Stanford Mathematics Department Math 205A Lecture Supplement #4 Borel Regular & Radon Measures

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1 2 1 Borel Regular Measures We now state and prove an important regularity property of Borel regular outer measures: Stanford Mathematics Department Math 205A Lecture Supplement #4 Borel Regular & Radon Measures 1 Borel Regular Measures Recall that a Borel measure on a topological space is a measure defined on the collection of Borel sets, and an outer measure on is said to be a Borelregular outer measure if all Borel sets are -measurable and if for each subset A there is a Borel set B A such that (B) D (A). (Notice that this does not imply (B n A) D 0 unless A is -measurable and (A) < 1.) Also if is a Borel measure on, then we get a Borel regular outer measure on by defining (Y ) D inf (B) where the inf is taken over all Borel sets B with B Y, and this outer measure coincides with on all the Borel sets. (See Q.1 of hw2.) Also, if is a Borel regular outer measure on and if A is -measurable with (A) < 1, then we claim A is also Borel regular. Here A is the outer measure on defined by ( A)(Y ) D (A \ Y ): To check this claim first observe that if E is -measurable and Y is arbitrary then ( A)(Y ) D (A \ Y ) D (A \ Y \ E) C (A \ Y n E) D ( A)(Y \ E) C ( A)(Y n E), hence E is also ( A)-measurable. In particular all Borel sets are ( A)-measurable, so it remains to prove that for each Y there is a Borel set B Y with ( A)(B) D ( A)(Y ). To prove this, first use the Borel regularity of and the fact that A is measurable of finite measure to pick Borel sets B 1 Y \ A and B 2 A with (B 1 ) D (Y \ A) and (B 2 n A) D 0, and then pick a Borel set B 3 B 2 n A with (B 3 ) D 0. Then Y B 1 [ ( n A) B 1 [ ( n B 2 ) [ B 3 (which is a Borel set) and ( A)(Y ) ( A)(B 1 [( nb 2 )[B 3 ) D ((A\B 1 )[(A\B 3 )) (B 1 ) D (Y \ A) D ( A)(Y ). 1.1 Theorem. Suppose is a topological space with the property that every closed subset of is the countable intersection of open sets (this trivially holds e.g. if is a metric space), pose is a Borel-regular outer measure on, and pose that D [ 1 j D1 V j, where (V j ) < 1 and V j is open for each j D 1; 2; : : :. Then (1) inf (U ) U open; U A for each subset A, and (2) (C ) C closed; C A for each -measurable subset A. 1.2 Remark: In case is a locally compact separable metric space (thus for each x 2 there is > 0 such that the closed ball B (x) D {y 2 W d (x; y) } is compact, and has a countable dense subset), the condition D [ 1 j D1 V j with V j open and (V j ) < 1 is automatically satisfied provided (K) < 1 for each compact K. Furthermore in this case we have from (2) above that K compact; KA (K) for each -measurable subset A K with (A) < 1, because under the above conditions on any closed set C can be written C D [ 1 id1 K i, K i compact. Proof of 1.1. We assume first that () < 1. By Borel regularity of, for any given A we can select a Borel set B A with (B) D (A), so it clearly suffices to check (1) in the special case when A is a Borel set. Now let A D { Borel sets A W (1) holds}: Trivially A contains all open sets and one readily checks that A is closed under both countable unions and intersections, as follows: If A 1 ; A 2 ; : : : 2 A then for any given " > 0 there are open U 1 ; U 2 ; : : : with U j A j and (U j n A j ) 2 j ". Now one easily checks [ j U j n ([ k A k ) [ j (U j n A j ) and \ j U j n (\ k A k ) [ j (U j n A j ) so by subadditivity we have ([ j 1 D1 U j n ([ k A k )) < " and lim N!1 (\ j N D1 U j n (\ k A k )) D (\ j 1 D1 U j n (\ k A k )) < ", so both [ k A k and \ k A k are in A as claimed. In particular A must also contain the closed sets, because any closed set in can be written as a countable intersection of open sets and all the open sets

