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This research was supported in part by the Air Force Office of Scientific Research under Contract AFOSR-68-14l5 and by the Office of Naval Research under Contract N00014-67-A-032l-0006 (NR042-69). -e ZERO-oi~E ~!S FOR GAUSS INJ r1EAsURES ON BANAQi SPACE Charles R. Baker Department of Statistics Uni versi ty of North Cam Una at Chape Z Hi ZZ Institute of Statistics Mimeo Series No. 785 November, 1971 ZERO-oNE lAws FOR GAUSS IAN l''1EAsURES ON BANACH SPACE Charles R. Baker ABSTRACT B be a real separable Banach space t Let Borel a-field of If G B [B] € ~ B, and B [B] -e 1 for the case where a Gaussian measure on the the completion of the Borel a-field under is a subgroup, we show that due to Kallianpur and Jain. ~(G) = ~ ~ ~(G) =0 or 1, ~. extending a result Necessary and sufficient conditions are given for G is the range of a bounded linear operator. These results are then applied to obtain a number of 0-1 statements for the sample function properties of a Gaussian stochastic process. The zero-one law is then extended to a class of non-Gaussian measures, and applications are given to some non-Gaussian stochastic processes. 2 L INTRODUCTION Kallianpur [1] and Jain [2] have proved the following result. a complete separable metric space, on T, and B[X] that P function X sets generated by sets of the form tl, •.. ,t n E T and C a Borel set in is a Gaussian probability measure on K and zero mean, and that Hilbert space of K. Let BO[x] B[X] n R • Suppose with continuous covariance X contains the reproducing kernel denote the completion of B[X] With these assumptions, Kallianpur and Jain have shown that for every T be X a linear space of real-valued functions the a-field of {x: (x(tl), ••• ,x(tn»EC}, Let BO[x]-measurable subgroup of under P(G) =0 P. or 1 X. This result does not immediately apply to the case of a Gaussian measure on the Borel sets of an arbitrary real separable Banach space, even in the case of ·e L, P since on We first show L p the co-ordinate maps 1T t '. X -+ xt are not well-defined. that this zero-one law holds for Gaussian measures on a Banach space, and without the assumption of zero mean. Some necessary and sufficient conditions for the alternatives are also given. is extended to a class of non-Gaussian measures. The 0-1 law A number of applications are given on path properties of Gaussian stochastic processes and some nonGaussian processes. As noted in [1], Cameron and Graves [3] first considered this problem for the case of Wiener measure. In another direction, Pitcher [4] essentially proved the following result. separable Hilbert space for Sand T H Let ~ be a zero-mean Gaussian measure on a real with covariance operator Hilbert-Schmidt linear operators. K. Then Suppose ~[range(T)] Our interest in this problem was motivated by Pitcher's work. result to any bounded operator law holds for any measurable T, K = TST* =0 or 1. We extended his and conjectured from this that a sioilar 0-1 linear manifold in,, Bilbert space. '~Je are indebted to the appearance of [1] for the key step in completing our proof (Lemma 2, below), and to [2] for enabling us to extend our result to subgroups. 3 2. ilEFIN IT! ONS B will denote a separable Banach space, with norm Hilbert space with inner product (.,.). Borel a-field of A sets. A Gaussian measure All linear spaces are defined over IF is the characteristic function of the set F. ~ on B[B] is by definition a probability measure such that all bounded linear functionals on B are Gaussian with respect S* is the space of all bounded linear functionals on ~. separable For a given topological space A, B[A] will denote the the real numbers. to 11·1 I, H a B• We will usually not distinguish between Hand H*• ·e 3. THE ZERO-(}JE 1A1 FOR BANAa-I SPACE In this section we obtain a zero-one law for any Gaussian measure on a real and separable Banach space. In particular, the measure need not have zero mean element. We first obtain the zero-one law for H. Let ~l be a Gaussian measure on B[H] with null mean element and covariance operator K; = f (x,u)<x,v)d~l(x), all u,v in H. K is linear, bounded, nonH negative, self-adjoint, trace-class, and uniquely determines ~l [5]. Since (Ku,v) H is a linear topological space, A is a Borel set, any z ~(A) = ~l{x: {e }, n n x + m = 1,2, ••• € A}. € {x: x + z € A} is a Borel set in H if H. We define the Gaussian measure ~ ~ on B[H] by has covariance operator K and mean element m. will denote a complete orthonormal set of eigenvectors for K; we require only that I IKen II # 0 for at least one value of n. 4 \l • Let We will obtain the zero-one law for B[H] \l. under For any v \l{x: x+Kv EA}, H, in and = exp{<x,v> Comparing L2 [\l])· G E B [H]. \l define the Gaussian measure by \lv = \l (A) v are mutually absolutely continuous [d\l /d\l] - l and have Radon-Nikodym derivative (= be the completion of The following result is well-known. Lemma 1 [6]: lv(x) H, G is a subgroup of B [H] \l Since definition of \1 v with given by v ~I IK~vl 12 }. -(m,v) - 12v v 12v' \lv ~ \l, lv E L 2 [H,B\l[H],\l] it is clear that = B\1[H], B\1v[H] = L2 [\1V], L2 [\1] {x: x+Kv E A} E B [H] \1 we obtain that and from the A E B [H]. \1 for Lemma 2: -e 0, l-l , vEH} is dense in L [\1]· 2 v Proof: Suppose IHg(x)d\l(x) = I Hg(x)[l-lv(x)]d\1(x) = for all v in H, some g in L [\1]· Then IHg(x)exp{(x-m,v)-~1 I~vl ,2}d\1(x) = 0, all 2 vEH. Let v - L~ai e , where the {a. } are scalars. Proceeding as in the i 1 N N proof of Lemma 3 of [1], one obtains E[gIB[H ]] = a.e. d\1, where B[H ] is ° ° the a-field generated by f : x + <x,e >, n n smallest a-field such that E[gIB[H]] = ° a.e. d\l, f n = 1, ••. ,N. is measurable for all n and thus = ° a.e. g(x) e, is the one obtains n d\1(x) B[H] Since (see Appendix 1 for details) • Lemma 3: For any are scalars such that \1(F) Proof: If 0, \1 (Fa ) n > that = 1, all l a v n n, FEB [H], \1 lim n \l(F) = 0, = IFl a (x) + 1 n a n = 0. then \l(F (x)d\l(x). v as we see that a n {l + a v F let 0, Then a v n a v n = {x: lim ) = \1 n a v n x+a Kv E F}, n \1(F a v Hence {IF1 a v} n {a } n = \l(F). n (F) = 0, Noting that la v since \l ~ \la v If n is positive for each n for all xEH, and that f.P H-a n v (x) d\1 (x) } is uniformly integrable with respect to \l n [7, p. 18]. ) where is uniformly integrable, 5 Ip(x)la v(x) Ip(x) + a as n fHIp(x)la v(x)d~(x) ~(p) + + 0, as a n n Lemma 4: If ~(G) > all + n x H, in and this implies O. 0, then G ~ range(K). Proof: We apply a proof used to establish Theorem 1 of [2]. Suppose ~(G) > l/s, s o ~ k ~ s, Ku~G. a positive integer. by Por each n, This implies that (s+l)~(G) ~ = G, G(n)o G(n) I:=o =0 ~(G) Proof: Suppose -_ v (G) = ~{x: G(n) j = 1,2, ••• define G(n)k' 1 ~ k ~ s, x+(slkn)-lKuEG }, j # k are disjoint if where (see [2]). ~(G(n)k) ~ 1, all n, and hence by Lemma 3 or 1. ~(G) > O. = ~(G), x+KvEG} We note that and k = {x: n a contradiction. 1, Lemma 5: ~ G(n)k Por From Lemma and 1 J. {l-l , vEH} , Span{l} @ fHIG(x)[l-lv(x)]d~(x) L2[~] so that by Lemma 2 v decomposition 4, G ~ Kv for all vEH, so that Span{l-l , vEH}. v = 0, all v has the orthogonal IG(x) = constant a.e. Hence, H. in d~(x). Corollary: If ~(G) = 1, then G ~ range(K~). Proof: Let ~ , ~' v (A) = ~{x: X+K~VEA}. then ~(G) = 1~~' ~ Now let B [B] ~ v (G) ~' v ..... = 0, ~ [6], so that B~t U1] v defined by =B ~ [H]. If K~V~G, a contradiction. be a Gaussian measure on be the completion of subgroup of B[H] be the Gaussian measure on v B[B] under B[B] ~, with mean element m, GEB [B] and suppose that ~ let is a B. Theorem 1: ~(G) =0 or 1. Proof: We can construct a separable Hilbert space H such that the e1ements of B[B] B constitute a dense linear manifold in = BnB[H]; see [8]. Let H, BEB[H], v be the Gaussian measure on and B[H] defined by 6 v(A) = l.I(BnA). sets in B[B] be the completion of v of l.I-measure zero belong to is clear that group in B [H] Let B [H] v B, G then B [B]. H, suIt now follows from Lemma 5 and the definition of v. 1.I Remarks: 1. Theorem ment. G is a sub- Hence, if 1.I G is a subgroup in € Since all v. under and have v-measure zero, it B[H] contains all elements of B [8], B[H] G € B [H]. The re- v 1 holds for Gaussian measures with non-zero mean ele- This does not appear to be obvious if one has only the zero-one law for zero-mean Gaussian measures. 2. If one limits attention to manifolds, there are several ways to prove Lemma 4. KV~G. Inl ·e + For example, let Then 00. G n Also, and l.I(Gn ) = {x: xt~v n Gn G are disjoint if m = l.I v / n (G) , uniformly integrable. Now suppose such that while l.I(G) n < 0, € G} n ~ n m, and the family l.I(G) > O. = ±1,±2, ••• so that l.I(G) n {l vn /} For any 0 and suppose + 0 as is easily seen to be > 0, there exists f G l -v / n (x)dl.l(x) = 1.I-v / n (Gn ) = l.I(G). n This n contradicts the uniform integrability of 4. {l vn / }. f'1EAsURABLE SUBGROUPS The following two lemmas can often be used to show that a given subgroup is measurable. Lemma 6 [9]: Let T be a map from a complete separable metric space Ml into a complete separable metric space B[Ml]/B[M ] Z measurable. Then T[A] € M2• Suppose T is one-to-one and B[M Z] when A € B[M l ]. 7 In our applications, we deal with a bounded linear operator two Banach spaces. T between The following lemma extends Lemma 6 (Kuratowski's theorem) NT to the case where (the null space of T) contains elements other than the null element. Lemma 7: flexive. Let Suppose T[A] E B[B ] 2 Proof: B l B2 be two separable Banach spaces, with Bl and T: B l B2 is a bounded linear operator. Then + A E B[B ]. if l See Appendix 2. We note that Lemma 7 can be used if p Bl is lp or Lp (Lebesgue) for E (1,00). 5. ·e re- NECESSARY AND SUFFICIENT CONDITIONS FOR ~(G) = 1 ~(G)· There are many problems in which one will wish to determine if for some measurable subgroup ditions for =1 ~(G) Theorem 2: Theorem 2 gives necessary and sufficient con- for a large class of subgroups. Suppose a Gaussian measure on G. T: H + B is a bounded linear operator. B[B], with mean element m, If ~ is then (a) ~[range(T)] = 0 or 1. (b) ~[range(T)] = 1 if and only if there exists a Gaussian measure v on B[H] such that (c) Let HI ~[A] = v{x: TXEA} for all A in B[B]. be a separable Hilbert space containing linear manifold and such that be the Gaussian measure on AEB[H ]. l 1 Then ~[range(T)] B E B[H], B[H ] l =1 B[B] defined by B as a dense = BnB[H]. ~l Let ~l[A]. ~[BnA], if and only if m E range(T) and there exists a covariance operator S in H such that K l = TSTl *, 8 where Kl ~l is the covariance operator of restriction of T* to and :1* is the HI' Proof: Part (a) follows from Theorem 1 and Lemma 7. To prove (b), we first note that if such a measure ~[range(T)] Thus, suppose v· and {x: x=Tv, ..l. iJm Yx = 0 : J. v € A n {x: xfrange(T)} • hence Y is ~; ~{X: Yx J. N }. v(A) if T Define a map xfrange(T). measurable. = v(H) ~[range(T)] range(T*); Y: B For O€A, y-l(A) In both cases, T If OfA. = ~{xIYxe:A}. any element of ·e if N J.r B[B]/B[H] from = 1. H by + then Yx =v y-l(A) = T[A Y thus induces a measure = ~{x:Yx = 1. € H} = 1; To see that = ~{x: Thus v{y: (y,u)<k} = ~{x: (Yx,u)<k} ~{x: lim u (TYx) <k} nn = ~{x: lim u (x)<k}, n n n NTJ.] u on v B[H] is Gaussian, let B* c such that limn(YX,T*un)<k} u T*u n + be u. = range(T) = {x: TYx=x} • since = Tv also, v {u } n then there exists x by Lemma 7, and B[B], belongs to if = y-l(A) AEB[H], = 1. ~[range(T)] exists, then obviously v As the a.e. limit of a sequence of Gaussian random variables, the random variable fu: x (x,u) + bility measure. ~{x: To see that ~ = ~{x:x e: Anrange(T)} = x€A} To prove (c), we note that if ~1 ~[range(T)] = vaT -1 v. is Gaussian with respect to = 1, • Let let v Sand y We first note that Ty = m, v is thus a Gaussian proba- is induced from ~{x:TYx v by T, one notes that € A} - v{x:Tx € Ale ~[range(T)] =1 <~ ~l[range(T)] be the Gaussian measure on B[H] = 