A GENERAL FATOU LEMMA Peter A. Loeb, Yeneng Sun SETUP Let � be a non-empty internal set, A 0 an internal algebra on � , and A the � -algebra generated by A 0 . Let J be a �nite or countably in�nite set. 8 j 2 J , let (� ; A 0 ; � 0 j ) and (� ; A ; � j ) be internal and Loeb probability spaces. From these generate � � so that 8 j , � j << � � . We may assume A is � � -complete. Let Y be a separable Banach lattice, and X is its dual Banach space with the natural dual order (denoted by � ) and lattice norm (i.e., j x j � j z j ) k x k � k z k ). Let P be any probability measure on (� ; A ) : 1
De�nition. A sequence f g n g 1 n =1 of functions from (� ; A ; P ) to X is said to be weak* P -tight, if for any " > 0 , there exists a weak* compact set K in X such that for all n 2 N , P ( g � 1 n ( K )) > 1 � " . De�nition. For each x 2 X , y 2 Y , the value of the linear functional x at y will be denoted by h x; y i . A function f from (� ; A ; P ) to X is said to be Gelfand P - integrable if for each y 2 Y , the real-valued function h f ( � ) ; y i is integrable on (� ; A ; P ) . Proposition. If f : (� ; A ; P ) 7� ! X is Gelfand P -integrable, then there is a unique x 2 X such that h x; y i = R � h f ( ! ) ; y i P ( d! ) for all y 2 Y . (That element x , called the Gelfand integral, R will be denoted by � f dP .) 2
Proof. (Well-known): Let T ( y ) be the element of L 1 ( P ) given by ! 7� ! h f ( ! ) ; y i . By Closed Graph Theorem, k T k < 1 , so � � Z Z � � � � h f ( ! ) ; y i P ( d! ) � � jh f ( ! ) ; y ij P ( d! ) � � � � k T k k y k : � Simplifying Assumption: 9 an increasing (perhaps constant) sequence y m � 0 in Y with lim m !1 h x; y m i = k x k 8 x � 0 in X . The assumption is valid when X = ` 1 or X = M ( S ) , the space of �nite, signed Borel measures on a second-countable, locally compact Hausdorff space S . The main result, stated here for a sequence of functions g n � 0 , is generalized with the assumption that each n 2 N , g n � f n where the sequence h f n i has appropriate properties. 3
Theorem. Let f g n g 1 n =1 be a sequence of nonnegative functions from � to X . Suppose 8 j 2 J , each function g n is Gelfand integrable on (� ; A ; � j ) , R and the Gelfand integrals � g n d� j have a weak* limit a j 2 X as n ! 1 . Then 9 g : � 7! X such that 1. for � � -a.e. ! 2 � , g ( ! ) is a weak* limit point of f g n ( ! ) g 1 n =1 , 2. the function g is Gelfand � j -integrable R with � gd� j � a j for each j 2 J ; R 3. the integral � h g; y i d� j = h a j ; y i for any y 2 Y and j 2 J for which fh g n ; y ig 1 n =1 is uniformly � j -integrable; R 4. In particular, � g d� j = a j for any j 2 J for which the sequence fk g n kg 1 n =1 is uniformly � j -integrable. 4
Corollary. Let f f n g 1 n =1 be a sequence of A -measurable functions from � to a complete separable metric space Z . Assume 8 j 2 J , f � j f � 1 n =1 converges n g 1 weakly to a Borel probability measure � j . Then, there is an A -measurable function f from � to Z such that f ( ! ) is a limit point of f f n ( ! ) g 1 n =1 for � � -a.e. ! 2 � , and � j f � 1 = � j for each j 2 J . Corollary. A simpli�ed version of our theorem holds for functions taking values in R p , where the norm of each x = ( x 1 ; : : : ; x p ) � � x i � in R p is given by � p � . i =1 For a more general theorem, the following consequence of the Simplifying Assumption about X must be added to the hypotheses. Claim. 8 j 2 J , f g n g 1 n =1 is weak* � j -tight. 5
Proof. By an argument of H. Lotz using the Monotone Convergence Theorem, � � �R � 8 j 2 J; 8 n 2 N , � g n ( ! ) d� j �Z � = lim g n ( ! ) d� j ; y m m !1 Z � = lim h g n ( ! ) ; y m i d� j m !1 Z Z � = m !1 h g n ( ! ) ; y m i d� j = lim k g n ( ! ) k d� j : R � � The Gelfand integrals � g n d� j converge in the weak*-topology, so by the Uniform Boundedness Principle 9 M j > 0 such that � � �R � � M j . Since 8 n 2 N , � g n d� j R 8 n; k 2 N , k g n k d� j � M j , fk g n ( ! ) k� k g � j ( f ! 2 � : k g n ( ! ) k � k g ) � M j =k: � 6
