaluation map (x, f ) 7→ f (x) from X × C(X, Y )
to Y need not be continuous. If we consider TU C = TCO = Tσ on C(X, Y ), then the evaluation
map (x, f ) 7→ f (x) from X × C(X, Y ) to Y is continuous. [Hint: (i) Consider a sequence (fn ) in
C([0, 1], R) such that (fn ) → 0 pointwise and fn (1/n) = 1 for every n ∈ N. (ii) Let (xn ) → x
and (fn ) → f . To show (fn (xn )) → f (x), note that (fn ) → f uniformly on the compact set
{x} ∪ {xn : n ∈ N}.]
From now onwards, when X is LCSC and (Y, ρ) is a metric space (with ρ ≤ 1), the topology on
C(X, Y ) will be TU C = TCO = Tσ in our discussions.
[114] Let X be LCSC and (Y, ρ) be a metric space. If Y is SC, then C(X, Y ) is SC and hence
separable.
Proof. It is convenient to work with TCO . Note that it is enough to produce a countable subbase.
Let BX be a countable base for X consisting of open balls with compact closures. Let {An : n ∈ N}
be the collection of all finite unions of closures of balls from BX . Let BY be a countable base of
open balls for Y and let {Vm : m ∈ N} be the collection of all finite unions of balls in BY . We
claim that {S(An , Vm ) : n, m ∈ N} is a countable subbase for (C(X, Y ), TCO ). Consider S(K, V )
and f ∈ S(K, V ). Cover f (K) with finitely many balls from BY contained in V and let Vm be their
union. Choose ² > 0 so that N² (f (K)) ⊂ Vm . For each x ∈ K choose Bx ∈ BX such that x ∈ Bx
and f (Bx ) ⊂ Bρ (f (x), ²) ⊂ N² (f (K), ²). Cover K with finitely many such balls and let An be the
union of their closures. Check that f ∈ S(An , Vm ) ⊂ S(K, V ).
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[115] Let X be LCSC and (Y, ρ) be a complete metric space. Then (C(X, Y ), d) is a complete
metric space.
Proof. Let X =
S∞
j=1 Kj ,
where Kj is compact and Kj ⊂ int[Kj+1 ], and let d =
P∞
−j
j=1 2 dj ,
where dj is the supremum metric on C(Kj , Y ). Consider a Cauchy sequence (fn ) in (C(X, Y ), d).
If x ∈ X, there is j such that x ∈ Kj and then ρ(fn (x), fm (x)) ≤ dj (fn , fm ) ≤ 2j d(fn , fm ) and hence
(fn (x)) is Cauchy in the complete space (Y, ρ). Define f : X → Y as the pointwise limit of (fn ).
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T.K.SUBRAHMONIAN MOOTHATHU
Since dj ≤ 2j d, (fn ) is Cauchy with respect to dj . Hence for a fixed j and ² > 0 there is n0 such that
dj (fn , fm ) < ²/2 for every n, m ≥ n0 . If n ≥ n0 and x ∈ Kj choose m ≥ n0 with ρ(f (x), fm (x)) <
²/2 by pointwise convergence. Then ρ(f (x), fn (x)) ≤ ρ(f (x), fm (x)) + dj (fm , fn ) < ². Hence
S
(fn ) → f uniformly on Kj and therefore f |Kj is continuous. Since X = ∞
j=1 int[Kj+1 ], it follows
that f ∈ C(X, Y ). To verify that (d(fn , f )) → 0, consider ² > 0 and let k ∈ N be such that
P
−j < ²/2. Choose n such that d (f, f ) < ²/2 for n ≥ n . Then d(f, f ) < ²/2 for j ≤ k
0
n
0
n
k
j>k 2
Pk
P
−j
−j
and n ≥ n0 so that d(f, fn ) ≤ ²/2( j=1 2 ) + j>k 2 < ² for n ≥ n0 . [Slightly easier argument
is to prove the completeness of C(X, Y ) first when X is compact and then to observe in the general
Q
case that C(X, Y ) is embedded as a closed subspace of ∞
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j=1 C(Kj , Y ).]
[116] Corollary: If X is LCSC, then C(X, R) and C(X, C) are complete separable metric spaces.
Remark : C(X, C) is not LC in general. For example, since C([0, 1], C) is an infinite dimensional
Banach space with respect to the supremum norm and since the supremum metric on C([0, 1], C)
comes from this suprem norm, C([0, 1], C) is not LC. Compact subsets of C(X, C) are described by
[117] below when X is a compact metric space.
Exercise-6 : Let (X, d) be a metric space and A ⊂ X. Then, (i) A is totally bounded iff A is totally
bounded, (ii) A is totally bounded iff every sequence in A has a Cauchy subsequence. [Hint: If
A is not totally bounded, there is δ > 0 such that A cannot be written as a finite union of open
balls of radius δ. Inductively choose x1 , x2 , . . . in A such that d(xj , xn ) ≥ δ for j < n. Then (xn )
has no Cauchy subsequence. Conversely, if (xn ) is a sequence in A, cover A with finitely many
balls of radius 1/2. One of them, say B1 , contains infinitely many xn ’s. Let xn1 ∈ B1 . At the
(k + 1)th step, cover A ∩ Bk with finitely many balls of radius 2−(k+1) . One of them, say Bk+1 ,
contains infinitely many xn ’s contained in Bk . Let nk+1 > nk be such that xnk+1 ∈ Bk+1 ∩ Bk .
Since d(xnk , xnk+1 ) < diam[Bk ] ≤ 2−k , it follows that (xnk ) is Cauchy.]
[117] [Arzela-Ascoli Theorem] Let X be a compact metric space and Γ ⊂ C(X, C). Then Γ is
compact iff Γ is pointwise bounded and equicontinuous.
Proof. Since Γ is closed in the complete space C(X, C), Γ is always complete. Hence Γ is compact
iff Γ is totally bounded iff Γ is totally bounded. We can show that:
(i) If Γ is totally bounded, the Γ is equicontinuous.
(ii) If Γ is pointwise bounded and equicontinuous, then Γ is uniformly bounded.
(iii) If Γ is uniformly bounded and equicontinuous, then Γ is totally bounded (where the crucial
point to use is that bounded subsets of C are totally bounded).
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Remark : The above result is no longer true if X is only LCSC. If X = R and Γ = {f } ⊂ C(X, C),
f (x) = x2 , then Γ is compact but not equicontinuous since f is not uniformly continuous on R. It
is not difficult to see that if X is compact and Γ ⊂ C(X, C) is pointwise equicontinuous, then Γ is
(uniformly) equicontinuous. The correct generalization of [117] to the case where X is LCSC is:
TOPOLOGICAL GROUPS - PART 1/3
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[1170 ] [Arzela-Ascoli Theorem] Let X be LCSC and Γ ⊂ C(X, C). Then Γ is compact iff Γ is
pointwise bounded and pointwise equicontinuous.
