Ex4-21

Chapter 4 Question 21 
Question: Show that the necessity part of Lusin's theorem is not true for \varepsilon=0, that is, find a measurable set E and a finite measurable function f on E such that f is not continuous relative to E-Z for any Z with |Z|=0. 
Solution: We first construct a type of Cantor set but with positive measure (See Chapter 3, Exercise 5, with \delta=1/2). On the unit interval I=[0,1] we cancel the open interval of length \dfrac1{2\cdot 3} on the center, hence become D_1=[0,5/12]\cup [7/12,1]. On the remaining 2 subintervals, we cancel open intervals of length \dfrac1{2\cdot 3^2} from their center, then get D_2 which is a union of 4 disjoint closed intervals. Then we keep doing this to get a decreasing sequence of closed sets \{D_n\}, each D_n is a disjoint union of 2^nclosed intervals. Define D(1)=\bigcap_{n=1}^\infty D_n, it has been proved (in Chapter 3, Exercise 5) that D(1) has measure 1-\frac12=\frac12. On each subinterval of complement of D_n, we construct D(1) in respective proportion, so as to fill in the gaps of D(1). After taking union with these newly constructed sets, we call it D(2). Note that D(2) is a countable union of measurable sets, it is also measurable. There are still gaps existing in each smaller gap of D(2), fill it up with the same method again, we get a measurable set D(3), and so on. The collection \{D(n)\}_{n=1}^\infty is increasing, so we can define E=\bigcup_{n=1}^\infty D(n) which is also measurable. It has been proved (in Chapter 3, Exercise 25) that given any subinterval J=[a,b]\subset [0,1], both J\cap E and J-E has positive measure. 
Now let f:I\to \mathbb R be defined as the characteristic function of E, that is 
f(x)=\begin{cases}1&\text{ if }x\in E,\\
0&\text{ if }x\in I-E.\end{cases}Thus, f is a measurable function. We also note that f is not constant on each subinterval [a,b] of I. Indeed, we know [a,b]\cap E and [a,b]-E both has positive measure, hence there exists some points p\in [a,b]\cap E and q\in [a,b]-E, hence f(p)=1, f(q)=0. 
We claim that for any Z\subset I with |Z|=0, f is not continuous on I-Z. Assume the contrary that f is continuous on I-Z for some set Z of measure zero, then since I-Z has measure 1, it is nonempty. We can choose x\in I-Z then f is continuous at x. Using \varepsilon-\delta definition of continuity, choosing \varepsilon=1/2, there exist \delta>0 such that f is constant on (x-2\delta,x+2\delta)\cap (I-Z). Note that ([x-\delta,x+\delta]\cap E) -Z and ([x-\delta,x+\delta]-E)-Z both has positive measure, they are nonempty, hence there are p,q\in(x-2\delta,x+2\delta)\cap (I-Z) such that f(p)=1, f(q)=0 , a contradiction.