1. Thus, such that be the covariance operator and mean element of since I H (x,u)ld~l(x) = I H(x,T 1*u)dv(x), all v. u 1 in HI' where product on H. (0'·)1 is the inner product on Now, for all u,v in HI' (K1u,v)1 = I H(x-y,T 1*u)(x-y,T l *V)dv(x) = (TSTl*u,v)l. if K 1 define = TST 1*, v S a covariance operator in to be the Gaussian measure on and mean element y. The map T: H + HI HI and (0,0) is the inner B[H] Hence, H, = I H (x-m,u)l(x-m,v)ld~l(x) 1 = TST l *. Kl and m = Ty for Conversely, y€H, with covariance operator induces from v we S a Gaussian measure 9 having covariance operator TST1 * and mean element m; on ~ Gaussian measure on HI since a is uniquely specified by its covariance operator and mean element, this: measure must be Some applications of Theorem 2 are given in the next section. it may be of interest to note here the following. measure on that B[H], with covariance operator ~{x:JC+m € range(K~)} .. arbi trary Gaussian example, if ~ lJ, for all 0 This property is not shared, for m€H. ~l{x:x-~€ range(T)} .. 1, where then ~l is the translate of ~ and T: If + H ~ [range (T)] € is bounded and linear, then range(T)} =0 for all m€H. this holds for all subgroups in For 1 for lJl[range(T)] .. 0, However, it is clear from Theorem 2 that if ~{x:X+m lJ (G) .. O. ~[range(T)]:II is a zero-mean Gaussian measure such that H, is a Gaussian Part (c) of Theorem 2 shows by all subgroups G € B[ H] such that T a bounded and linear map in ·e K. ~ Suppose However, ~ by m(range(T) • m, is a zero-mean Gaussian measure .. 0 <=0;> It would be interesting to know whether B[H]. If so, the method used to prove Theorem 1 shows that a similar result would hold for any subgroup in B[B]. Part (c) of Theorem 2 is an extension of a result by Pitcher [4], who proved the following. ~ Suppose with covariance operator K. is a zero-mean Gaussian measure on Suppose K = TST, where T and S are linear, bounded, non-negative, self-adjoint, and Hilbert-Schmidt operators in T strictly positive. and only if that K Then ~[range(T)] S is trace-class. = TST, that range(T) = 0 or 1, and B[H], ~[range(T)]" H, with 1 if His method of proof requires the assumptions = range(S) = H~ and that ~ is self-adjoint La'J.d compact. The following theorem generalizes (a) and (b) of Theorem 2 to two Banach spaces. 10 Theorem 3: ~ T: Bl that Suppose B 2 + T on B[B l ] Proof: to-one. Then ~[range(T)] Moreover, v is one-to-one and B[B ]. 2 Gaussian measure on sure ~ is a bounded linear operator. reflexive, or else or 1. such that Suppose that either = Bl *. range(T*) range(T) B[B ], 2 E ~ there exists a probability measure B 3 T = Bl * range(T*) v ;nduces • Since on ~ T -1 and B[B ] l f rom Define such that B ; range(T) exists ([10], p. 44). belonging to range(T*); easy to see that v{x: y(x) < k} y(x) tion that y say y = Tl*v l , VI = Il{x: = T*v. If is Gaussian with respect to v, has a norm- Since be any element of B1 * is not the null element, it is B : y(Tl 3 E vI all v -1 y to x) < k} B • 3 Now = Il l {x: range(T*). E vl(x) < k}. The assump- B1* thus implies that v is a Gaussian measure. For the case where suppose that y y the restriction of yeT-Ix) < k} = Il {x l range(T*) = Let and the be the restriction of ~l is thus a Gaussian measure. 2 as defined i s Gauss~an. . v then by the Hahn-Banach theorem Hence, v is one- v(A) = We s h ow v. Let preserving extension to T exists, it is obvious that denote the separable Banach space consisting of norm. =0 is induced from v by T. = 1. 11 [range (T)] and that is be a ~[range(T)] and It is clear that we need only prove the existence of "{X·, T-lx E A}, ~ ~ Let B l = 1 if and only if there exists a Gaussian mea- First consider the case where in the theorem when Let is a real separable Banach space, and B. (i=1,2) Bl ll[range(T)] is reflexive, with = 1. T not necessarily one-to-one, The proof of Lemma 7 (Appendix 2) shows that one can imbed B as a dense Borel-measurable linear manifold in a separable Hilbert space H. in such a way that and T i =v B[B ] i C B[H ], i B[B ] i = Bi n B[H ], can be extended by continuity to a bounded linear operator It is clear that Yx ~ if range(T) x = T v 1 and = Tl[B l ] v.L NT ' while Yx = 1 .1 NT 1 0 ]. Define Y: H2 if x ~ i T l :H l +H 2 • + range(T ). l HI by Define 11 e ~ on B[B 2 ] by ~ (A) = ~(B2nA). The procedure used to prove (b) of Theorem 2 shows that there exists a Gaussian measure v on B[H ] such that l , (A) ~ = A} for A in B[B ]. v(B ) = v{x: Tlx €B } = ~ (B ) = 1. Let va Z 2 2 l * then u(x) is a be the restriction of v to B[B ]. If u € B * belongs to HI' l l v{x: Tlx € Gaussian random variable with respect to vo' since v Hence va is Gaussian. provided H * is dense in 3 *. Suppose there exists f in 8 ** 1 l 1 such that f(v) = 0 for all v in H *. Since 3 is dense in HI and 8 is 1 1 l is Gaussian ** ~ ** ** 1 ' 0 1 is dense in HI and hence f must be the null element of Bl ** • Thus H1* is dense in 8 *' so that va is Gaussian, 1 isometrically isomorphic to B with ~(A) = vO{x: Tx € A} for A € B[B Z]. The remainder of the proof is obvious. Theorems 2 and 3 extend the following well-known result: measure on B[B ], and T: 3 1 -e 1 ~ If ~ is a Gaussian B is bounded and linear, then the measure on 2 B[B 2 ] induced from ~ by T is also Gaussian. Thus, if B is reflexive, and T- l l exists, then Theorem 3 shows that for a Gaussian measure ~ on B[B 2 ], ~[domain(T-l)] = a or 1; if ~[domain(T-l)] = 1, then the measure induced from ~ by T-1 is Gaussian. if and only if by T. If T-1 does not exist, we still have that ~[range there exists a Gaussian measure on B[B ] from which I ~ ( T)] =1 is induced 12 o. PfPLICATIONS TO GAUSSIAN rEASURES AND GAuSSIAN STOCHASTIC PROCESSES In this section we apply the preceding theorems to obtain a number of zero-one statements for Gaussian measures and Gaussian stochastic processes. Statements regarding analytical properties of sample functions are contained in Theorems 4, 7, and 8. Theorems 5 and 6 