2 Math 205A Lecture Supplement 3 are in A. Thus if we let za D {A 2 A W n A 2 A} then za is a -algebra containing all the closed sets, and hence za contains all the Borel sets. Thus A contains all the Borel sets and (1) is proved in case () < 1. To check (2) in case () < 1 we can just apply (1) to n A: thus for each i D 1; 2; : : : there is an open set U i n A with () (A) C 1=i D ( na)c1=i > (U i ) D () (C i ), where C i D nu i n( na) D A. Thus C i A is closed and (C i ) > (A) 1=i. In case () D 1, to prove (1) we first take a Borel set B with (B) D (A) and then apply the above result for finite measures to the Borel regular measure V j, j D 1; 2; : : :, thus obtaining, for given " > 0, open U j B with ( V j )(U j n B) < "2 j and hence (U j \ V j n B) "2 j and by subadditivity this gives ([(U j \ V j ) n B) < ": Since [(U j \ V j ) is an open set containing B (hence A), this is the required result. Similarly to check (2) in case () D 1, we apply (2) for the finite measure case to the Borel regular measure V j, giving closed C j such that C j A and ( V j )(AnC j ) < "2 j for each j D 1; 2; : : :, and hence (A\V j n [ 1 C kd1 k) < "2 j, so by subadditivity and the fact that [V j D we have (An ([ 1 C kd1 k)) < " and hence (A) ([ 1 C kd1 k) C " D lim N!1 ([ N C kd1 k) C ". Since [ N C kd1 k is closed for each N this is the required result. Using the above lemma we now prove Lusin s Theorem: 1.3 Theorem (Lusin s Theorem.) Let be a Borel regular outer measure on a topological space having the property that every closed set can be expressed as the countable intersection of open sets (e.g. is a metric space), let A be -measurable with (A) < 1, and let f W A! R be -measurable. Then for each " > 0 there is a closed set C with C A, (A n C ) < ", and f jc continuous. Proof: For each i D 1; 2; : : : and j D 0; 1; 2; : : : let A ij D f 1 ((j 1)=i; j=i]; so that A ij ; j D 1; 2; : : :, are p.w.d. sets in A and [ 1 j D 1 A ij D A. By the remarks preceding Theorem 1.1 we know that A is a Borel regular outer measure and since it is finite we can apply Theorem 1.1 to it, and hence for given " > 0 there are closed sets C ij in with C ij A ij with ( A)(A ij n C ij ) D (A ij n C ij ) < 2 i jj j 1 ", hence (A ij n ([ 1`D C 1 i`)) < 2 i jj j 1 " and 4 2 Radon Measures, Representation Theorem hence (A n ([ 1`D 1 C i`)) < 2 i " so for each i D 1; 2; : : : there is a positive integer J (i) such that (A n ([ jj jj (i) C ij )) < 2 i " Since A n (\ 1 id1 ([ jj jj (i)c ij )) D [ 1 id1 (A n ([ jj jj (i)c ij ) (by De Morgan), this implies (A n C ) < "; where C D \ 1 id1 ([ jj jj (i)c ij ) is a closed subset of A. Now define g i W [ jj jj (i) C ij! R by setting g i (x) (j 1)=i on C ij, jj j J (i). Then g i is clearly continuous and its restriction to C is continuous for each i; furthermore by construction 0 f (x) g i (x) 1=i for each x 2 C and each i D 1; 2; : : :, so that g i jc converges uniformly to f jc on C and hence f jc is continuous. 2 Radon Measures, Representation Theorem In this section we work mainly in locally compact Hausdorff spaces, and for the reader s convenience we recall some basic definitions and preliminary topological results for such spaces. Recall that a topological space is said to be Hausdorff if it has the property that for every pair of distinct points x; y 2 there are open sets U ; V with x 2 U, y 2 V and U \ V D. In such a space all compact sets are automatically closed, the proof of which is as follows: observe that if x K then for each y 2 K we can (by definition of Hausdorff space) pick open U y ; V y with x 2 U y, y 2 V y and U y \V y D. By compactness of K there is a finite set y 1 ; : : : ; y N 2 K with K [ j N D1 V y j. But then \ j N D1 U y j is an open set containing x which is disjoint from [ j V yj and hence disjoint from K, so that K is closed as claimed. In fact we proved a bit more: that for each x K there are disjoint open sets U ; V with x 2 U and K V. Then if L is