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 $$2^n$$closed 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.