Proof. Suppose Γ is compact. The set {f (x) : f ∈ Γ} is the continuous image of the compact
set {x} × Γ under the evaluation map (x, f ) 7→ f (x) from X × C(X, C) to C [∵ Exercise-5]. So
{f (x) : f ∈ Γ} is compact and hence bounded for each x ∈ X. Thus Γ is pointwise bounded. To
show Γ is pointwise equicontinuous, consider x ∈ X. Choose compact K ⊂ X with x ∈ int[K].
Since f 7→ f |K from C(X, C) to C(K, C) is continuous, {f |K : f ∈ Γ} is compact in C(K, C), and
then by [117], {f |K : f ∈ Γ} is equicontinuous. In particular, Γ is equicontinuous at x.
Conversely, suppose that Γ is pointwise bounded and pointwise equicontinuous. Write X =
S∞
j=1 Kj , where Kj is compact and Kj ⊂ int[Kj+1 ]. Let Γj = {f |Kj : f ∈ Γ}. Then Γj is pointwise
bounded and (uniformly) equicontinuous. Hence by [117], Γj is totally bounded for every j ∈ N.
If Γ is not totally bounded, there is δ > 0 and (fn ) in Γ such that d(fn , fm ) ≥ δ for n 6= m [∵
P
Exercise-6]. If k ∈ N is such that j>k 2−j < δ/2, then we have dk (fn , fm ) ≥ δ/2 for n 6= m, a
contradiction to the total boundedness of Γk . Thus Γ must be totally bounded and therefore Γ is
compact.
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Example: Let Γ = {fn : n ∈ N} ⊂ C([0, 1], C), where fn (x) = xn . Note that if xn is the positive nth root of 1/n, then (xn ) → 1, fn (1) = 1 and (fn (xn )) = (1/n) → 0. Hence Γ is not equicontinuous
and therefore Γ is not compact by [117].
2. Measure Theory on LCSC spaces
We need some abstract measure theory, especially on LCSC spaces. If X is LCSC, the σ-algebra
B on X generated by open subsets of X is called the Borel σ-algebra on X and the members of B are
called Borel subsets of X. If X is LCSC, let Cc (X, R) = {f ∈ C(X, R) : f has compact support},
where the support of f is defined as supp(f ) = {x ∈ X : f (x) 6= 0} (thus Cc (X, R) = C(X, R) when
X is compact). It may be noted that Cc (X, R) is a vector subspace (in fact, a subalgebra) of
C(X, R), and the following says that Cc (X, R) has plenty of elements.
[118] Let X be LCSC and K ⊂ U ⊂ X, where K is compact and U is open. Then there exists
f ∈ Cc (X, R) such that 0 ≤ f ≤ 1, f |K ≡ 1 and supp(f ) ⊂ U .
Proof. Choose open W ⊂ X with compact closure such that K ⊂ W ⊂ W ⊂ U . If d is an
admissible metric on X, let f : X → [0, 1] be f (x) = d(x, X \ W )/[d(x, K) + d(x, X \ W )].
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Remark : The above result is true, but non-trivial to prove, when X is only LC Hausdorff.
Exercise-7 : Let X be LCSC, K ⊂ X be compact and let {Uα : α ∈ I} be an open cover for K.
Then there is ² > 0 such that N² (K) is compact and is also covered by {Uα : α ∈ I}. [Hint: If
x ∈ K, there is α with x ∈ Uα and then there exists an open ball B with compact closure such that
x ∈ B ⊂ B ⊂ Uα .]
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T.K.SUBRAHMONIAN MOOTHATHU
A concept called partition of unity is useful in obtaining global results by patching together
various local arguments. Generalizing [118], we have:
[119] [Partition of unity] Let X be LCSC, K ⊂ X be compact and let {Uα : α ∈ I} be an open
cover for K. Then there exists finitely many functions f1 , . . . , fn ∈ Cc (X, R) such that 0 ≤ fi ≤ 1
P
for every i, ni=1 fi (x) = 1 for every x ∈ K, and for each i there is α such that supp(fi ) ⊂ Uα (we
say fi ’s are subordinate to {Uα : α ∈ I}).
Proof. There is open W ⊂ X with compact closure such that K ⊂ W and {Uα : α ∈ I} is an
open cover for W also [∵ Exercise-7]. Since K, X \ W are disjoint closed sets, there is continuous
h : X → [0, 1] such that h ≡ 0 on K and h ≡ 1 on X \ W .
Now, we can cover W with finitely many open balls B1 , . . . , Bn having compact closures such
that for each i, there is αi with Bi ⊂ Uαi . Let gi ∈ Cc (X, [0, 1]) be such that gi ≡ 1 on Bi and
P
P
supp(gi ) ⊂ Uαi for each i. Note that ni=1 gi (x) > 0 for every x ∈ W and hence ni=1 gi (x)+h(x) >
P
0 for every x ∈ X. Let fi (x) = gi (x)/[ nj=1 gj (x) + h(x)].
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Coming back to σ-algebras, let X be LCSC and consider the following σ-algebras on X:
A1 = the σ-algebra generated by compact subsets of X,
A2 = the σ-algebra generated by compact Gδ subsets of X,
A3 = the smallest σ-algebra such that every f ∈ C(X, C) is measurable (with respect to the Borel
σ-algebra on C),
A4 = the smallest σ-algebra such that every f ∈ Cc (X, C) is measurable.
Remark : A3 is sometimes called the Baire σ-algebra, and the members of A3 are called Baire subets
of X.
[120] Let X be LCSC, let B be the Borel σ-algebra on X and let Ai ’s be as given above. Then,
B = A1 = A2 = A3 = A4 .
Proof. We have A2 ⊂ A1 ⊂ B clearly. In a metric space every closed set and every hence every
T
compact is Gδ [∵ A = ∞
n=1 N1/n (A)], and closed subsets of LCSC spaces are σ-compact [∵
S∞
S∞
A = j=1 [A∩Kj ] if X = j=1 Kj ]. Thus B = A1 = A2 . If K ⊂ X is compact, then f ∈ Cc (X, R) ⊂
Cc (X, C) given in the proof of [118] satisfies f (x) = 1 iff x ∈ K and thus K = f −1 ({1}) ∈ A4 . So
A1 ⊂ A4 . Since A4 ⊂ A3 ⊂ B = A1 , we get B = A3 = A4 .