another compact set disjoint from K we can repeat this for each x 2 L thus obtaining disjoint open U x ; V x with x 2 U x and K V x, and then compactness of L implies 9x 1 ; : : : ; x M 2 L such that L [ j M D1 U x j and then [ j M D1 U x j and \ j M D1 V x j are disjoint open sets containing L and K respectively. By a simple inductive argument (left as an exercise) we can extend this to finite pairwise disjoint unions of compact subsets: 2.1 Lemma. Let be a Hausdorff space and K 1 ; : : : ; K N be pairwise disjoint compact subsets of. Then there are pairwise disjoint open subsets U 1 ; : : : ; U N

3 with K j U j for each j D 1; : : : ; N. Math 205A Lecture Supplement 5 Notice in particular that we have the following corollary of Lemma 2.1: 2.2 Corollary. A compact Hausdorff space is normal: i.e. given closed disjoint subsets K 1 ; K 2 of a compact Hausdorff space, we can find disjoint open U 1 ; U 2 with K j U j for j D 1; 2. Most of the rest of the discussion here takes place in locally compact Hausdorff space: A space is said to be locally compact if for each x 2 there is a neighborhood U x of x such that the closure U x of U x is compact. An important preliminary lemma in such spaces is: 2.3 Lemma. If is a locally compact Hausdorff space and V is a nhd. of a point x, then there is a nhd. U x of x such that U x is a compact subset of V. Proof: First pick a neighborhood U 0 of x such that U 0 is compact and define W D U 0 \ V. Then W is compact and hence so is the closed subset W n W. Then W n W and {x} are disjoint compact sets so by Lemma 2.1 we can find disjoint open U 1 ; U 2 with x 2 U 1 and W n W U 2. Without loss of generality we can assume U 1 W (otherwise replace U 1 by U 1 \ W ). Then U 1 n U 2 n (W n W ) and hence U 1 W. Thus the lemma is proved with U x D U 1. Remark: In locally compact Hausdorff space, using Lemmas 2.1 and 2.3 it is easy to check that we can select the U j in Lemma 2.1 above to have compact pairwise disjoint closures. The following lemma is a version of the Urysohn lemma valid in locally compact Hausdorff space: 2.4 Lemma. Let be a locally compact Hausdorff space, K compact, and K W, W open. Then there is an open V K with V W, V compact, and an f W! [0; 1] with f 1 in a neighborhood of K and spt f V. Proof: By Lemma 2.3 each x 2 K has a neighborhood U x with U x W and U x compact. Then by compactness of K we have K V [ j N D1 U x j for some finite collection x 1 ; : : : ; x N 2 K and V D [ j N D1 U x j W. Now V is compact, so by Corollary 2.2 it is a normal space and the Urysohn lemma can be applied to give f 0 W V! [0; 1] with f 0 1 on K and and f 0 0 on V n V. Then of course the function f 1 defined by f 1 f 0 on V and f 1 0 on n V is continuous ( check!) because f jv is continuous and f is identically 6 2 Radon Measures, Representation Theorem zero (the value of f j n V ) on the overlap set V n V V \ ( n V ). Finally we let f 2 min{f 1 ; 1 } and observe that f is then identically 1 in the set 2 where f 1 > 1, which is an open set containing K, and f evidently has all the 2 remaining stated properties. The following corollary of Lemma 2.4 is important: 2.5 Corollary (Partition of Unity.) If is a locally compact Hausdorff space, K is compact, and if U 1 ; : : : ; U N is any open cover for K, then there exist continuous ' j W! [0; 1] such that spt ' j is a compact subset of U j for each j, and j D1 ' j 1 in a neighborhood of K. Proof: By Lemma 2.3, for each x 2 K there is a j 2 {1; : : : ; N } and a neighborhood U x of x such that U x is a compact subset of this U j. By compactness of K we have finitely many of these neighborhoods, say U x1 ; : : : ; U xn with K [ N id1 U x i. Then for each j D 1; : : : ; N we define V j to be the union of all U xi such that U xi U j. Then the V j is a compact subset of U j for each j, and the V j cover K. So by Lemma 2.4 for each j D 