deal with the convergence of certain series, and with integrability properties of the sample functions of measurable Gaussian stochastic processes. In addition to the results on Gaussian measures and processes presented in this section, Theorem 10 of Section 7 contains results on the convergence of sequences and series of Gaussian random variables. We use L or L [T] to denote the set of equivalence classes of all p -e p real-valued Lebesgue-measurable functions f on T (an interval) such that fTlftlPdt < ~, together with the n~rm I If I Ip = -/ [fTlftIPdt]~ p. We do not distinguish between a function and its equivalence class. l p x = (x l ,x 2 ' ••• ) together with the norm IIxl Ip = [L n Ixn IP]~/p. denotes the set of all sequences of real numbers such that L n Ixn IP < ~, We use either C or CrT] to denote the space of real-valued functions defined and continuous on the compact interval T, together with the norm I If II = sup t E T Iftl. 13 Theorem 4. ~ C[T]. Suppose = {x: ACP[T] any fixed (2) (a. ,x) , a. [a,b]. denote the real separable Banach space with elements P C[T]. = ACP[T], (2) Cn[T] I Ixl I = I~=o + in x C[T] P be defined by Noting that P SUPt€T IX t (i)l. > 1 s-1 f = = where x(s)ds, range(S) is a linear manifold and that =0 or 1. The natural injection of and C[T], )J(AC P) (f(a),f(l» Thus, Cn[T] Cn[T] into C[T] is is a Borel-measurable and the result follows. = 1, differentiation induces from that = a+f at t II(a,x)11 is a real separable Banach space under the norm linear manifold in If S(a.,x) range(S) ~(ACP[T]) one sees that a bounded, linear, and one-to-one map. .tit and with norm defined by L [T], is a one-to-one bounded linear map, so that by Lemma 6 S is a Borel set in range(S) x is n-times continuously differentiable on T}. S: Q Let zero or one: derivative}, p a real scalar, 1a.1+llxll· p ~-measure and have x is absolutely continuous with L Let (1) B[C[T]] p€[l,oo) = {x: Cn[T] Proof: = is a Gaussian measure on the Borel a-field of The following sets belong to (1) T ~ for induces a Gaussian measure on then Theorem 3 shows that the operation of a Gaussian measure on ~ f € AC P , B[Qp]' S L p . For this, one notes as in the proof of Theorem 4. and projection into L p This gives a Gaussian (1) € A} for A€ B[L]. B[L ], \) (A) = ].l{x € AC P : x p P The following theorems deal with various properties of the sample functions measure \) on of Gaussian stochastic processes. real-valued. Lemma 8: In each case the process is assumed We first establish a lemma Suppose (X t ) , t€T essentially used by Pitcher [4]. (a compact interval), is a measurable Gaussian stochastic process on a probability space (n,B,p). If almost all 14 sample paths of (X ) t Gaussian measure belong to on ~ is the sample path of evaluated at e. -+ L {y: duces a probability measure ~ L , S/B[L] on every bounded linear functional on X(w) > belongs to form a neighborhood base 0, for A€B[L]. Hence, P X in- By a result of Doob [11, pp.64-65]. p p where X(w)€A}, = X-l{x: I Ixl Ip<E} £ B[L l. L induces a t measurable, one notes that p X-l[A]€e P (X) w. I Iyl Ip <£}, at zero for the norm topology in then = p{w: ~(A) {w: [!T1Xt(w)IPdt]1/p<E} Since sets of the form that is p P < 00, S defined by p (X ) t 1 p B[L], Proof: To see that X: n by Fubini's theorem L [T], so is Gaussian with respect to p is a Gaussian measure. The problems treated in Theorem 5 have a long history (see ell, and the references given there). The proof is remarkably simple, given Theorem 1. Theorem 5: Suppose (X t ) , t€T (a compact interval), is a measurable Gaussian stochastic process on a covariance function (X ) , t sample path of (1) (2) K(t,s) probability space and mean element evaluated at w. met). Let (n,e,p), with denote the X(w) Then P{w: X(w)€L } = 0 or 1, any fixed p ~L p Let {en } denote a set of functions continuous on T, with {f } n p{w: X(w)€L } = 1, then Z p{w: Ien (t)!Tfn (s)X s (w)ds converges uniformly} = 0 or 1. Proof: As shown in [12], one can define a function g€L l such that 2 gX(w)€L a.e. dP(w); e.g., get) = 1 if K(t,t)+m (t) S 1, and get) = l 2 -1 2 [K(t, t)+m (t)] i f K(t,t)+m (t) > 1. Let (Y t) = (gtXt); (Y ) is a meast urable Gaussian process with almost all sample paths in L • Thus Y induces l from P a Gaussian measure ~ on B[L ]. Now let G: Ll -+ Ll be defined l any set of functions belonging to L • 2 If IN by Gv = gv, {w: X(w)€L } l g as above. = {w: G is bounded, linear, and one-to-one. Y(w)€range(G)}. Hence, {w: X(w)€L } € e, l and Clearly, 15 p{w: X(w)€L l } = ~y{range(G)} = 0 or 1. For and that the natural injection of p > 1, L P one uses the fact that into is a bounded, linear, and one-to-one operator. The set of (2) A= w for which the series converges uniformly is ~~1 U~l nm>n>M nt€S {w: IL~=n+l ek(t) IT fk(s) Xs(w)dsl is a countable dense subset of T. But A = X- 1 [D], 1 < -} N ' where s where D = nun n {x: IL~=n+l ek(t) IT fk(s) xsdsl < .!} N~l M~l m>n>M N • D is obt€S viously in B[L ], since {x: IIk€I Yk IT fk(s)xsdsl < .!} 2 M € B[L 2 ] for any finite index set I and scalars {Y }. D is also a linear manifold. Hence, k P[A] = ~X[D] =0 ~X or 1, the measure induced from P by X. This completes the proof. Part (1) of Theorem 5, for zero-mean (X ), has been previously t announced by B. Rajput. Part (2) contains the orthogonal series . ~ (Karhunen-Loeve) representation of (X t ) as a special case. It is shown in [13,14] that the orthogonal series converges uniformly with probability one when (X t ) has continuous paths. The following theorem generalizes Part (2) of Theorem 5, and gives the analogue of Part (1) for l Theorem 6: p spaces. Let {Z }, n = 1,2, ••• be a family of jointly Gaussian n random variables on a probability space (n,8,p). Let {e } be a set of n functions defined and continuous on the interval T. 1 ::; p < Define, for 00, A = {w: IN1 Z (w)e (t) converges uniformly on T}, n n B = {w: INl[z n (w)]Pen (t) converges absolutely and uniformly on T}, P C P = {w: I~lzn(w)IP < oo}. Then A, B , and C each belongs to 8, each has probability one or zero, p p and P[C ] = 1 if and only if p I~[E(zn2(w»]P/2 < 00 • 16 Proof: g Let n .. n- 1 if E Z; 2 (w) n :5 and 1, g .. n- 1[E Z 2(w)]-1 n . n then Y(w) = (Y (w),Y (w), ••• ) € i a.e. Y .. g z· l 2 2 n n n' dP(w). Let n = {w: Y(w)€i }. Y: Q + i induces a probability measure ~Y l 2 l 2 on B[i 2 ]; ~Y is Gaussian, since if LI~12 < 00, L~Yk is a Gaussian random variable as the a.e. limit of a sequence of Gaussian random variables. If \' n -2' {I x 12 :5 sup I x 12 L\' n -2: :5 2 II x 11 2 • Hence we Ln IxIP<oo, p ~ 1, then L n n n "Il'm n p by (G x) .. define the bounded 9 linear, and one-to-one operator G : ip + i 2 p n p gnxn for x€ip. Define the continuous function en' by e~'(t) .. g n -Pen (t) ' otherwise. Let and define e" n l A .. y- {x€i 2 : l Bp = y- {x€i 2 : by e "(t) .. g -1 e (t). n n n L~ xn en"(t) converges uniformly on T}, L~ Ixnl P en:(t) converges absolutely and cniformly on T}, and Cp" y-l[range(G )]. Using the method of proof for part (b) of Theorem 5, one p sees easily that -e '/a + bl P :5 Note that A, is in Band P(A) .. 0 or 1. - Similarly, using the inequality 2P (la1 P + fbI P ), P(B ) p il,nd that P (C ) = 0 or 1. p =0 or 1. It is clear that C belongs toB. p Finally, P.(C ) .. 1 if and only 1'f the paths of (Zn) p induce a Gaussian measure on B[i ]. the results of [15, 16] show that this occurs p if and only if ~ [EZ 2(w)]p/2 n n < 00 • For the next result, we make the following definitions. set of functions on cn eq [T] and T that are equal a.e. dt AC P [T] eq are defined similarly. Theorem 7: Suppose (X t ), t€T to an element of (1) (2) (3) B and have {w: X(w)€C [T]} eq n {w: X(w)€C [TD eq {w: X(w)€AC P [TD. eq is the CrT]; (a compact interval), is a measurable Gaussian stochastic process on a complete probability space following sets belong to Ceq [T] (n,B,p). probability zero or one: The 17 Proof: It is clear that Ceq [T], Cn eq [T] Ll[T]. From Theorem 5, either almost all paths of or almost all do not belong to Ll[T]. l then X- [A]€8 for all subsets A of AC P [T] eq and (X ) t are subsets of belong to Ll[T], If almost all paths lie outside Ll , L , since (n,8,p) is complete. l Thus, to prove the theorem, it is sufficient to show that the sets indicated in (1)-(3) of the theorem are elements of one or zero, when almost all paths of Thus, suppose that p{w: X(w)€L } l Gaussian measure induced on be the map taking fTlxtldt Thus S x in T SUPt€T Ixtl, range(W) € B[L l ] range(W) =C ~ [T], B[L ] l by 8, and that each set has probability (X ) t belong to = 1, and let ~X(A) (X ); t Ll[T]. ~X be the = p{w: X(w)€A}. C into its equivalence class in W is bounded and linear; X-l(C ~ )€8 W Since W is also one-to-one. ~X[range(W)] by Lemma 6, and hence this shows that L • l Let and that =0 or 1. P[X-l(C ~ )] Since =0 or 1. To prove the remainder of the theorem, we note that in the proof of Theorem . ~ 4 both Cn[T] and Defining + AC Peq W: C = W[ACP ] , L l ACP[T] were shown to be Borel linear manifolds in as above, noting that Cn eq = W[Cn] CrT]. and that the remainder of the proof is clear. The preceding theorems contain various zero-one laws for Gaussian processes. The problem of determining necessary and sufficient conditions for the alternatives is usually more difficult. rection are given in Theorem 2. Some general results in this di- The following theorem is an application of Theorem 1 an1 Theorem 2. Theorem 8: Suppose that (X t ) , t€(-oo,oo), is a measurable mean-square continuous Gaussian process on a complete probability space (n,8,p). Suppose that (X ) is stationary with rational spectral density function t mean. Let f and with zero 18 A = {w: ~ X(w) is absolutely continuous on (-00,00), with derivative belong:lng to L2 [T] for every compact interval T} A - {w: X(w) is equal a.e. (Lebesgue) to an element of A}. eq Then, and P(X-l[A ]) eq or 1. x-leA (2) (X ) is separable, then t l X- [A]€8, and P(X-l[A]) = 0 or 1. (3) P(X-l[Aeq ]) ... 1 eq ]€s, =0 (1) If separable~. 2 f(A)dA < 00 foo-A0 0' if and only if If is then this condition is also necessary and sufficient for P(X-l[A]) ... 1. Proof: A function is absolutely continuous on (-00,00) if and only if it is absolutely continuous on every compact interval. X-leA ]€8 eq ·e Let The fact that and has probability zero or one thus follows from Theorem 7. T be any fixed compact interval. gral operators in L [T] having kernels 2 R(t,s) J~oo ... Let and R.r and R(t,s) f(A) exp(iA(t-s» RO(t,s), defined by dA, 2 -1 -00 (A +1) exp(iA(t-s» ... be the inte- ROT OO f dA • Using a result of Hajek [17, Sec. 7], one can show that W trace-class if and only if the range of (R~OT) f:00 (A 2+l)f(A)dA consists of all elements of dt to an absolutely continuous function with concludes that P(X-l[A ]) ... 1 eq (n,8,p) c A, In [18] L [T] 2 L [T] it is shown that that are equal a.e. derivative. 2 if and only if To prove the statements regarding x-leA] 00. < for foo-00 A2f(A)dA < Hence, one 00. we first note that f: 2 ]. Hence, if = 00, then X-1 [A]€8, since oo A f(A)dA eq :I.s complete, and P(X-l[A]) = o. To complete the proof, it is suf- x-leA ficient to show that f: oo 2 A f(A)dA < 00 and (X ) t separable imply that almost 19 e (X ) t all paths of are absolutely continuous. This has been proved by Doob [11, pp. 535-537]. As noted in the proof of the theorem, this result is an extension in one (X ) has absolutely contint F the spectral distribution function of direction of a result of Doob, which states that uous paths if (X t ) , and f: (X t ) oo 2 A dF(A) < 00, How~ver, separable [11, pp. 535-537]. the result given here (X ) has a spectral density, and this t spectral density must be rational. The requirement that (X ) be Gaussian is t is restricted to the case where not assumed by Doob; the process need only be separable, measurable, meansquare continuous, and wide-sense stationary. Under these same assumptions, with