¤
Exercise-8 : Let (X, A), (Y, A0 ) be measure spaces and suppose that A0 is generated by S ⊂ A0 (for
example, the collection of open intervals generates the Borel σ-algebra on R). If f : (X, A) → (Y, A0 )
is a function such that f −1 (S) ∈ A for every S ∈ S, then f is measurable. [Hint: Verify that
A00 = {A ∈ A0 : f −1 (A) ∈ A} is a σ-algebra.]
TOPOLOGICAL GROUPS - PART 1/3
9
Remark : Let X be LCSC. Any measurable f : (X, BX ) → (C, BC ) is called a (complex) Borel
function. The family of all complex Borel functions on X is characterized as the smallest family containing C(X, C) and closed under the operation of taking pointwise limit of sequences of
functions. This follows from Exercises 9,10 and 11 below.
Exercise-9 : Let X be LCSC and (fn ) be a sequence of complex Borel functions on X converging
pointwise to a function f : X → C. Then f is also Borel. [Hint: Enough to show that f −1 (K) ∈ BX
T
S∞
−1
for every compact K ⊂ C. Let Um = N1/m (K) and show that f −1 (K) = ∞
m=1 n=m fn (Um ).
Here the inclusion ⊃ is proved as follows. If x belongs to the right-side, then for every m there is
n ≥ m with d(fn (x), K) < 1/m. Since K is compact, there is y ∈ K and a subsequence (nk ) such
that limk→∞ fnk (x) = y. Then y = f (x) so that x ∈ f −1 (K).]
Exercise-10 : Let X be LCSC and let F be the smallest collection of complex functions on X such
that C(X, C) ⊂ F and F is closed under the operation of taking pointwise limits of sequences of
functions. Then F contains all complex-valued simple Borel functions (i.e., finite linear combinations of characteristic functions of Borel sets). [Hint: Since F is a vector space, it suffices to show
χB ∈ F for every Borel B ⊂ X. Let A = {B ⊂ X : χB ∈ F }. If K ⊂ X is compact, then there is
continuous fn : X → [0, 1] such that fn |K ≡ 1 and fn ≡ 0 on X \ N1/n (K). Then (fn ) → χK pointwise, and thus A contains all compact subsets of X. Now it suffices to show that A is a σ-algebra.
Observe the following: (i) (fn ) → χB implies (1 − fn ) → χX\B , (ii) Since C(X, R) is closed under
the operations of taking maximum and minimum of finitely many functions, the same is true for
real valued functions in F and consequently A is closed under finite unions and finite intersections,
S
(iii) A, B ∈ A implies A\B = A∩(X \B) ∈ A, (iv) If An ∈ A, B1 = A1 and Bn+1 = An+1 \ ni=1 Ai ,
S∞
P
S
then Bn ∈ A; and if fn = ni=1 χBi , then (fn ) → χA , where A = ∞
n=1 Bn .]
n=1 An =
Exercise-11 : Let X be LCSC and f : X → C be Borel. Then f is the pointwise limit of a sequence
of simple Borel functions from X to C. [Hint: Assume f is real valued. For −n2n ≤ j < n2n , let
P(n−1)2n
A(n, j) = f −1 ([ 2jn , j+1
j=−n2n χA(n,j) .]
2n )), which is clearly Borel and define fn : X → R as fn =
Exercise-12 : Let f : R → R be differentiable. Then f 0 is the pointwise limit of a sequence of
continuous real valued functions on R and hence f 0 is Borel. [Hint: gn (x) = (f (x + 1/n) −
f (x))/(1/n).]
Aside: BR and R have the same cardinality (proof by transfinite induction), whereas the cardinality
of the Lebesgue σ-algebra on R is same as that of P(R) [∵ every subset of the middle-third Cantor
set is Lebesgue measurable]. Hence there are non-Borel, Lebesgue measurable sets A ⊂ R.
Locally finite regular Borel measure: Let X be LCSC. Any measure on (X, B) is called a Borel
measure. A Borel measure is locally finite if µ[K] < ∞ for every compact K ⊂ X (equivalently if
every point has a neighborhood with finite measure). Note that every locally finite Borel measure
10
T.K.SUBRAHMONIAN MOOTHATHU
µ on a LCSC space X is σ-finite; and µ[X] < ∞ when X is compact. A Borel measure µ on a
LCSC space X is regular if for every Borel B ⊂ X, we have:
(i) µ[B] = sup{µ[K] : K ⊂ B compact}
[inner regularity],
(ii) µ[B] = inf{µ[U ] : B ⊂ U, U open in X}
[outer regularity].
S
Exercise-13 : Let (X, µ) be a measure space, let En , Fn ⊂ X be measurable, and let E = ∞
n=1 En ,
T∞
F = n=1 Fn . (i) If En ⊂ En+1 and µ[E] < ∞, then for every ² > 0 there is k ∈ N such that
µ[E \ Ek ] < ². (ii) If Fn+1 ⊂ Fn and µ[F1 ] < ∞, then for every ² > 0 there is k ∈ N such that
µ[Fk \ F ] < ². (iii) The finiteness assumption ‘µ[·] < ∞’ cannot be dropped from (i) and (ii).
S
P∞
[Hint: (i) Let A1 = E1 and An+1 = En+1 \ En . Then, µ[E] = µ[ ∞
n=1 An ] =
n=1 µ[An ] < ∞
S∞
P∞
by assumption. Hence µ[E \ Ek ] = µ[ n=k+1 An ] =
n=k+1 µ[An ] → 0 as k → ∞. (ii) Take
En = F1 \ Fn and apply (i). For (iii), consider X = R, En = (−n, n) and Fn = (n, ∞).]
[121] Let X be a compact metric space and let µ be a Borel measure on X with µ[X] < ∞. Then
µ is regular.
Proof. Let A be the collection of all B ∈ B such that for every ² > 0, there exist compact K² ⊂
X and open U² ⊂ X with K² ⊂ B ⊂ U² and µ[U² \ K² ] < ². If K ⊂ X is compact, then
limn→∞ µ[N1/n (K)] = µ[K] by Exercise-13. So A contains all compact sets. Now we show A is a
σ-algebra. If B ∈ A and K² ⊂ B ⊂ U² , then X \ U² ⊂ X \ B ⊂ X \ K² so that X \ B ∈ A. If
Bn ∈ A and ² > 0, choose compact Kn,² ⊂ B and open Un,² ⊃ B with µ[Un,² \ Kn,² ] < ²2−(n+1) .
S
Sk
Let U² = ∞
n=1 Un,² . Next, by Exercise-13, there is k ∈ N such that for K² =
n=1 Kn,² , we have
S∞
S∞
S
µ[ n=1 Kn,² \ K² ] < ²/2. If B = n=1 Bn , then K² ⊂ B ⊂ U² and µ[U² \ K² ] ≤ µ[U² \ ∞
n=1 Kn,² ] +
S∞
P∞
−(n+1)
µ[ n=1 Kn,² \ K² ] < n=1 ²2
+ ²/2 = ². Thus B ∈ A.
