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Unformatted text preview: ASSIGNMENT 8 SOLUTIONS
MAT 473 SPRING 2012 Exercise 14.1. For A Rn and x Rn , prove that A + x L if and only if A L, in which case (A + x) = (A). Proof. For any x Rn , translation by x is continuous on Rn with continuous inverse (namely, translation by x). It follows that a set G is open if and only if G + x is open. Also, recall from Exercise 12.2 that (G + x) = (G) for all open G Rn . Now consider any set A Rn . Since a set G is open with A + x G if and only if G  x is open with A G  x, it follows that (A + x) = inf{(G)  G is open and A + x G} = inf{(G)  G  x is open and A G  x} = inf{(H)  H is open and A H} = (A). = inf{(G  x)  G  x is open and A G  x} Since a set K is compact if and only if K +x is compact, it follows from this that (K +x) = (K) for all compact K. Arguing as above, for any A Rn we now have (A + x) = sup{(K)  K is compact and K A + x} = sup{(K)  K  x is compact and K  x A} = sup{(L)  L is compact and L A} = (A). Now it follows that M + x L0 (M + x) = (M + x) < (M ) = (M ) < M L0 , = sup{(K  x)  K  x is compact and K  x A} in which case (M + x) = (M + x) = (M ) = (M ). Date: March 1, 2012. S. Kaliszewski, School of Mathematical and Statistical Sciences, Arizona State University.
1 Finally, for any A Rn and any M we have (A + x) M = (A (M  x)) + x, so A + x L (A + x) M L0 for all M L0 (A (M  x)) + x L0 for all M L0 (A (M  x)) L0 for all M L0 (A (M  x)) L0 for all M  x L0 A N L0 for all N L0 A L, (A + x) = sup{((A + x) M )  M L0 } in which case = sup{((A (M  x)) + x)  M L0 } = sup{(A (M  x))  M L0 } = sup{(A N )  N L0 } = (A). = sup{(A (M  x))  M  x L0 } Remark. An alternative to the measurability argument given here would be to use the Carathodory Theorem, once we know outer measure is translationinvariant: if A L, then e for any E Rn , (E (A + x)) + (E (A + x)c ) = (E (A + x)) + (E (Ac + x)) = ((E (A + x))  x) + ((E (Ac + x))  x) = ((E  x) A) + ((E  x) Ac ) = (E  x) = (E), so A + x L. (B) = (B \ A) + (A). Exercise 15.1. Prove that for every measurable set B Rn and every A B, Proof. For any open set G which contains A, we have (B G) (G) and (B \ G) = (B \ G) (B \ A) by monotonicity. Thus, Taking the inf over all such G gives (B) (A) + (B \ A). For the reverse inequality, let K be a compact set contained in B \ A. Then A B \ K, so (A) (B \ K) = (B \ K) by monotonicity. Thus Taking the sup over all such K gives (A) + (B \ A) (B). (A) + (K) (B \ K) + (K) = (B). (B) = (B G) + (B \ G) (G) + (B \ A). 2 ...
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 Spring '09

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