the process also assumed to have a spectral density function, which is rational, the proof of Theorem 7 can be used to show that almost all sample -e path derivatives belong to L [T] for all compact intervals 2 co 2 f -00 A f(A)dA < 00; i.e., normality is not required. Theorem 8 is proved for zero-mean Gaussian processes. mean function m, A eq (or to have probability one is that each compact interval T, sufficient condition for oo 1~(A)12 A2dA < 00 where ~ [19], A, oo J -00 if separability of 2 A f(A)dA ffir(t) to be in where < = met), and 00 t€T, 1.range(R~OT) (X t ) is assumed) ~ ffir€range(R OT) for = 0, A ~(t) t4T. is that is the fourier transform of We discuss an application to information theory. (Nt)' t€T If the process has then application of Theorem 2 shows that the necessary and sufficient condition for J: T whenever Suppose that ffir' (St)' (compact interval), are measurable zero-mean Gaussian stochastic processes on a probability space (Q,e,p). We assume that (St) and (Nt) are statistically independent, and that almost all sample paths belong to for each process. Let denote the Gaussian measures respectively; e.g. ~S(A) = 20 p{w: S(w)eA}, ~ the sample path of S(w) be the Gaussian measure induced on mutual information (AMI) of (St) evaluated at B[L2xL2] and by (St+Nt) (St,St+Nt). if Let llS,S+N The average is defined as [20] = AMI (S,S+N) w. dllS,S+N (x,y) and equal to +00 otherwise. This quantity is of much interest in many Communication Theory problems, Where, it represents the average information about the "signal" noise" (St+Nt). AMI(S,S+N) ~ < 00 ·e obtained by observing "signal plus Hajek [17] has shown that for the case considered here, <~ KS = ~~G~~ for a trace-class operator G, where KS and denote the covariance operators of 2 above, one sees that of (St)' (St) AMI(S,S+N) < belong to the range of 00 (St) and (Nt). From this and Theorem if and only if almost all sample paths ~~. ?his result has been previously obtained by Pitcher [4]. 7 I ZERO-oNE LAWS FOR NON-GAUSS I N..J MEASURES Our results so far involve only Gaussian measures. Gaussian measure on that o or II and ll' B[B], and II v However, if is any probability measure on are mutually absolutely continuous, then obviously 1 for any subgroup G belonging to B [B]. II II B[B] is a such ll'(G) = In this section, we consider 21 a class of non-Gaussian measures for which this holds. ~ on the following lemma. Lemma 9 [21]: Suppose ~x @ ~ Let {(x,y): x+y€A} ~X+y measure v in B, ~2 and ~x belongs to a meas ure is absolutely are two probability measures on ~y (BxB, B[B]xB[B]). denote product measure on B[B]xB[B] ~X+y(A) on B[B] by ~l means that ~2' continuous with respect to B[B]. ~l« The notation Our results are based for all A€B[B]. = ~X0~y{(X,y):x+y Define a probability €A}, and for each fixed ~X+v (A) = llX{x: x+v EO A}. llX+v by The set We then have the following results: (1) If llX+Y « (2) If llX « (3) If llX ~ a.e. dlly(y), then ~X+y « llX+Y a.e. dlly(Y), then llX « ~x d~(y), llX+Y a.e. then We thus consider Gaussian measures defined as in the lemma, where ~y case where to imbed space H, K, then ~ llX+Y ~X+Y' and the class of measures need not be Gaussian. quence of this lemma is the following. has a covariance operator llX llX llX' If llX+Y An obvious conse- B is a Hilbert space, so that ~y[range(K~)] = 1. For the if ~X+y '" ~X llX B is not a Hilbert space, one can use the methods of Kuelbs [8] B as a dense measurable linear manifold in a separable Hilbert and work with the covariance operator of the extension of ~X. There are thus obvious extensions of the results of the preceding section. Two such extensions are given below. n = 1,2, ••• Theorem 9: Suppose {Zn}' n = 1,2, ••• and {V}, n are two families of real random variables defined on a probability space (Q,S,P). {V}. n Let Suppose that K(n,m) {Zn} is Gaussian, and that denote the covariance of denote the correlation of V n and V. m Let Z {Zn} and n {e} n functions. defined and continuous on the interval T. is independent of Z, m and let R(n,m) be any set of real-valued Define 0 e A' B' C' l.~ = {w: p p (z n (w)+Vn (w» = {w: l.~ (Z n (w)+Vn (w»p = {w: l.~ Izn (w)+Vn (w)I P where converges absolutely and uniformly on T} e n (t) ' < oo}, is a real scalar, p converges uniformly on T}, e n (t) p ~ 1. g = (gl,gZ' ••• ) be a sequence of real numbers such tha t gi > 0, all i, ,00 2 2 and li=l gi E[Zi (w)] < 00. Suppose that one of the following conditions is Let satisfied: (1) (l.i giVi(w)xi)z k, (Z) x€iZ' all for a fixed random variable k(w) li;j x.x.K(i,j) 1.' J a.e. dP(w); For some matrix operator R' = K"~SK'~ 1 l' by '- ~ (3) l. KZ(i,j) i~j K(i,i)K(j,j) ~ Kii such that where the matrix operators = gigjR(i,j) R'(i,j) (a) S in iZ 0, ~ Kjj < 0) and Kl(i,j) R(i,i) and ~ and K l R' < 00, are defined = gigjK(i,j). such that i,j (summation over 1 l.i Sii for all kK(i,i) i and some finite scalar k, (b) V i 2 (w) , li K( i,i ) < 00 a.e. dP(w). A', B'p' and C'p Then, zero, and P(C'p) = 1 Proof: Let (gi) induces from P each has probability one or 'r liLE(Zn 2 (w)] P/2 if and only if < .00. be defined as in the theorem, and set Then Y(w) gnZn(w). a, all belong to = (Y l (w),Y 2 (w), •.• ) ~Y a Gaussian measure Yn(w) = is in iZ a.e. dP(w), B[iZ]. on ~Y and hence has covariance oper- condition implies = gigjK(i,j). Consider condition (1) of the theorem. This that V'(w) = (glVl(w),g2VZ(w), .•• ) belongs to iZ a.e. dP(w) , V' ator Ky, Ky(i,j) and thus induces from P a probability measure ~' on Moreover, condition (1) is a necessary and sufficient condition that range(Ky~) duced on for almost all B[iZ] by Y+V'. w [~]. Since Let ~l k B[iZ]. V' (w) € be the probability measure in- ~'[range(K2y)] = 1, ~l ~~, by Lemma 9. 