¤
[122] Let X be LCSC and µ be a locally finite Borel measure on X. Then µ is regular.
S
Proof. Write X = ∞
j=1 Kj , where Kj is compact and Kj ⊂ int[Kj+1 ]. If B ∈ B, let Bj = B ∩ Kj .
Given ² > 0, applying [121] to the restriction of µ to Kj+1 , find compact Aj ⊂ Bj and relatively
open Vj ⊂ Kj+1 with Bj ⊂ Vj such that µ[Bj ] − µ[Aj ] < ²2−j and µ[Vj ] − µ[Bj ] < ²2−j . If
Uj = Vj ∩ int[Kj+1 ], then Uj is open in int[Kj+1 ] and hence in X. Moreover, Bj ⊂ Uj and
S
µ[Uj ] − µ[Bj ] < ²2−j . If µ[B] < ∞, then arguing as in the proof of [121], for open U² = ∞
j=1 Uj
Sk
and compact A² = j=1 Aj (k large), we have A² ⊂ B ⊂ U² and µ[U² \ A² ] < 2². If µ[B] = ∞,
S
then for compact Dk = kj=1 Aj ⊂ B, we have µ[B] = supk∈N µ[Dk ]; and µ[B] = inf{µ[U ] : U ⊂
X open and B ⊂ U } trivially.
¤
Integration: Recall that
Riemann integration: integration of step functions first, then approximation.
Lebesgue integration: integration of simple functions first, then approximation.
Let X be LCSC and µ be a locally finite Borel measure on X. A real valued simple Borel
P
P
function f = ni=1 ci χBi ≥ 0 is integrable with respect to µ if ni=1 ci µ[Bi ] < ∞. If this happens
TOPOLOGICAL GROUPS - PART 1/3
we write
R
RX
f dµ =
Pn
i=1 ci µ[Bi ].
11
A non-negative Borel f : X → R is integrable with respect to
µ if sup{ X gdµ : g is simple Borel, 0 ≤ g ≤ f } < ∞. If this happens, this finite quantity is the
R
value of X f dµ. A general real valued Borel function on X is integrable if its positive and negative
R
R
R
parts are integrable; and X f dµ = X f+ dµ − X f− dµ. Note that every f ∈ Cc (X, R) is integrable
with respect to µ. A complex Borel function on X is integrable if its real and imaginary parts are
R
R
R
integrable; and X f dµ = X Re(f )dµ + i X Im(f )dµ. Analogues of standard results of Lebesgue
integration theory are true here - we skip the details.
[123] Let X be LCSC and let µ, β be locally finite Borel measures on X.
(i) If µ[K] = β[K] for every compact K ⊂ X, or if µ[U ] = β[U ] for every open U ⊂ X, then µ = β.
R
R
(ii) If X f dµ = X f dβ for every f ∈ Cc (X, R) with f ≥ 0, then µ = β.
Proof. (i) is a corollary to [122]. To prove (ii), consider compact K ⊂ X and choose a sequence
(fn ) in Cc (X, R) such that 0 ≤ fn+1 ≤ fn ≤ 1 and (fn ) → χK pointwise. Then by Dominated
R
R
R
R
Convergence Theorem, µ[K] = X χK dµ = limn→∞ X fn dµ = limn→∞ X fn dβ = X χK dβ =
β[K]. Now use (i).
¤
Let L1R (X, µ) be the space of (equivalence classes of) all µ-integrable functions f : X → R (one
has to enlarge the class of Borel functions first by completing the Borel σ-algebra, etc.). L1R (X, µ)
R
is a normed space with respect to the norm kf k1 = X |f |dµ.
[124] Let X be LCSC and µ be a locally finite Borel measure on X. Then (L1R (X, µ), k · k1 ) is
separable and Cc (X, R) is dense in it.
Proof. The collection of all µ-integrable simple Borel functions on X is dense in L1R (X, µ) by the
definition of integration. Now let U be a countable base of open balls with compact closures in
X and let V be all finite unions of members of U. Consider a µ-integrable simple Borel function
P
g = ki=1 ci χBi . Since µ is regular by [122], each Bi we can approximate in measure by an open
superset W , and then there is V ∈ V such that µ[W ∆V ] is sufficiently small. Also each ci can be
P
approximated by a rational.. Hence the countable collection F = { ki=1 ci χVi : k ∈ N, ci ∈ Q, Vi ∈
V} is dense in L1R (X, µ). Now to show Cc (X, R) is dense in L1R (X, µ), it is enough to show that
χV , V ⊂ X open with µ[V ] < ∞, can be approximated by members of Cc (X, R) since Cc (X, R) is
a vector subspace and since F is dense in L1R (X, µ). Let Kn ’s be compact with Kn ⊂ Kn+1 and
S
V = ∞
n=1 Kn . Let fn ∈ Cc (X, R) be such that 0 ≤ fn ≤ 1, fn ≡ 1 on Kn , and fn ≡ 0 on X \ V .
Then, kχV − fn k1 ≤ µ[V \ Kn ] → 0.
¤
Similar argument applies for LpR (X, µ), LpC (X, µ) (1 ≤ p < ∞). It is also true that Lp (X, µ) is
complete for 1 ≤ p ≤ ∞.
12
T.K.SUBRAHMONIAN MOOTHATHU
[1240 ] Let X be LCSC and µ be a locally finite Borel measure on X. Then for 1 ≤ p < ∞,
(LpC (X, µ), k · kp ) is a separable Banach space and Cc (X, C) is dense in LpC (X, µ). Also L2C (X, µ) is
R
a Hilbert space with respect to the inner product hf, gi = X f gdµ.
3. Measure Theory and Functional Analysis
Let µ, β be two measures on an abstract measurable space (X, A). We say β is absolutely
continuous with respect to µ if µ[A] = 0 ⇒ β[A] = 0 for every A ∈ A (notation: β << µ); and we
say µ, β are singular (or orthogonal ) if there is A ∈ A such that β[A] = 0 = µ[X \ A] (notation:
µ ⊥ β).
Examples: (i) Let β be the Lebesgue measure on [0, 1], µ[A] = β[A ∩ [0, 1/2]], β2 [A] = [A ∩ [1/2, 1]],
and β1 = µ. Then β1 << µ, β2 ⊥ µ and β = β1 + β2 .
