PMATH450_S2015.pdf

# C n consists of 2 n closed intervals of length 3 n

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C n consists of 2 n closed intervals of length 3 - n . The Cantor set C is defined as C = T 1 n =0 C n . C is closed and bounded, and therefore compact. Each endpoint of a Cantor interval is in C . Consider the ternary expansion x 2 [0 , 1]: x = P 1 i =1 a i 3 - i where a i 2 { 0 , 1 , 2 } . The Cantor set consists of all numbers whose ternary expansion consists of only 0s and 2s. (Step n removes all numbers with a 1 in the n th place after the decimal). This shows that the cantor set it uncountable. Every point in the Cantor set is an accumulation point, and the Cantor set is closed, so the Cantor set is perfect. Proof: Let x 2 C . Then x 2 C n 8 n . Take an endpoint y 6 = x such that | y - x | < 3 - n and note that y 2 C . This can be done for any n so each x 2 C is an accumulation point. The Cantor set has empty interior (totally disconnected): If I C was an interval of length δ , then pick n such that 3 - n < δ , so I 6✓ C n . m ( C ) = 0, since C C n and m ( C n ) = 2 n 3 - n ! 0. Alternatively, C n & C so by downward continuity of measure, m ( C ) = lim n !1 m ( C n ) = 0. All subsets of C are measurable, having outer measure 0, so the cardinality of Lebesgue measurable sets will be at least |P ( C ) | = |P ( R ) | , which is clearly also an upper bound. If instead of keeping outer thirds, we keep outer intervals of length r parent interval length, where r < 1 2 , we get a homeomorphic set of measure 0. If you vary the lengths in each step, you can get a Cantor-like set with any measure 2 (0 , 1).

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2 INTRODUCTION TO LEBESGUE INTEGRATION 17 Lecture 11: May 29 Theorem. If f is Riemann integrable over [ a, b ], then f is also Lebesgue integrable over [ a, b ], with R R b a f = R b a f . Proof. Suppose f is Riemann integrable. Then f is bounded. First, we show that f is measurable. Recall that f is Riemann integrable i m ( E ) = 0 where E is the set of discontinuities of f . Then f = f χ E | {z } g + f χ E C | {z } h . Notice that g = 0 a.e. , since m ( E ) = 0, and hence g is measurable. Now h is a continuous function on E C , so h is measurable. Thus f is measurable, so | f | is measurable, so R b a | f | is well-defined. Now since f is bounded, there exists M such that | f | M 8 x . By monotonicity, R b a | f | R b a M = M ( b - a ) < 1 . Take a partition P of [ a, b ]. Then U ( f, P ) = X M i ( x i - x i - 1 ) = R Z b a X M i χ [ x i - 1 ,x i ] = Z b a X M i χ [ x i - 1 ,x i ] Z b a f since on [ x i - 1 , x i ], f ( x ) M i . Similarly, L ( f, P ) = X m i ( x i - x i - 1 ) = R Z b a X m i χ [ x i - 1 ,x i ] = Z b a X m i χ [ x i - 1 ,x i ] Z b a f since f ( x ) m i for x 2 [ x i - 1 , x i ]. Thus we have U ( f, P ) R b a f L ( f, P ) for any partition. We also have U ( f, P ) R R b a f L ( f, P ). But inf U ( f, P ) = sup L ( f, P ) since f is Riemann integrable, so R R b a f = R b a f . 2.6 L p Spaces We will start with 1 p < 1 . Definition. Let f : E ! C , with E and f measurable. We define the p -norm as k f k L p ( E ) = k f k p = ✓Z E | f | p 1 /p .
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• Spring '12
• N.Spronk
• FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

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