1; : : : ; N we can select j W! [0; 1] with j 1 on V j and j 0 on n W j for some open W j with W j a compact subset of U j and W j V j. We can also use Lemma 2.4 to select ' 0 W! [0; 1] with ' 0 0 in a neighborhood of K and ' 0 1 outside a compact subset of [ j N D1 V j. Then by construction id0 i > 0 everywhere on, so we can define continuous functions ' j by ' j D j ; j D 1; : : : ; N : id0 i Evidently these functions have the required properties. We now give the definition of Radon measure. Radon measures are typically used only in locally compact Hausdorff space, but the definition and the first two lemmas following it are valid in arbitrary Hausdorff space: 2.6 Definition: Given a Hausdorff space, a Radon measure on is an outer measure on having the 3 properties: is Borel regular and (K) < 1 8 compact K (U ) D inf (U ) for each subset A U open; U A K compact; KU (K) for each open U : (R1) (R2) (R3) Such measures automatically have a property like ( R3) with an arbitrary - measurable subset of finite measure:

4 Math 205A Lecture Supplement Lemma. Let be a Hausdorff space and a Radon measure on. Then automatically has the property KA; K compact for every -measurable set A with (A) < 1. (K) Proof: Let " > 0. By definition of Radon measure we can choose an open U containing A with (U na) < ", and then a compact K U with (U nk) < " and finally an open W containing U n A with (W n (U n A)) < " (so that (W ) " C (U n A) < 2"). Then we have that K n W is a compact subset of U n W, which is a subset of A, and also (A n (K n W )) (U n (K n W )) (U n K) C (W ) 3"; which completes the proof. The following lemma asserts that the defining property ( R1) of Radon measures follows automatically from the remaining two properties (( R2) and ( R3)) in case is finite and additive on finite disjoint unions of compact sets. 2.8 Lemma. Let be a Hausdorff space and assume that is an outer measure on satisfying the properties (R2), (R3) above, and in addition assume that (K 1 [ K 2 ) D (K 1 ) C (K 2 ) < 1; K 1 ; K 2 compact, K 1 \ K 2 D. Then is Borel regular, hence (R1) holds, hence is a Radon measure. Proof: Note that (R2) implies that for every set A we can find open sets U j such that A \ j U j and (\ j U j ). So to complete the proof of (R1) we just have to check that all Borel sets are -measurable; since the - measurable sets form a -algebra and the Borel sets form the smallest -algebra which contains all the open sets, we thus need only to check that all open sets are -measurable. Let " > 0 be arbitrary, Y an arbitrary subset of with (Y ) < 1 and let U be an arbitrary open subset of. By (R2) we can pick an open set V Y with (V ) < (Y ) C " and by (R3) we can pick a compact set K 1 V \ U with (V \ U ) (K 1 ) C ", and then a compact set K 2 V n K 1 with 8 2 Radon Measures, Representation Theorem (V n K 1 ) (K 2 ) C ". Then (V n U ) C (V \ U ) (V n K 1 ) C (K 1 ) C " (K 2 ) C (K 1 ) C 2" D (K 2 [ K 1 ) C 2" ((V n K 1 ) [ K 1 ) C 2" D (V ) C 2" (Y ) C 3"; (by (i)) hence (Y n U ) C (Y \ U ) (V n U ) C (V \ U ) (Y ) C 3" which by arbitrariness of " gives (Y n U ) C (Y \ U ) (Y ), which establishes the -measurability of U. Thus all open sets are -measurable, and hence all Borel sets are -measurable, and so (R1) is established. The following lemma guarantees the convenient fact that, in a locally compact space such that all open subsets are -compact, all locally finite Borel regular outer measures are in fact Radon measures. 