(ii) Let µ be the Lebesgue measure on [0, 1], and let β1 , β2 be Borel measures on [0, 1] defined by:
R
β1 [A] = 1 if 0 ∈ A, β1 [A] = 0 if 0 ∈
/ A, and β2 [A] = A f dµ where f : [0, 1] → R is µ-integrable and
≥ 0 (here, the countable additivity of β2 is checked using the Monotone Convergence Theorem).
Then β1 << µ and β2 ⊥ µ.
(iii) Let f : R → R be a Lipschitz continuous homeomorphism, µ be the Lebesgue measure on R
and define β[A] = µ[f (A)]. Then β is a Borel measure on R. Let λ > 1 be a Lipschitz constant for
f . If B ⊂ R is Borel with µ[B] = 0 and ² > 0, then there are countably many intervals Jn such that
S
P∞
P∞
B⊂ ∞
n=1 Jn and
n=1 µ[Jn ] < ²/λ. If Kn = f (Jn ), then µ[Kn ] ≤ λµ[Jn ] so that
n=1 µ[Kn ] < ².
S∞
Also f (B) ⊂ n=1 Kn . Hence β[A] = µ[f (A)] = 0. Thus β << µ.
Exercise-14 : Let µ, β be finite measures on (X, A). Then β << µ iff for every ² > 0 there is
δ > 0 such that β[A] < ² for every A ∈ A with µ[A] < δ. [Hint: Let ² > 0. If there is no δ > 0
with the required property, then there are An ∈ A such that µ[An ] < n−2 and β[An ] ≥ ². If
T
S∞
B= ∞
k=1 n=k An , then µ[B] = 0 and β[B] ≥ ² > 0. To obtain ‘⇐’, note that if β[A] > ² > 0 and
if δ > 0 is chosen for this ², then µ[A] ≥ δ > 0.]
Exercise-15 : Let µ, β be measures on (X, A) with β << µ and β ⊥ µ. Then β ≡ 0.
Recall the following elementary geometric fact: if u, v are two vectors in the plane, then v can
be written uniquely as v = v1 + v2 so that the vector v1 is parallel to u and v2 ⊥ u. A result of the
same spirit for measures is:
[125] [Lebesgue Decomposition Theorem] (true for σ- finite measures also) Let β, µ be two finite
measures on a measurable space (X, A). Then there are unique measures βa , βs on (X, A) such
that β = βa + βs , βa << µ and βs ⊥ µ.
Proof. Note that beta << β + µ. The idea is to obtain a suitable g such that β[A] =
for every A ∈ A. Then βa , βs can be defined with the help of g.
R
A gd(β
+ µ)
TOPOLOGICAL GROUPS - PART 1/3
13
We have (β + µ)[X] < ∞ and hence L2R (X, β + µ) ⊂ L1R (X, β + µ). Consider the linear functional
R
φ : L2R (X, β + µ) → R given by φ(h) = X hdβ. By Cauchy-Schwarz,we have
Z
Z
|φ(h)| ≤
|h|dβ ≤
|h|d(β + µ) = khk1 = kh · 1k1 ≤ khk2 k1k2 = khk2 [(β + µ)[X]]1/2 ,
X
X
and hence φ is bounded. By Riesz representation theorem, there is g ∈ L2R (X, β + µ) such that
φ(h) = hh, gi. That is,
R
R
(i) X hdβ = X hgd(β + µ) ∀h ∈ L2R (X, β + µ), and
R
(ii) β[A] = A gd(β + µ) ∀A ∈ A (taking h = χA in (i)).
We use (ii) for the rest of the proof; (i) will be used to prove the next theorem. First we claim
that 0 ≤ g ≤ 1. Let An = {x ∈ X : g(x) ≤ −n−1 } and Bn = {x ∈ X : g(x) ≥ 1 + n−1 }. Using
A = An in (ii), β[An ] ≤ −n−1 (β + µ)[An ] and hence (β + µ)[An ] = 0. Similarly, A = Bn in (ii) gives
β[Bn ] ≥ (1 + n−1 )(β[Bn ] + µ[Bn ]) which implies β[Bn ] = 0 and then µ[Bn ] = 0. Thus 0 ≤ g ≤ 1
(β + µ)-almost everywhere.
Let S = {x ∈ X : g(x) = 1}, T = X \ S. Then, T =
S∞
n=1 Tn ,
Tn = {x ∈ X : g(x) ≤ 1 − n−1 }.
Taking A = S in (ii), we get µ[S] = 0. For Y ∈ A with µ[Y ] = 0, taking A = Y ∩ Tn in (ii), we
P
have β[Y ∩ Tn ] ≥ (1 − n−1 )β[Y ∩ Tn ] implying β[Y ∩ T ] ≤ ∞
n=1 β[Y ∩ Tn ] = 0. Thus if we define
βs [Y ] = β[Y ∩ S], βa [Y ] = β[Y ∩ T ], then βs ⊥ µ, βa << µ and β = βa + βs .
Uniqueness: Suppose β = αa + αs is another decomposition with respect to µ. Then αa [S] = 0
since µ[S] = 0. Hence βs [Y ] = β[Y ∩ S] = αa [Y ∩ S] + αs [Y ∩ S] = αs [Y ∩ S] ≤ αs [Y ]. Thus
βs ≤ αs and hence βa ≥ αa . Then the measure ν = αs − βs = βa − αa satisfies ν ⊥ µ and ν << µ.
Therefore ν ≡ 0.
¤
[126] [Radon-Nikodym Theorem for finite measures] Let β, µ be finite measures on a measurable
space (X, A) and suppose that β << µ. Then there exists a non-negative µ-integrable function
R
f : X → R such that β[A] = A f dµ for every A ∈ A. Moreover, f is unique µ-almost everywhere.
(Thus β << µ ⇔ β is obtained by integrating against µ.)
Proof. Let φ, g, S, T be as in the proof of [125]. Let β = βa + βs be the Lebesgue decomposition
of β with respect to µ. Since β << µ, we have βa = β and βs = 0 by the uniqueness of the
decomposition. Now, let us try to make a guess about f . Taking h = f χA in (i) of the proof
R
R
R
of [125] we see that the required f must satisfy A f dβ = A f gd(β + µ), or (adding A f dµ to
R
R
both sides and rearranging,) A f (1 − g)d(β + µ) = A f dµ. On the other hand we must have
R
R
A f dµ = β[A] = A gd(β + µ) by (ii) from the proof of [125], so that we may expect the identity
f (1 − g) = g.
R
Define f : X → R as f ≡ 0 on S and f = g/(1 − g) on T . Then A f (1 − g)d(β + µ) =
R
R
A∩T f (1 − g)d(β + µ) = A∩T gd(β + µ) = (by (ii) from [125] )β[A ∩ T ] = βa [A] = β[A]. But we
R
R
R
have already derived above that A f (1 − g)d(β + µ) = A f dµ. Thus β[A] = A f dµ for every
A ∈ A.
