By Tom Lindstrom

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11 N large enough, any element u tells in F- {x } by a strict 1-walk of length less p is the multiplicity of E, we get P^{w: Tu(u))>N}0-(k/p)N By induction P"{GJ: for all Tu(a))>mN}<(l-(k/p)N)m m£ IN and from this the lemma follows easily. 4 Corollary. ,p ) is continuous on y. Proof: Given P vx, w y For each =p x,y6F iI N s and n€ IN, define B i i (t x v (w),u))=y i

V y But then . , F. x V^'V as a fixed point. If . , then " ',3n x x=cp. "" for ^ ' ' ##:3n belongs to four 2n-cells and also . (y) 3 also belongs to two n-cells x . C; just let F. D F K, • • , K F. , 11 • •# l i V •*' n, T "'n . Repeating the argument, we can make x an element of as many N-cells as we wish by just choosing N large enough. 3. 1 that denotes topological closure). For each closed ball with center z ECV (as usual, the bar z€F, let and radius 1. Since B z z€V, be the the 38 TOM LINDSTR0M intersection B flV must h a v e p o s i t i v e Vol (B flV) z V o l (B ) volume, and hence > j^ K z for some positive integer belongs to and let If B z.

1 , t. l+l s i' s i+i 7i. ,s leaves or enters H at i 1 n otherwise. All that remains is the following simple calculation, where t=(t , t , . . , t ), s=(s, ,s_, . ,s ) 1 2 n 1 2 n and C(t) is the set of critical indices: -i(i *x,y lit s =y t€C0 ^x,y' . , , s ) = n 2 t • , t . z - 7 1 1+1 t . _,_l' >o, 1+1 if and only if A 7i =7t ^ for an odd number of indices i. ,s t. , t . , . l l+l l l+l To prove that p is continuous, we shall need the following simple lemma. 3 Lemma. ,p )6j) of basic transition probabilities is used in the construction of Bi.