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By P. Walters

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Example text

Closed) Let subspace of and satisfying L2(m) as + ~IIfll2. n ~ ®. and let containing Hf f denote the smallest and the constant functions UTH f c Hf. Ff is a subspace of f E L2(m) + ~ + (llfll2 + ~ ) ~ = {g ( L2(m): n (UTf,g) ~ (f,l)(l,g)} L2(m), is closed, contains f and the constant 44 functions (U f,g) Hence and is : 0 for If invariant n ~ 0 Ff = L2(m). Definition and so it contains (1,g) : 0 Hf. If and therefore g ~ Hf then Hf ¢ Ff. 5: T: X ~ X (T×T)(x,y) on UT is measure-preserving, = (T(x),T(y)).

T. e. this is true. (f,l) = 0 ~ O. Hence, all we need to show is implies 1 n-i ~- I(U~f,f) I2 i:O ~ 0. By the spectral theorem it suffices to show that if ous (non-atomic) measure on K ~f is a continu- then 1 nil if kid~f(k)12 i:O -~ 0. We have i n~l (I kid~f(k) %1 if id f( )12 : i:O I k_id~f(k)) ~ i=0 1 n-1 I - 1 nf 1 - ~i:O : II (k~)id(~f×~f)(k,~) (by Fubini's Theorem) K×K ;f(inl i) ~ K×K _id~f (~) ) ~- (k~) i=0 d(~f×Zf)(k,~). 48 If (k,T) is n o t in the diagonal of K×K n i=0 as n ~ -. e. the b o u n d e d investigate The (~f×~f)(diagonal) integrand convergence the m i x i n g : 0 has m o d u l u s theorem to o b t a i n properties of the is e r g o d i c iff all and t h e r e - ~ i, so that the result.

V(fg) = 56 Proof: Let B 2 { B2. We have V(X~2) and we see that 3 B 1 E B1 (Bl,ml) then V(×B2) such that by is induced by @ • 1 V(XB2) = XBI 0 as its only values. e. We define This is unambiguous so that IIV(XB2) @ = m2(B 2) = @: (B2,m2) since if m2(B2AA 2) = 0 V(XA2)II = 0. Clearly, V = (×~(B2),X~(B2)). # First note that since = 1 X@B 2 + X @ ( X ~ B 2 ) = 1 Also, = ~I(~B2 ). preserves complements V is norm-preserving teristic functions to characteristic ×#2 + X~2\#2 V -I. XB2XB2 dm 2 = (XB2,XB2 ) It remains to show that Since Thus is an isomorphism of measure algebras.

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