14
T.K.SUBRAHMONIAN MOOTHATHU
Uniqueness µ-almost everywhere: If h is another such function and if An = {x ∈ X : f (x)−h(x) ≥
R
R
R
then An f dµ = β[An ] = An hdµ and hence 0 = An (f −h)dµ ≥ n−1 µ[An ] implying µ[An ] = 0
n−1 },
and therefore µ[{x ∈ X : f (x) > h(x)}] = 0. Similarly µ[{x ∈ X : h(x) > f (x)}] = 0 also.
¤
[1260 ] [Radon-Nikodym Theorem] Let β, µ be σ-finite measures on a measurable space (X, A)
and let β << µ. Then there exists a non-negative measurable function f : X → R such that
R
β[A] = A f dµ for every A ∈ A. Moreover, f is unique µ-almost everywhere.
Proof. Write X =
S∞
i=1 Xi ,
a disjoint union of measurable sets Xi with (β + µ)[Xi ] < ∞. Get
fi : Xi → R by [126] and let f : X → R be f |Xi = fi .
¤
We know that locally finite Borel measures on LCSC spaces are σ-finite. Therefore, we have:
[12600 ] [Radon-Nikodym Theorem for LCSC spaces] Let β, µ be locally finite Borel measures on a
LCSC space X and let β << µ. Then there exists a non-negative Borel function f : X → R such
R
that β[A] = A f dµ for every Borel A ⊂ X. Moreover, f is unique µ-almost everywhere.
Remark : The function f given by the Radon-Nikodym Theorem is called the Radon-Nikodym
derivative of β with respect to µ and is denoted by
dβ
dµ .
This notation is justified by the following
R
R
observation. If g : (X, A) → R is a non-negative measurable function, then X gdβ = X g dβ
dµ dµ [∵
R
R
R dβ
dβ
if g = χA , then X gdβ = β[A] = A dµ µ = X g dµ µ; extend this to nonnegative simple functions
by linearity and then use approximation].
Execise-16 : Let µ be the Lebesgue measure on [0, 1], let f : [0, 1] → [0, 1] be a twice differentiable
homeomorphism with f 0 > 0 on (0, 1) (example: f (x) = xn ) and define β[A] = µ[f (A)] for Borel
A ⊂ [0, 1]. We already know that β << µ (since f is Lipschitz by Mean value theorem). Show that
dβ
dµ
= f 0 . [Hint: Since f 0 > 0, f is increasing on (0, 1) and hence on [0, 1]. If [a, b] ⊂ [0, 1], then
R
Ra 0
R
dβ
0
[a,b] f dµ = b f (t)dt = f (b) − f (a) = µ[f ([a, b])] = β[[a, b]] = [a,b] dµ dµ. Since closed intervals
R 0
R dβ
generate the Borel sigma algebra on [0, 1], we deduce A f dµ = A dµ dµ for every Borel A ⊂ [0, 1].
By uniqueness,
dβ
dµ
= f 0. ]
Exercise-17 : Let Y, X be measurable spaces, g : Y → X be a measurable surjection and let
β be a finite measure on Y . Define µ[A] = β[g −1 (A)] for measurable A ⊂ X. Then µ is a
R
R
measure on X and Y f ◦ gdβ = X f dµ for integrable f : X → R. [Hint: If f = χA , then
R
R
R
−1
Y f ◦ gdβ = g −1 (A) 1dβ = β[g (A)] = µ[A] = X f dµ.]
Locally finite Borel measure = positive linear functional: Let X be LCSC. If µ is a locally finite
R
Borel measures on X and if we define φ : Cc (X, R) as φ(f ) = X f dµ, then it is easy to see that φ
is a positive linear functional (here, φ positive means φ(f ) ≥ 0 for every f ∈ Cc (X, R). Our aim
is to prove the converse that every positive linear functional on Cc (X, R) has this form. We need
some preparation.
TOPOLOGICAL GROUPS - PART 1/3
15
Let X be a set. We say A0 ⊂ P(X) is an algebra if ∅ ∈ A0 and if A0 is closed under finite
unions and complementation. For example, if X = [0, 1] and A0 is the collection of all finite unions
of subintervals of [0, 1], then A0 is an algebra and the σ-algebra A generated by A0 is the Borel
σ-algebra on [0, 1].
By a measure on an algebra A0 we mean a function µ : A0 → [0, ∞] such that µ[∅] = 0
S
P∞
0
and µ[ ∞
n=1 An ] =
n=1 µ[An ] whenever A1 , A2 , . . . ∈ A are pairwise disjoint subsets such that
S∞
0
0
n=1 An ∈ A (note that the union need not be in A sometimes, in which case the condition holds
automatically). We omit the proof of the following.
[127] [Caratheodory’s Extension Theorem (special case)] Let X be a set, A0 be an algebra on X
and let µ be a finite measure on A0 . If A is the σ-algebra generated by A0 , then µ extends to a
unique finite measure on (X, A).
Example: Let X = [0, 1] and A0 be the collection of all finite unions of subintervals of [0, 1]. If
S
P
A ∈ A0 , write A = nn=1 Ai , a disjoint union of intervals, and define µ[A] = ni=1 length[Ai ]. Then
µ extends to the Lebesgue measure on the Borel σ-algebra on [0, 1].
Another preparation needed is the following topological fact about the Cantor space {0, 1}N .
[128] If X is a compact metric space, then there is a continuous surjection f : {0, 1}N → X.
Proof. Let A(1), . . . , A(n1 ) ⊂ X be finitely many nonempty compact sets with diam[A(i)] < 1 and
Q
S 1
A(i). At the (k+1)th step, for each (i1 , . . . , ik ) ∈ kj=1 {1, 2, . . . , nj }, choose finitely many
X = ni=1
nonempty compact sets A(i1 , . . . , ik , 1), . . . , A(i1 , . . . , ik , n(k+1) ) having diameter < 1/(k + 1) such
Q
nk
that their union equals A(i1 , . . . , ik ). Since {0, 1}N = ∞
k=1 {0, 1} , it is easy to find a continuous
Q
surjection g : {0, 1}N → ∞
k=1 {1, 2, . . . , nk }. Hence it suffices to find a continuous surjection h :
Q∞
Q∞
k=1 {1, 2, . . . , nk }
k=1 {1, 2, . . . , nk } → X (for then we may take f = h◦g). For y = (y1 , y2 , . . .) ∈
T∞
¤
define h(y) to be the unique point in k=1 A(y1 , . . . , yk ). Verify that this does the job.
Now we prove Riesz Representation Theorem for Measures [RRTM] in three steps (adapting the
proof due to V.S. Sunder): first for {0, 1}N , then for compact metric spaces, and then for LCSC
spaces.