2.9 Lemma. Let be a locally compact Hausdorff space and pose that each open set is the countable union of compact subsets. Then any Borel regular outer measure on which is finite on each compact set is automatically a Radon measure. Proof: First observe that in a Hausdorff space the statement each open set is the countable union of compact subsets is equivalent to the statement is -compact (i.e. the countable union of compact sets) and every closed set is the countable intersection of open sets as one readily checks by using De Morgan s laws and the fact that a set is open if and only if its complement is closed. Thus we have at our disposal the facts that is -compact and every closed set is a countable intersection of open sets. The latter fact enables us to apply Theorem 1.1 (1), and we can therefore assert that (1) inf (U ) whenever A has the property U open; AU that 9 open V j with A [ j V j and (V j ) < 1 8j. Also, by applying Theorem 1.1 (2) to the finite Borel regular measure A, (2) C closed; C A (C ); provided A is -measurable and (A) < 1. Now observe that by the first part of the conclusion in Lemma 2.4 there is an open set V K such that V (the closure of V ) is compact. So since we are given D [ j 1 D1 K j, where each K j is compact, we can apply this with K j in place of K, and we deduce that there are open sets V j in such that [ j V j D

5 Math 205A Lecture Supplement 9 and (V j ) < 1 for each j, and so in this case (when is -compact) the identity in (1) holds for every subset A ; that is inf (U ) for every A ; U open; AU which is the property (R2). Next we note that if A is -measurable, then we can write A D [ j A j, where A j D A \ K j (because D [ j K j ) and (A j ) (K j ) < 1 for each j, so (2) actually holds for every -measurable A in case is -compact (i.e. in case D [ j 1 D1 K j with K j compact), and for any closed set C we can write C D [ j C j where C j is the increasing sequence of compact sets given by C j D C \ ([ j id1 K i ) and so (C ) D lim j (C j ) and hence (C ) D KC ; K compact (K). Thus in the -compact case (2) actually tells us that KA; K compact (K) for any -measurable set A. This in particular holds for A D an open set, which is the remaining property (R3) we needed. The following result is one of the main theorems related to Radon measures, asserting that for a Radon measure on a locally compact Hausdorff space, the continuous functions with compact port are dense in L p (), 1 p < 1. Here and subsequently we use the notation C c () D {f W! R W f is continuous with sptf compact}; where sptf D port of f D closure of {x 2 W f (x) 0} Theorem. Let be a locally compact Hausdorff space, a Radon measure on and 1 p < 1. Then C c () is dense in L p (); that is, for each " > 0 and each f 2 L p there is a g 2 C c () such that kg f k p < ". In view of Lemma 2.9 and the fact that to every Borel measure on a topological space (i.e. every map W {all Borel sets of }! [0; 1] with ( ) D 0 and ([ j 1 D1 B j ) D P 1 j D1 (B j ) for every pairwise disjoint collection of Borel sets B 1 ; B 2 ; : : :), there is a Borel regular outer measure on defined by inf B Borel; BA (B), we see that Theorem 2.10 directly implies the following important corollary: 2.11 Corollary. If is a locally compact Hausdorff space such that every open set in is the countable union of compact sets, and if is any Borel measure on which is finite on each compact set, then the space C c () is dense in L 1 () and is the restriction to the Borel sets of a Radon measure. Proof of Theorem 2.10: Let f W! R be -measurable with kf k p < 1 and let " > 0. Observe that the simple functions are dense in L p () (which 10 2 Radon Measures, Representation Theorem one can check using the dominated convergence theorem and the fact that both f C and f can be expressed as the pointwise limits of increasing sequences of non-negative simple functions), so we can pick a simple function ' D j D1 a j Aj, where the a j are distinct non-zero reals and A j are pairwise disjoint -measurable subsets of, such that kf 'k p < ". Since k'k p k' f k p C kf k p < 1 we must then have (A j ) < 1 for each j. Pick M > max{ja 1 j; : : : ; ja N j} and use Lemma 2.7 to select compact K j A j with (A j n K j ) < " p =(2 pc1 M p N ). Also, using the definition