EXAMPLES We have an example showing that even for a single measure � , there may be no function g if � is Lebesgue measure on [0 ; 1] . Here, we let X = ` 1 . An example of Liapounoff constructs an h : [0 ; 1] ! ` 1 such that for no E � [0 ; 1] is it true that for coordinate-wise integration, R R E h ( t ) dt = 1 [0 ; 1] h ( t ) dt . 2 We use the Liapounoff Theorem and 8 n the �rst n components of h , to construct a sequence g n � 0 satisfying the conditions of our theorem, but g can not exist by the Liapounoff example. A modi�cation of this �rst example shows that the corollary, even for R 2 , can fail when the measures � j are multiples of Lebesgue measure on [0 ; 1] . 7
Lemma 1. Let X be a standard, separable metric space with metric � and the Borel � -algebra B . Fix x 0 2 X . Let P 0 be an internal probability measure on (� ; A 0 ) with Loeb space (� ; A ; P ) . Let h be an internal, measurable map from (� ; A 0 ) to ( � X; � B ) . Let � be the internal probability measure on ( � X; � B ) such that � = P 0 h � 1 . Fix a standard tight probability measure � on ( X; B ) such that � � ' � in the nonstandard extension of the topology of weak convergence of Borel measures on X . Then the standard part � h ( ! ) exists for P -almost all ! 2 � (where h ( ! ) is not near- standard, set � h ( ! ) = x 0 ). This function � h is measurable, and � = P ( � h ) � 1 . 8
Proof. For every standard, bounded, continuous real-valued f on X , Z Z Z � f d� ' � f d � � = f d�: � X � X X Let K 0 = ? , and 8 n 2 N , let K n � K n � 1 be compact in X with � ( K n ) > 1 � 1 2 n . 8 j 2 N , n := f x 2 X : � ( x; K n ) < 1 V j j g has the property that � ( � V j n ) > 1 � 1 n , whence 9 H 2 � N 1 , with � ( V H n ) > 1 � 1 n . Now the monad m ( K n ) := \ j 2 N � V j n , and h � 1 [ m ( K n )] = h � 1 � � = \ j 2 N h � 1 � � V j � � V j \ j 2 N n n � � h � 1 [ m ( K n )] is measurable and P � 1 � 1 n . The standard part � h is de�ned on h � 1 [ m ( K n )] , is measurable there, and takes values in K n . 9
Therefore, � h de�nes a measurable mapping from [ n h � 1 [ m ( K n )] to [ n K n , and � � [ n h � 1 [ m ( K n )] P = 1 : Set � h = x 0 on � � [ n h � 1 [ m ( K n )] . With this extension, � h is a measurable mapping de�ned on (� ; A ; P ) . Finally, given a bounded, continuous, real- valued function f on X , Z Z f dP ( � h ) � 1 = f � � h dP Z X � st ( � f � h ) dP = Z Z � � f � h dP 0 = � f d� ' Z Z � X � � f d � � = ' f d�: � X X It follows that � = P ( � h ) � 1 on X . � 10
Lemma 2. Let ( X; � ) be a separable metric space with the Borel � -algebra B . Let P 0 be an internal probability measure on (� ; A 0 ) with Loeb space (� ; A ; P ) . Fix an internal sequence f h n : n 2 � N g of measurable maps from (� ; A 0 ) to ( � X; � B ) . Fix a nonempty compact K � X . Then 9 H 2 � N 1 and a P -null set S � � such that if n � H in � N 1 , while ! = 2 S , and h n ( ! ) has standard part in K , then for any standard " > 0 , there are in�nitely many limited k 2 N for which � � ( h k ( ! ) ; h n ( ! )) < " . 11
Proof. Given l 2 N cover K with n l open balls of radius 1 =l . Let B ( l; j ) denote the nonstandard extension of the j th ball. For each i 2 � N , set A i ( l; j ) := f ! 2 � : h i ( ! ) = 2 B ( l; j ) g : 8 k 2 N , choose m k ( l; j ) 2 � N 1 so that � � � � \ m k ( l;j ) \ 1 P A i ( l; j ) = P i = k;i 2 N A i ( l; j ) : i = k Set � � � \ m k ( l;j ) \ 1 S k ( l; j ) := i = k;i 2 N A i ( l; j ) A i ( l; j ) : i = k Fix H 2 � N 1 with H � m k ( l; j ) 8 l 2 N , 8 j � n l , and 8 k 2 N . Let S be the P -null set formed by the union of the set S k ( l; j ) 8 l 2 N , 8 j � n l , 8 k 2 N . 12
Fix n 2 � N 1 with n � H , and suppose st ( h n ( ! )) 2 K but 9 l 2 N for which there are at most �nitely many limited k 2 N for which � � ( h k ( ! ) ; h n ( ! )) < 2 =l . Then for some j � n l , h n ( ! ) 2 B ( l; j ) , and by assumption there is a limited k 2 N such that for all limited i � k , h i ( ! ) = 2 B ( l; j ) . It follows that ! 2 S k ( l; j ) � S . � Idea of Parts of Theorem's Proof. Replace sequence f g n g with a subsequence so 8 j 2 J , � j g n � 1 converges weakly to � j . Lift and extend f g n g to f h n g and work with measures � 0 j h n � 1 . Use Lemma 1 to show 9 H 2 � N 1 so g ( ! ) := ( � h H ) ( ! ) exists for � -a.e. ! 2 � and � j = � j ( � h H ) � 1 8 j 2 J . Use Lemma 2 to show that for � � -a.e. ! 2 � , g ( ! ) is a weak* limit point of f g n ( ! ) g 1 n =1 . 13
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