[129] [RRTM for {0, 1}N ] Let X = {0, 1}N and let φ : C(X, R) → R be a positive linear functional.
R
Then there is a unique finite Borel measure µ on X such that φ(f ) = X f dµ for every f ∈ C(X, R).
Proof. For F = {n1 < n2 < · · · < nk } ⊂ N and w = w1 w2 · · · wk ∈ {0, 1}k , let A(F, w) = {x ∈ X :
xni = wi for 1 ≤ i ≤ k}. Each A(F, w) is both compact and open. Let A0 be the collection of all
finite unions of A(F, w)’s, where F ⊂ N is finite and w ∈ {0, 1}|F | . Then A0 is an algebra and every
member of A0 is both compact and open; in particular, χA ∈ C(X, R) for A ∈ A0 . If µ : A0 → [0, ∞)
is defined as µ[A] = φ(χA ), then µ[A ∪ B] = φ(χA∪B ) = φ(χA + χB ) = φ(χA ) + φ(χB ) = µ[A] + µ[B]
S
for disjoint A, B ∈ A0 . Since every member of A0 is both compact and open, ∞
/ A0 if
n=1 An ∈
16
T.K.SUBRAHMONIAN MOOTHATHU
An ∈ A0 are pairwise disjoint nonempty sets. Hence µ is countably additive on A0 trivially. Also
µ[X] = φ(1) ∈ [0, ∞). Thus µ is a finite measure on A0 . Since A0 is also a base for the topology of
X, the algebra generated by A0 is the Borel σ-algebra. So µ extends to a finite Borel measure on X
R
by [127]. Clearly φ(χB ) = µ[B] = X χB dµ for every Borel B ⊂ X. By linearity and approximation,
R
φ(f ) = X f dµ for every f ∈ C(X, R).
Uniqueness: If β is another such measure, then β[A] = φ(χA ) = µ[A] for every A ∈ A0 and then
β[U ] = µ[U ] for every open U ⊂ X by approximation. Then β = µ by [123].
¤
Lemma: Let X be a compact metric space and consider C(X, R) with the supremum norm. If
φ : C(X, R) → R is linear, then the following are equivalent:
(i) φ is positive.
(ii) φ(f ) ≤ φ(g) for f, g ∈ C(X, R) with f ≤ g.
(iii) φ is bounded and kφk = φ(1).
Proof. (i) ⇔ (ii) is easy. To see (ii) ⇒ (iii), consider f ∈ C(X, R) with |f | ≤ 1. Then f ≤ |f | ≤ 1
and −f ≤ |f | ≤ 1 so that φ(f ) ≤ φ(1) and −φ(f ) = φ(−f ) ≤ φ(1). Hence |φ(f )| ≤ φ(1), and
equality holds when f = 1. To prove (iii) ⇒ (i), it suffices to show φ(f ) ≥ 0 for f ∈ C(X, R) with
0 ≤ f ≤ 1. Since 0 ≤ 1 − f ≤ 1, φ(1) − φ(f ) = φ(1 − f ) ≤ kφk ≤ φ(1) and so φ(f ) ≥ 0.
¤
Remarks: (i) The above lemma is true for linear φ : C(X, C) → C, but the proof is more involved.
(ii) If X is only LCSC, a positive linear functional φ : (Cc (X, R), sup norm) → R need not be
R
bounded. For instance, consider φ : Cc (X, R) → R given by φ(f ) = R f (x)dx.
[1290 ] [RRTM for compact metric spaces] Let X be a compact metric space and let φ : C(X, R) → R
be a positive linear functional. Then there is a unique finite Borel measure µ on X such that
R
φ(f ) = X f dµ for every f ∈ C(X, R).
Proof. Let Y = {0, 1}N and let p : Y → X be a continuous surjection given by [128]. Define
T : C(X, R) → C(Y, R) as T (f ) = f ◦ p. Then T is a linear isometry and T (f ) ≥ 0 whenever f ≥ 0.
So C(X, R) may be identified with a vector subspace of C(Y, R) and φ may be considered as a
positive linear functional on this subspace. By the lemma, φ is bounded and kφk = φ(1). By Hahne (f )) = φ(f )
Banach Theorem, there is a bounded linear functional φe : C(Y, R) → R such that φ(T
e = kφk = φ(1X ) = φ(T
e (1X )) = φ(1
e Y ). So φe is positive by the
for every f ∈ C(X, R) and kφk
R
e
lemma. Therefore, by [129] there is a unique finite Borel measure β on Y such that φ(g)
=
gdβ
Y
β[p−1 (A)]
for every g ∈ C(Y, R). Define µ[A] =
for Borel A ⊂ X. Then µ is a finite Borel measure
R
R
e (f )) =
on X and Y f ◦ pdβ = X f dµ for every f ∈ C(X, R) by Exercise-17. Hence φ(f ) = φ(T
R
R
e ◦ p) =
φ(f
f ◦ pdβ =
f dµ for every f ∈ C(X, R). Uniqueness of µ follows from [123].
¤
Y
X
TOPOLOGICAL GROUPS - PART 1/3
Remark : It may be noted that kφk = φ(1) =
R
X
17
1dµ = µ[X], so that positive linear functionals of
norm 1 on C(X, R) correspond to Borel probability measures on X, when X is a compact metric
space.
[12900 ] [Riesz Representation Theorem for Measures for LCSC spaces] Let X be LCSC and let
φ : Cc (X, R) → R be a positive linear functional. Then there is a unique locally finite Borel
R
measure µ on X such that φ(f ) = X f dµ for every f ∈ Cc (X, R).
Proof. Step-1: Let K ⊂ X be compact and Un = N1/n (K). Note that Un+1 ⊂ Un for all n and Un
is compact for all large n. Let gn ∈ C(X, [0, 1]) be such that gn ≡ 1 on Un+1 and gn ≡ 0 on X \ Un .
Then 0 ≤ gn+1 ≤ gn ≤ 1.
Let C+ (K, R) = {f ∈ C(K, R) : f ≥ 0}. If f ∈ C+ (K, R), we see that there is fe ∈ Cc (X, R) with
fe ≥ 0 and fe|K = f (∵ choose open U with K ⊂ U , let fe = f on K, fe ≡ 0 on X \ U and apply
Tietze). Since 0 ≤ fegn+1 ≤ fegn , (φ(fegn )) is a decreasing sequence of nonnegative reals and hence
converges. Define φK : C+ (K, R) → R as φK (f ) = limn→∞ φ(fegn ).
