continuous function can map a set with outer measure zero onto a set with outer

# Continuous function can map a set with outer measure

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continuous function can map a set with outer measure zero onto a set with outer measure one. Solution: Choose E = C to be the Cantor set contained in [0 , 1] and let f ( x ) be the continuous Cantor Lebesgue function de fi ned on [0 , 1] (see book p. 35) . We know that when restricted to C, f ( x ) is still a continuous function on C. Moreover, one can easily see that f ( C ) = f ([0 , 1]) = [0 , 1] (for example, we have f ¡£ 1 3 , 2 3 ¤¢ = f ¡© 1 3 , 2 3 ª¢ , etc.). ¤ 3. (10 points) Let E 1 and E 2 be two subsets of R n such that E 1 E 2 and E 2 E 1 is countable. Show that | E 1 | e = | E 2 | e . Solution: Clearly we have | E 1 | e | E 2 | e . Also | E 2 | e | E 1 | e + | E 2 E 1 | e = | E 1 | e which implies | E 1 | e = | E 2 | e . ¤ 4. (10 points) Find a continuous function f ( x ) de fi ned on [0 , 1] such that f ( x ) is di ff erentiable on a subset E [0 , 1] with | E | e = 1 and f 0 ( x ) = 0 for all x E, but f ( x ) is not a constant function. 1
Solution : Let f ( x ) be the Cantor Lebesgue function de fi ned on [0 , 1] as given in p. 35 of the book. We know f ( x ) is di ff erentiable on the open set O = O 1 O 2 O 3 · · · , where O 1 = μ 1 3 , 2 3 , O 2 = μ 1 9 , 2 9 μ 7 9 , 8 9 , O 3 = μ 1 27 , 2 27 · · · , O 4 = · · · The total length of these open intervals is given by 1 3 " 1 + 2 3 + μ 2 3 2 + μ 2 3 3 + · · · # = 1 . ¤ The example below shows you how to obtain a set which is measurable, but not Borel measurable.