By W. B. Jones

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**Extra resources for Analytic theory of continued fractions Proc. Loen**

**Sample text**

A) We have {x} = n:=i (x - (l/n), x], and therefore all singletons are in F. But then all sets are in F since Q is countable. (b) Since Q is a countable union of singletons, µis u-finite on F. But every nonempty set in F 0 has infinite measure, so µ is not u-finite on F 0 • (c) If A is any finite nonempty subset of n, then µ(A 6. F 0 must contain infinitely many points not in A. F. (A) =µ(A)= oo, A e F 0 (except for A= 0). I Problems 1. F relative to µ. F, B :::>A}. Fµ, show that µ 0 (A) = µ 0 (A) =µ(A).

3(a) (1) and since µ*(H 1 u H 2Y +µ*(Hi n H 2 Y::;; µ*(Hie)+ µ*(H/). 3(b), and the sum of the right sides is 2. Thus the sum of the left sides is 2 as well. If a= µ*(H1 u H 2 ) + µ*(H1 u H 2 Y, b = µ*(H, n H 2 ) + µ*(H 1 n H 2 Y, then a + b = 2, hence a ::;; I or b ::;; I. If a ::;; I, then a I, so b = I also. Consequently H 1 u H 2 E :If and H 1 n H 2 E :If. We have therefore shown that :If is a field. Now equality holds in (I), for if not, the sum of the left sides of (I) and (2) would be less than the sum of the right sides, a contradiction.

N}. 6b,01 ... 6bnOnF(x1, ... ' Xn), where + '" +(-l)"Fn; F(c1o ... 6bnonF(x,, ... ,xn)=Fo-F1 +F2-F3 F1 is the sum of all ('l) terms of the form for exactly i integers in {I, 2, ... , n}, and ck= bk for the remaining n - i integers. PROOF. 7. 4 LEBESGUE-STIELTJES MEASURES AND DISTRIBUTION FUNCTIONS 29 We know that a distribution function on R determines a corresponding Lebesgue-Stieltjes measure. This is true in n dimensions if we change the definition of increasing. Let F: Rn-+ R, and, for a :s; b, let F(a, b] denote 6b,a 1 ''' 6b"a" F(Xto · · ·, Xn).