To verify that φK (f ) is independent of the choice of the extension fe, consider another extension
f ∗ ∈ Cc (X, R) of f with f ∗ ≥ 0. Suppose that Un is compact for every n ≥ n0 . Given ² > 0
let An = {x ∈ Un : |fe(x) − f ∗ (x)| ≥ ²} for n ≥ n0 . Then An ’s are compact, An+1 ⊂ An and
T
T
T
∗
n≥n0 An ⊂
n≥n0 Un = K so that
n≥n0 An = ∅ (∵ f = f = f on K). So there is n(²) ≥ n0
such that An(²) = ∅. Then |fe(x) − f ∗ (x)| < ² for every x ∈ Un(²) . Now gn ≡ 0 on X \ Un(²) gor
n ≥ n(²). Hence |fegn − f ∗ gn | ≤ ²gn ≤ ²gn0 for every n ≥ n(²) ≥ n0 . That is, fegn − f ∗ gn ≤ ²gn0
and f ∗ gn − fegn ≤ ²gn0 for n ≥ n(²). Applying φ, |φ(fegn ) − φ(f ∗ gn )| ≤ ²φ(gn0 ) for every n ≥ n(²)
which implies limn→∞ φ(fegn ) = limn→∞ φ(f ∗ gn ).
Next note that φK has the following properties: φK (f + h) = φK (f ) + φK (h) and φK (αf ) =
αφK (f ) for every f, h ∈ C+ (K, R) and α ≥ 0. Hence by the following result φK extends uniquely
to a linear functional φK on C(K, R).
Result: Let Y be a compact metric space and ψ : C+ (Y, R) → R be such that ψ(f +h) = ψ(f )+ψ(h)
and ψ(αf ) = αψ(f ) for every f, h ∈ C+ (Y, R) and α ≥ 0. Then ψ has a unique extension to a linear
functional on C(Y, R). [Proof : Let f1 , f2 , h1 , h2 ∈ C+ (Y, R) and suppose f1 − f2 = h1 − h2 . Then
f1 + h2 = h1 + f2 so that by applying ψ and rearranging we obtain ψ(f1 ) − ψ(f2 ) = ψ(h1 ) − ψ(h2 ).
For f ∈ C(Y, R), define ψ(f ) = ψ(h1 ) − ψ(h2 ), where h1 , h2 ∈ C+ (Y, R) are such that f = h1 − h2
(for example, f = f+ − f− ). Then this extension ψ is well-defined by the above argument. Check
other properties.]
Note that the extended linear functional φK : C(K, R) → R is positive (∵ if f ≥ 0, then
φK (f ) = limn→∞ φ(fegn ) ≥ 0 since φ is positive). So by [1290 ], there is a unique finite Borel
R
measure µK on K with φK (f ) = K f dµK . Also observe that φK (f |K ) = φ(f ) for nonnegative
f ∈ Cc (X, R) with supp(f ) ⊂ K (∵ taking fe = f , φK (f |K ) = limn→∞ φ(f gn ) = φ(f ) since
f gn = f ).
18
T.K.SUBRAHMONIAN MOOTHATHU
Step-2: Write X =
S∞
j=1 Kj ,
where Kj ’s are compact and Kj ⊂ int[Kj+1 ]. Step-1 gives a positive
linear functional φj : C(Kj , R) → R defined using φ and let µj be the unique finite Borel measure
on Kj corresponding to φj . We claim that µj+1 |Kj = µj .
Choose gn ’s for Kj as in Step-1 and for f ∈ C+ (Kj , R), let fe ∈ Cc (X, R) be an extension with
fe ≥ 0 and supp(fe) ⊂ Kj+1 . Then supp(fegn ) ⊂ Kj+1 and hence φj+1 (fegn |Kj+1 ) = φ(fegn ) for all
R
large n by the observation at the end of Step-1. Hence φ(fegn ) = Kj+1 fegn dµj+1 for all large n
R
R
and therefore, Kj f dµj = φj (f ) = limn→∞ φ(fegn ) = limn→∞ Kj+1 fegn dµj+1 , which is equal, by
R
R
Dominated convergence theorem, to Kj+1 f χKj dµj+1 = Kj f d(µj+1 |Kj ). Hence the claim holds
by the uniqueness part in [1290 ].
For Borel A ⊂ X, define µ[A] = limj→∞ µj [A ∩ Kj ] ∈ [0, ∞]. It may be checked that µ is a
Borel measure. For countable additivity, note the fact that if (xn,j ) is a double sequence of reals
increasing in each variable, then limn→∞ limj→∞ xn,j = limj→∞ limn→∞ xn,j , and use this with
xn,j = µj [An ∩ Kj ]. If K ⊂ X is compact, then K ⊂ Kj for some j and this implies µ[K] ≤
µj [K] < ∞ and so µ is locally finite. Also, if f ∈ Cc (X, R) is nonnegative, then supp(f ) ⊂ Kj for
R
R
R
R
some j so that φ(f ) = φj (f |Kj ) = Kj f dµj = Kj f dµ = X f dµ, and therefore φ(f ) = K f dµ for
every f ∈ Cc (X, R) since f = f+ − f− .
Uniqueness: if β is another such measure and if βj = β|Kj , then βj = µj by the uniqueness in
[1290 ] and hence β = µ.
¤
Remark : RRTM has other formulations when X is a compact metric space: (i) bounded linear
functional on C(X, R) correspond to finite signed real-valued Borel measures, (ii) bounded linear
functional on C(X, C) correspond to ‘finite’ complex-valued Borel measures, etc.
Remark : Let X be a compact metric space. Then the dual of the Banach space (C(X, C), sup norm)
can be identified with M (X) := {all complex-valued ‘finite’ Borel measures on X} by Riesz Theorem. Now recall from Functional Analysis that if Y is a Banach space, then the weak* topology
on the dual Y ∗ is the smallest topology such that the evaluation maps φ → φ(y) from Y ∗ to C
are continuous for every y ∈ Y ; the closed unit ball of Y ∗ is weak* compact (Alaoglu’s Theorem)
and it is weak* metrizable if Y is separable. Hence the closed unit ball of M (X) = C(X, C)∗ is
weak* compact and weak* metrizable. For µ, µn ∈ M (X), (µn ) → µ in the weak* topology iff
R
R
R
( X f dµn ) → X f dµ in C for every f ∈ C(X, C), since φ 7→ φ(f ) = X f dµ is the evaluation
map if φ ∈ C(X, C)∗ corresponds to µ ∈ M (X). If (µn ) is a bounded sequence in M (X), then
it is contained in some closed ball (which is compact and metrizable in the weak* topology) and
hence there exist µ ∈ M (X) and a subsequence (µnk ) such that (µnk ) weak* converges to µ, or
R
R
equivalently ( X f dµnk ) → X f dµ for every f ∈ C(X, C).
*****