By Munkres J.R.

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**Example text**

Ai) < 00. /2 i ;::: > O. (Bij). *(Ai) + f. (A) and if f.! (A). Proof: The first statement is clear. Suppose f.! °(A). 0 is finitely additive on 5. (Ai) so 42 ClfAPTER 2. MEASURE TlfEORY BE S with An B Proof: Let < assume that f 00. + t. fI(A,) < Let. E 0 ana set C = AU R. l1y Theorem 2 we may > O. There exists <::; S such that Ai ::;> C and Then {Ai n A} [{ Ai n B}] covers A [B] so since S is an algebra i=l /1*(A) + /1*(B) S I: (/1(A; n f /1(A; n + /1C4, n Bl) i=-l S £=1 /1(A,) < /1*(C) +t Hence, fI*(A) + It*(B) S I,*(e) and Theorem 2 the reverse im>quality.

We show that translation invariance along with regularity characterizes Lebesgue measure. 5. LEBESGUE MEASURE 53 Theorem 6 If 11 is a translation invariant regular measure on B(Rn), then 11 for some positive constant c. = cm Proof: Let 1= [0, 1] x .. x [0, 1] be the unit "cube" in R n and set c 11(1). I is the pairwise disjoint union of 2nk bricks of side length 2- k for any kEN, and since by translation invariance each of these bricks has the same Il-measure, for any brick B with side length 2- k so Il(B) = cm(B) for any such brick B.

Proof: Partition N into a pairwise disjoint sequence of infinite sets {Kj}~l' By the observation above, ItJl( U E j ) - t 0 as i - t 00. So 3i such that ItJl( U E j ) < 1/2. jEK~ jEK! Let N j = Kl and nl = inf N J • Now partition NJ \{nd into a pairwise disjoint sequence of infinite sets {Kj}~J' As before 3i such that ItJl( U E j ) < 1/22. Let jEK: N2 Kl and n2 = inf N 2. Note n2 > nJ and N2 ~ N j • Continuing produces a subsequence nj 1 00 and a sequence of infinite subsets of N, {Nj }, such that N j +1 ~ N j and ItJI( U Ed < 1/2j .

### Analysis of manifolds by Munkres J.R.

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