of Radon measure, we can find open U j K j with (U j n K j ) < " p =(2 pc1 M p N ) and by Lemma 2.7 we can assume without loss of generality that these open sets U 1 ; : : : ; U N are pairwise disjoint (otherwise replace U j by U j \ Uj 0, where U1 0; : : : ; U N 0 are pairwise disjoint open sets with K j Uj 0 ). By Lemma 2.4 we have g j 2 C c () with g j a j on K j, {x W g j (x) 0} contained in a compact subset of U j, and jg j j ja j j, and hence by the pairwise disjointness of the U j we have that g j D1 g j agrees with ' on each K j and jgj D j'j < M. Then ' g vanishes off the set [ j ((U j n K j ) [ (A j n K j )) and we have R j' gjp d P R j (U j nk j )[(A j K j ) j' gjp d (2M ) pp j ((A j n K j ) C (U j nk j )) " p, and hence kf gk p kf 'k p Ck' gk p 2", as required. We now state the Riesz representation theorem for non-negative functionals on the space K C, where, here and subsequently, K C denotes the set of nonnegative C c (; R) functions, i.e. the set of continuous functions f W! [0; 1) with compact port Theorem (Riesz for non-negative functionals.) Suppose is a locally compact Hausdorff space, W K C! [0; 1) with (cf ) D c(f ), (f C g) D (f ) C (g) whenever c 0 and f; g 2 K C, where K C is the set of all nonnegative continuous functions f on with compact port. Then there is a Radon measure on such that (f ) D R f d for all f 2 K C. Before we begin the proof of 2.12 we the following preliminary observation: 2.13 Remark: Observe that if f; g 2 K C with f g then g f 2 K C and hence (g) D (f C (g f )) D (f ) C (g f ) (f ), so () f; g 2 K C with g 1 on spt f ) (f ) ( f ) (g); f 2 K C ; spt f K: because f g f and f ( f )g. Proof of Theorem 2.12: For U open, we define (1) (U ) D (f ); f 2K C ;f 1;spt f U

6 and for arbitrary A we define Math 205A Lecture Supplement 11 (2) inf (U ): U A;U open Notice that these definitions are consistent when A is itself open, and of course the definitions ( 1),( 2) guarantee ( ) D 0 and that is monotone i.e. (3) A B ) (A) (B): Also if f 2 K C with f 1 and V is open with V spt f then by (1) (f ) (V ), and hence, taking inf over such V and using (2), we see (4) f 2 K C with f 1 ) (f ) (spt f ); and then for any open U we can use (1) and (4) to conclude (5) (U ) D (spt f ); f 2K C ;f 1;spt f U Notice next that if K is compact then, by Lemma 2.4, if W K is open there is g 2 K C with g 1 in a neighborhood V of K and with g 1 and spt g W. Then by (3),(1) and () we have, for any such g, (6) (K) (V ) D (f ) (g) (W ): f 2K C ;f 1;spt f V To prove that is an outer measure it still remains to check countable subadditivity. To see this, first let U 1 ; U 2 ; : : : be open and U D [ j U j, then for any f 2 K C with f 1 and spt f U we have, by compactness of spt f, that spt f [ j N D1 U j for some integer N, and by using a partition of unity ' 1 ; : : : ; ' N for spt f subordinate to U 1 ; : : : ; U N (see Corollary 2.5), we have (f ) D j D1 (' j f ) j D1 (U j ). Taking over all such f we then have (U ) P 1 j D1 (U j ). It then easily follows by applying definitions (1),(2) that ([ j A j ) P j (A j ). So indeed is an outer measure on. Finally we want to show that is a Radon measure. For this we are going to use Lemma 2.8 above, so we have to check the hypotheses of Lemma 2.8. Hypothesis (R2) needed for Lemma 2.8 is true by definition and (R3) is true by (5). Since we also have finiteness of (K) for compact K by (6), it remains only to prove the additivity property (7) K 1 ; K 2 disjoint compact sets in ) (K 1 [ K 2 ) D (K 1 ) C (K 2 ): To check this, let U be any open set containing K 1 [ K 2 and use Corollary 2.2 to choose disjoint open V j K j with V j U, j D 1; 2. Then by (3) and (1) 12 2 Radon Measures, Representation Theorem On the other hand g 1 C g 2 1 on (because V 1 \ V 2 D ) and so g j 2K C ;spt g j V j ;g j 1;j D1;2 (g 1 C g 2 ) f 2K C ;spt f U ;f 1 (f ) D (U ): Hence