hw4 sol_4

# hw4 sol_4 - R . By this method, it is possible to reduce...

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Tufts University Department of Mathematics Math 136 Homework 4 problem 4. 2008. 1. 2. 3. 4. (15 points) Let A R n be bounded and let g : A R be bounded. Let S 0 be a rectangle with sides parallel the axes that contains A . Let ǫ > 0 and let P be a partition of S 0 . Assume R P, x 1 R, x 2 R, | g ( x 1 ) - g ( x 2 ) | < ǫ . Prove that U ( g,P ) - L ( g,P ) ǫv ( S 0 ). (This proof is like one part of the proof of Lebesgue’s Theorem.) Extend g to be 0 on S 0 \ A . Given R P , we know x 1 ,x 2 R ( | g ( x 1 ) - g ( x 2 ) | < ǫ ). So, by Theorem 3 in HW4 HINTS, we have sup R g - inf R g ǫ . Thus U ( g,P ) - L ( g,P ) = s R P sup R g · v ( R ) - s R P inf R g · v ( R ) = s R P (sup R g - inf R g ) v ( R ) s R P ǫv ( R ) = ǫ s R P v ( R ) = ǫv ( S 0 ) . ( If one knows the epsilon condition above for all ǫ > 0 , then one can prove g is continuous. Then, since the closed, bounded rectangles R P are compact, we know by the Extreme Value Theorem that g attains its maximum and minimum on each
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Unformatted text preview: R . By this method, it is possible to reduce the diference to R P ( g ( x 1 ,R )-g ( x 2 ,R )) v ( R ) For some points x 1 ,R ,x 2 ,R R and not have to worry about suprema and inma.) I did not initially realize our deFnition required all rectangles in a partition to be closed. Some people asserted that the supremum of g on a set is actually a value of g and I subtracted a point for doing so. The reason for this is that you do not know g is continuous on the large rectangle S unless you know the epsilon condition for all &gt; 0 (and some set of rectangles). 1...
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