23 A', B' The assertions of the theorem regarding Theorem 6. For example, defining P(A' ) e "(t) n = gn-1 C' and p now follow from p e (t), n = () 'IN ~l { X€~2: l 1 x~ e~(t) converges uniformly on T} = 1 0 ~y { x€~z: l'IN converges uniformly on T } , because the latter quantity is e~(t) x n 0 or 1, ~l ~ ~. by Theorem 6, and Condition (2) implies condition (1) [18]. KyZ(i,j) Condition (3a) of the theorem is equivalent to 2 and ~ gi R(i,i) kKy(i,i), 2 gi R(i,i) ~ kKy(i,i) all i, some finite k. The assumption that V' (w)€lZ a.e. dP(w), implies that V'i(w) Kuelbs has shown [8] that the first part of (3a) implies that ~'y are mutually absolutely continuous, where B[lZ] trix Vi 2 (w) I i ---'-'---,K(i,i) for almost all w, ~'y is the Gaussian measure on but with diagonal covariance ma- R' , the range of First, V'(w) V'(w) To see this, we note that if 12) eigenvectors of R' , and hi ' eignevectors corresponding to the zero eigenvalues of R' , In Hence, I k j hi' hi' Vk'(w)Vj'(w)dP(w) k j V'(W)~hi'} all h.'. ~ . = 1, each hi" Noting that =0 for each so that hi'. range(R'). {w: V' (w)·frange(R')} is in the null space Suppose N of Q Q, Qii = O. and are {hi} are those then p{w: V'(w) is not orthogonal to hi'} thogonal to hi'}, we have that P{w:V'(w)€range(R')} :::> belongs to belongs almost surely to the closure of = gigjR(i,j). R'ij the complete orthonormal (in , [21] • We now show that if this sum is finite and if condition (3a) is satisfied, then k r,ange(Q) and [V'i(w)]2 = Ii ---~~ Ky(i,i) • range(Q2) a.e. dP(w). = 0, ~y ~ is that Q~ and K~Y have the same range space and Note that p{w: = giVi(w). A necessary condition for mutual absolute continuity Q, ~' y of ~Y' having the same mean element as < 1 Ii;j Ky(i,i)Ky(j,j) = Uh ,{w: V' (w) is not or- = Next we show that i 1. Then the element Ik,j e ij eikR'(j,k) = <Sij' since 24 R'(i,i) e ~ space of R' • i, and = Ky(i,i). Qii Hence range(R') c range(Q). e such that i Hence, e is in i v' (w)Erange(Q) (= range(~» for almost ,l; then xErange(Q) if Ii all w. We now note that if xEl and x Erange(Q), Z x 2 and only if I -!- <~. Hence, the assumption that QU i i, ,l; E range(Q ) a.e. dP(w). Ky~i,i) < ~ a.e. dP(w) Ii imply that \' KZ(i,j) < 1 li,j K(i,i)K(j,j) The assumption that = range(Ky,l;). range(Q,l;) R(i,i) ~ k!<y(i,i), all [V' (w)]2 and the assumption that that is in the null ei N span N , so that Q Q Q is contained in the null space of R' • This implies The elements the null space of that all k!<y(i,i), V'(w) implies Thus, the assumptions of (3) imply that V'(w) € range(Ky~) a.e. dP(w). Hence, assumption (3) implies assumption (1), and the proof is co~pleted. Remark: It is clear that Corollary: Suppose (X t ) and (Y t ) , .~ tET R(i,i) < ~ \' if li K(i,i) (a bounded interval), are (n,S,p). measurable stochastic processes on the complete probability space Suppose that (X ) t sample paths in and that (Y ) t (Y ) t that and L [T], 2 are independent, that both have almost all (X ) t is Gaussian with covariance operator has covariance operator R and mean element K, set of functions defined and continuous on {y} T. two sets of real scalars, and define, for any Bp " = {w: Let p m. and let be a set of complete orthonormal eigenfunctions for A" • n and Let K, {e } n {f} n {A} be any be any n € [l,~), IN1 A [(X(w)+Y(w),e \j+Y ] f (t) converges uniformly on T}, n n n n = {w: I~ on T}, A:«X(W)+Y(w),en)+yn)p fn(t) converges absolutely and uniformly and cp " = {w: I~l IA where (.,.) denotes the n I PI(x(w)+Y(w),e n )+Y n IP for all n < ~}, L inner product. Z and some finite scalar (Re ,e )+(m,en )2 ~ n n <Y(w),e n>2 k, and if I n <Ke ,e > < ~ If n a.e. dP(w), then A", B" p and C" p each belongs to S, n each has probability 25 one or zero, and p[e "] p =1 if and only if Proof: We note that (Ken,e m> = 0 if n f m; condition (3) of Theorem 7 is thus satisfied for the random variables <Y(w),e >2 \ _ n Ln <Ke ,e > < 00 a.e. dP and '(Ren,en ) ~ n n finite), are given in [ 21], where a number of examples Sufficient conditions for k ( Ken,e n ), all n (k are also presented. For instance, if tegra1 operator with kernel either Y(w) on T belongs to k range (K 2 ) T = [O,b], exp(-alt-sl), if and only if Y(w) to an absolutely continuous function with b < 00, and a 0, or > K is the inb-It-sl, then is equal a.e. (Lebesgue) L [T] 2 derivative. We also note that the corollary has straightforward extensions to processes whose sample paths belong (a.s.) to L p for any p~[l,oo). The following theorem gives additional results on the convergence of sequences and sums of random variables. Theorem 10. Let {Z }, n n = 1,2, •• , and {V}, n n = 1,2, ••• , or real random variables on a probability space (n) S, P). be two families Suppose Let {gn} be a set that {Z } is Gaussian, and that {Z n lis independent of {V}. n n g ZE [Z Z] < 00 L: g 2 < 00, and gi > 0 for i = 1,2, •• of real scalars such that I: nn n ' nn (1) T:te, fpllovitJ,S sets have IJro0P..bility zero A 1 = {w: or one: Z (w) converges to some real scalar} n AZ = {w: Zn (w) converges to k}, where k is a fixed real number. (2) Suppose that any of the conditions (1)-(3) of Theorem 9 is satisfied for {g } defined as above. n or one~ Then the following sets have probability zero 26 Z (w) + V (w) converges}, n n Z (w) + V (w) converges to k}. n n Moreover, P(B i ) =1 if and only if P(A ) i Define yew) by Yi(w) Pro0f: = g.Z.