we have proved that (K 1 ) C (K 2 ) (U ), and taking inf over all open U K 1 [ K 2 we have by (2) that (K 1 ) C (K 2 ) (K 1 [ K 2 ), and of course the reverse inequality holds by subadditivity of, hence the hypotheses of Lemma 2.8 are all established and is a Radon measure. Next observe that by (4) we have (h) (spt h) h; h 2 K C, and hence (h) D lim n!1 (max{h 1=n; 0}) ({x W h(x) > 0}) h; h 2 K since h is the uniform limit of max{h 1=n; 0} in and spt max{h 1=n; 0} {x W h(x) > 0} for each n. For f 2 K C (f not identically zero) and " > 0, we let M D f can select points 0 D t 0 < t 1 < t 2 < : : : < t N 1 < M < t N with t j t j 1 < " for each j D 1; : : : ; N and with ({f 1 {t j }}) D 0 for each j D 1; : : : ; N. Notice that the latter requirement is no problem because ({f 1 {t}}) D 0 for all but a countable set of t > 0, by virtue of the fact that {x 2 W f (x) > 0} (spt f ) < 1. Now let U j D f 1 {(t j 1 ; t j )}, j D 1; : : : ; N. (Notice that then the U j are pairwise disjoint and each U j K, where K, compact, is the port of f.) Now by the definition (1) we can find g j 2 K C such that g j 1, spt g j U j, and (g j ) (U j ) "=N. Also for any compact K j U j we can construct a function h j 2 K C with h j 1 in a neighborhood of K j [ spt g j, spt h j U j, and h j 1 everywhere. Then h j g j, h j 1 everywhere and spt h j is a compact subset of U j and so (9) (U j ) "=N (g j ) (h j ) (U j ); j D 1; : : : ; N : Since is a Radon measure, we can in fact choose the compact K j U j such that (U j n K j ) < "=N. Then, because {x W (f f j D1 h j )(x) > 0} [(U j n K j ), by (8) we have (10) (f f j D1 h j ) M j D1 (U j n K j ) "M : Then by using (9); (10) and the linearity of (together with the fact t j 1 h j f h j t j h j ) for each j D 1; : : : ; N ), we see that j D1 t j 1(U j ) "M (f P j h j ) (f ) (f P j h j ) C "M (K 1 ) C (K 2 ) (V 1 ) C (V 2 ) D D ((g 1 ) C (g 2 )) g j 2K C ;spt g j V j ;g j 1;j D1;2 (g 1 C g 2 ) g j 2K C ;spt g j V j ;g j 1;j D1;2 Since trivially j D1 t j j D1 t j (U j ) C "M : 1(U j ) R f d j D1 t j (U j );

7 we then have Math 205A Lecture Supplement 13 " ( (K) C M ) j D1 (t j t j 1 )(U j ) "M R f d (f ) j D1 (t j t j 1 )(U j ) C "M "((K) C M ); where K D spt f. This completes the proof of We can now state the Riesz Representation Theorem. In the statement, C c (; R n ) will denote the set of vector functions f W! R n which are continuous and which have compact port. (That is f 0 outside a compact subset of.) 2.14 Theorem (Riesz Representation Theorem.) Suppose is a locally compact Hausdorff space, and L W C c (; R n )! R is linear with L(f ) < 1 whenever K is compact. f 2C c (;R n );jf j1;spt f K Then there is a Radon measure on such that for each compact K there is a vector function W! R n with jj D 1 everywhere and j -measurable, j D 1; : : : ; n, and with L(f ) D f d for any f 2 C c (; R n ) with spt f K: In the cases when is -compact (i.e. 9 compact K 1 ; K 2 ; : : : with D [ j K j ) or L is bounded (i.e. f 2Cc (;R n );jf j1 jl(f )j < 1), can be chosen independent of K. Proof: We first define (h) D f 2C c (;R n ); jf jh L(f ) for any h 2 K C. We claim that has the linearity properties of Lemma Indeed it is clear that (ch) D c(h) for any constant c 0 and any h 2 K C. Now let g; h 2 K C, and notice that if f 1 ; f 2 2 C c (; R n ) with jf 1 j g and jf 2 j h, then jf 1 C f 2 j g C h and hence (g C h) L(f 1 ) C L(f 2 ). Taking over all such f 1 ; f 2 we then have (g C h) (g) C (h). To prove the reverse inequality we let f 2 C c (; R n ) with jf j g C h, and define f 1 D { g gch f if g C h > 0 0 if g C h D 0; f 2 D { h gch f if g C h > 0 0 if g C h D 0: Then f 1 C f 2 D f, jf 1 j g, jf 2 j h and it is readily checked that f 1 ; f 2 2 C c (; R n ). Then L(f ) D L(f 1 ) C L(f 2 ) (g) C (h), and hence taking 14 2 Radon Measures, Representation Theorem over all such f we have (g C h) (g) C (h). Therefore we have (g C h) D (g) C (h) as claimed. Thus satisfies the conditions of the Theorem 2.12, hence there is a Radon measure on such that (h) D h d; h 2 K C ; j D 1; : : : ; n: That is, we have ( ) L(f ) D h d; h 2 K C : f 2C c (;R n ); jf jh Thus if j 2 {1; : : : ; n} we have in particular (since jfe j j D jf j 2 K C for any f 2 C c (; R)) that jl(fe j )j jf j d kf k L 1 () 8f 2 C c (; R): Thus L j (f ) L(fe j ) extends to a bounded linear functional on L 1 (). In either of the 3 cases (i) K compact is given and we use Riesz Representation Theorem for L 1 ( K), or (ii) klk(d ()) < 1 and we use Riesz Representation Theorem for the finite measure case, or (iii) D [ j 1 D1 K j with K j compact for each j and we use Riesz Representation Theorem for the -finite case, we know that there is a bounded -measurable function j such that L(fe j ) D f j d; f 2 C c (; R); where in case (i) we impose the additional restriction spt f K. Since any f D (f 1 ; : : : ; f n ) can be expressed as f D P n j D1 f j e j, we thus deduce () L(f ) D f d; f 2 C c (; R n ); where D ( 1 ; : : : ; n ), and so by ( ) h d D f d D f 2C c (;R n );jf jh f 2C c (;R n );jf jd1;g2k C ;gh f 2C c (;R n );jf jd1;g2k C ;gh gf d for every h 2 K C, where in case (i) we assume spt h K. Now jf j jf jjj so we have gf d gjj d D hjj d g2k C ;gh Since C c () is dense in L 1 (), we can choose a sequence f k with jf k j D 1 and f k! jj on spt h, so the bound on the right of the previous inequality is attained and we have proved h d D hjj d and again using the density of C c () in L 1 () we have jj D 1 -a.e.

8 Math 205A Lecture Supplement 15 We conclude with an important compactness theorem for Radon Measures. Recall that Alaoglu s theorem ( see e.g. Royden, Real Analysis 3rd Edition, Macmillan 1988, p.237), which is a corollary of Tychonoff s theorem, tells us that the closed unit ball in the dual space of a normed linear space must be weak* compact: that is, given any normed linear space with dual space (i.e. is the normed space consisting of all the bounded linear functionals F W! R), then {F 2 W kf k 1} is weak* compact, meaning that for any sequence F j 2 with j kf j k < 1 there is a subsequence F jk and an F 2 with F jk (x)! F (x) for each fixed x 2. In particular if is compact and D C () (the continuous real-valued functions on ) { 2 W kk 1 } is weak* compact. That is, given a sequence { k } of bounded linear functionals on C () with k1 k k k < 1, we can find a subsequence { k 0} and bounded linear functional such that lim k 0(f ) D (f ) for each fixed f 2 C (). Using the above Riesz Representation 2.12, this implies the following assertion concerning sequences of Radon measures on, assuming is -compact Theorem (Compactness Theorem for Radon Measures.) Suppose { k } is a sequence of Radon measures on the locally compact, -compact Hausdorff space with the property k k (K) < 1 for each compact K. Then there is a subsequence { k 0} which converges to a Radon measure on in the sense that lim f d k 0 D f d; for each f 2 C c (): Proof: Let K 1 ; K 2 ; : : : be an increasing sequence of compact sets with D [ j K j and let F j;k W C (K j )! R be defined by F j;k (f ) D R K j f d k, k D 1; 2; : : :. By the Alaoglu theorem there is a subsequence F j;k 0 and a nonnegative bounded functional F j W C (K j )! R with F j;k 0(f )! F j (f ) for each f 2 C (K j ). By choosing the subsequences successively and taking a diagonal sequence we then get a subsequence k 0 and a non-negative linear F W C c ()! R with R f d k 0! F (f ) for each f 2 C c(), where F (f ) D F j (f jk j ) whenever spt f K j. (Notice that this is unambiguous because if spt f K j and ` > j then F`(f jk`) D F j (f jk j ) by construction.) Then by applying Theorem 2.12 we have a Radon measure on such that F (f ) D R f d for each f 2 C c(), and so R f d k 0! R f d for each f 2 C c ().