(w). ~ ~ convergent sequences of real numbers x = (xl' x , ••• ). Z I ([22], p. II xii = suplx n n Z Z Li(gixi)Z S (Lig )I Ixl I • Thus we can define i to-one operator G: ~ ~tz by (Gx)i = gixi. We ]J yew) range(G)}. £ o on B[-t.. Z]' Al assertion for i = 1,2. Let c denote the space of all space under the norm {w: = 1, 68). c is a separable Banach Suppose x £~. Then a bounded, linear, and onesee that {w: Z(w) £~} = Now we note that Y induces from P .a Gaussian measure = Y-1 [range(G)], and peAl) = ]J(range(G», proving the for AI. For the set A2 , we consider the space =0 of real sequences that are convergent to zero. We note that w £ A if and only if Zn(w) - k 2 ~o is a separable Banach space under the norM I Ixl I = sMPlxnl ~ o. ([22], p. 68). Defining the operator G: ~o ~t'2 by (Gx)i = gixi' one sees that G is bounded, linear, and one-to-one. Let yew) be defined by Yi(w) = gi(Zi(w) is in t 2 a. e. dP (w), thus induces a Gaussian measure on B[i 2] • Z (w) - k n ~ 0 if and only if yew) £ range (G). - k). yew) Moreover, The remainder of the proof is clear. We now consider B • l gi[Zi(w) + View)]. Theorem 9 Define yew) by Yi(w) = giZi(w) and Q(w) by Qi(w) = Under the assumption that one of the conditions (1) -(3) of . is satisfied, both yew) and Q(w) belong to t 2 for almost all w, and thus induce probability measures 1.ly and ]JQ on 9 also shows that of the previous result The assertion, on B 1 for AI. B~Z]. The proof of Theorem is an obvious consequence 27 The assertions on B follow 2 in the same way. As a corollary of this theorem, we note that if {Zn }, n is a family of Gaussian random variables, then n with probability one or zero. -1 n = 1,2, ••• i[Zt(w) - EZ i ] ~ 0 Similarly, an L~ Zi(w) converges with probability one or zero, and an L~ Zi(w) ~ k with probability one or zero, 0.. for any set of scalars {a }. n ACKNOWLEDGEMENT The author thanks S. Cambanis for helpful discussions on Theorem 5 • ...... .~.:. .,""'~, 28 10 J'PPENDIX PROOF OF LE~1J'AA 2 In this appendix, we supply the missing steps in the proof of Lemma 2, as outlined in the proof of Lemma 3 of [1]. Lemma 2 above; we suppose that I: =f H g(x)exp{2~ g The notation is as in the proof of L2[~]' € ai(x,ei)}d~(X) = 0 and that for any N and scalars (al, ••• ,aN). Now I: = I E[g(x)exp{2~ai(x,ei)} N IB[H ]] d~(x) H = I eXP{2~ai(x,ei)} H N The conditional expectation E[g(x)1 B[H ]] N from the definition of B[H ]. N can be written as E[g(x)1 B[H ] qN: x -- + «x,el),···,(x,eN» E[g(x)1 B[HN]] d~(x), = h[(x,el>' ••• '(x,~)] = h[qN(x)], where N and h is B[RN]/B[R] measurable, since B[H ] is generated by sets of the form {x: qN(x) € A}, A € N B[R ]. N Since qN is B[H]/B[R ] measurable, the usual change of variable formula gives I H h[qN(x)]exp{ r~ ai(x,e i )} d~(x) =I N h(rl,··,rN)exp{ R 2~airi}dP(rl,···,rN) N l where P is the Gaussian measure on B[R ] having covariance matrix K , K~j / >_ < > and mean vector \Kei,e j - 0ij Kej,e j 11 / ) m, mi _ - \m,e i • N family of probability measures {Q} on B[R ], defined by = We now have a a dQa(rl,.··,rN) where a = = C(a) (al' ••• '~) exp{ 2~airi} dP(rl, ••• ,rN) N is any element of R , and C(a) is a normalizing By Theorem 1, p. 132 of [23], the family {Qa} is complete; i.e., N N IN f(x)dQ (x) = 0, all a in R , implies f(x) = 0 a.e. dQ a (x), all a in R • R a factor. - 29 = ~{x: h[q~(x)]=O} = ~{x: h[(x,el), ••• ,(x,eN)]}~ 0, we see that h[(x,el), ••• ,(x,eN)] = 0 a.e. d~(x), or, E[g(x)IB[~]] = 0 a.e. d~(x). By the martingale convergence theorem ([7], pp. 84-85), exists and is equal to containing B[HN] E[g(x)IB[H]], for all N. E[g(x)IB [H]] = 0 a.e. d~(x), ~ -- Thus since B[H] E[g(x)IB[H]] and thus li~E[g(x)IB[tli]] is the smallest a-field = 0 a.e. d~(x), g(x) = 0 a.e. d~(x). 30 ;\PPENDIX 2 D PROOF OF LEM'1A 7 Let B be the separable Banach space consisting of range(T) and 3 the B norm. Define T : B1~ B by T x = Tx. Since range(T) is dense in 1 2 3 1 *-1 ~ B3' Tl exists (t 10 ], p.44). Let {xn}' n = 1,2, ••• be a dense subset of B • If there exists u in B * such that U(T x ) = 0 for all n, then l 3 1 n * [T ](x ) = 0 for all n; since {xn} is dense in B and T *-1 exists, this l n l l implies that u is the null element of B * • Hence, {Tx } is dense in B • 3 3 n For each xn' define an element F in B3* as follows. If II TXn 11 2 = 0, n Fn is the null element. II.Fn II = 1 and F (Tx) n n IITx n 11 2 • Let {a } be a set of real scalars such that a n n =0 while otherwise a n > 0, and Ln a n = 1. Define a norm 2 2 1 I 11 I lui 1 on 8 by 3 3 3 = Ln ahFn (u). This norm is obviously weaker than 2 2 2 the II • 11 2 norm, since I ITul 1 ~ sRP Fn (TU) ~ I ITul 1 2 • Let H3 denote 3 if -- = I ITxnl 1 2 = 0, the completion of B in the I 1.11 3 c B[H ], so that B[B ] = B 3 3 Consider T1*F. n II Txnl 1 2 Hence = 3 n II T *Fn II 1 3 norm. Kuelbs has shown [8] that B[B 3 ] B[H ]. 3 :; liT II since II Fn II = 1. Moreover, ITl*Fn(Xn )I ~ II Tl*Fn I I II xn 11 1 , so that IITII ~ IIT/Fnll. I IT1*Fnl I = I ITI I, each Fn • Since 8 1 is reflexive, there exists an B1 such that Ilyn 11 1 = 1 and T1*Fn (yn ) = I ITI I ([23], p.103). l Define G = I ITI l- T1*F ; IIG II = 1 and G (y ) = 1. Choose Ln in B *, 1 n n n n n n = 1,2, ••• such that IlLn II = 1 and Ln (xn ) = I Ixn I I. Let {Sn }, element y n in be a set of strictly positive real scalars such that an'- inner product (.'.)1 on 81 by L S L (u)L (v). n n n n (u,v)l = ~L L S n n =~. a G (u)G (v) n n n n Define + Let H be the completion of B under the norm obtained l l from this inner product. Using the results of [8], we conclude that 31 2 Consider T as a map from HI into H3 • For u in 8 , one has I ITlul 1 3 = l 1 2 LnanFn2(Tlu} S 21 ITI 1 <u,u)1. Thus Tl is a bounded linear operator from 81 (cHI) into H , and can be extended by continuity to a bounded linear operator 3 on HI. Denote this extension as T2 • We show that T maps Borel sets in HI into Borel sets in 2 range (T Z) is in B[H ], since TZ[H ] 3 l = T2 [NT! Z H3• Note that ] and T2 is a one-to-one bounded linear map of the separable Hilbert space NT! (with the HI inner product) into 2 H3 • Lemma 6 thus implies that TZ[NT~] is in B[H ]. 3 in HI. We have T2 [A] must belong to B[H3] • Finally, let A = TZ[A n NT ! 2 ]. ! Now let A be any Borel set ! ! Since AnNT is in B[NT ], T2 [AnNT ] J Z 2 2 j B[B ]. Then A e; B[fl ], and T [A] = T[A]. T[A] is thus I 2 l a Borel set in H , and since B[B ] = B nB[H ], this implies that T[A] € B[B 2 ]. 3 2 z 3 This concludes the proof. e; 32 REFERENCES [1] G. Ka11ianpur, Math. 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This research was supported in part by the Air Force Office of Sci- entific Research under Contract AFOSR-68-1415 and by the Office of Naval Research under Contract N00014-67-A-0321-0006 (NR042-69). ·e