Ex4-21

**Chapter 4 Question 21**

Question: Show that the necessity part of Lusin's theorem is not true for

# ε
0

{\displaystyle \varepsilon = 0}

, that is, find a measurable set

E

{\displaystyle E}

 and a finite measurable function

f

{\displaystyle f}

 on

E

{\displaystyle E}

 such that

f

{\displaystyle f}

 is not continuous relative to

E
−
Z

{\displaystyle E-Z}

 for any

Z

{\displaystyle Z}

 with

Z

=
0

{\displaystyle |Z|=0}

.

Solution: We first construct a type of Cantor set but with positive measure (See Chapter 3, Exercise 5, with

# δ
1

/

2

{\displaystyle \delta=1/2}

). On the unit interval

# I
[
0
,
1
]

{\displaystyle I = [0, 1]}

 we cancel the open interval of length

1

2
⋅
3

{\displaystyle \dfrac1{2\cdot 3}}

 on the center, hence become

D
1

=
[
0
,
5

/

12
]
∪
[
7

/

12
,
1
]

{\displaystyle D\_1=[0,5/12]\cup [7/12,1]}

. On the remaining 2 subintervals, we cancel open intervals of length

1

2
⋅

3
2

{\displaystyle \dfrac1{2\cdot 3^2}}

 from their center, then get

D
2

{\displaystyle 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

}

{\displaystyle \{D\_n\}}

, each

D
n

{\displaystyle D\_n }

 is a disjoint union of

2
n

{\displaystyle 2^n}

closed intervals. Define

D
(
1

# n
1

∞

D
n

{\displaystyle D(1)=\bigcap\_{n=1}^\infty D\_n}

, it has been proved (in Chapter 3, Exercise 5) that

D
(
1
)

{\displaystyle D(1)}

 has measure

1
−

1
2

=

1
2

{\displaystyle 1-\frac12=\frac12}

. On each subinterval of complement of

D
n

{\displaystyle D\_n }

, we construct

D
(
1
)

{\displaystyle D(1)}

 in respective proportion, so as to fill in the gaps of

D
(
1
)

{\displaystyle D(1)}

. After taking union with these newly constructed sets, we call it

D
(
2
)

{\displaystyle D(2)}

. Note that

D
(
2
)

{\displaystyle 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
)

{\displaystyle D(2)}

, fill it up with the same method again, we get a measurable set

D
(
3
)

{\displaystyle D(3)}

, and so on. The collection

{
D
(
n
)

}

# n
1

∞

{\displaystyle \{D(n)\}\_{n=1}^\infty}

 is increasing, so we can define

# n
1

∞

D
(
n
)

{\displaystyle 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
]
⊂
[
0
,
1
]

{\displaystyle J=[a,b]\subset [0,1]}

, both

J
∩
E

{\displaystyle J\cap E}

 and

J
−
E

{\displaystyle J-E}

 has positive measure.

Now let

f
:
I
→

R

{\displaystyle f:I \to \R}

 be defined as the characteristic function of

E

{\displaystyle E}

, that is

f
(
x
# )

{

1

 if
x
∈
E
,

0

 if
x
∈
I
−
E
.

{\displaystyle f(x)=\begin{cases}1&\text{ if }x\in E,\\
0&\text{ if }x\in I-E.\end{cases}}

Thus,

f

{\displaystyle f}

 is a measurable function. We also note that

f

{\displaystyle f}

 is not constant on each subinterval

[
a
,
b
]

{\displaystyle [a, b]}

 of

I

{\displaystyle I}

. Indeed, we know

[
a
,
b
]
∩
E

{\displaystyle [a,b]\cap E}

 and

[
a
,
b
]
−
E

{\displaystyle [a,b]-E}

 both has positive measure, hence there exists some points

p
∈
[
a
,
b
]
∩
E

{\displaystyle p\in [a,b]\cap E}

 and

q
∈
[
a
,
b
]
−
E

{\displaystyle q\in [a,b]-E}

, hence

f
(
p
# )
1
,
f
(
q
# )
0

{\displaystyle f(p)=1, f(q)=0 }

.
We claim that for any

Z
⊂
I

{\displaystyle Z\subset I}

 with

Z

=
0

{\displaystyle |Z|=0}

,

f

{\displaystyle f}

 is *not* continuous on

I
−
Z

{\displaystyle I-Z}

. Assume the contrary that

f

{\displaystyle f}

 is continuous on

I
−
Z

{\displaystyle I-Z}

 for some set

Z

{\displaystyle Z}

 of measure zero, then since

I
−
Z

{\displaystyle I-Z}

 has measure 1, it is nonempty. We can choose

x
∈
I
−
Z

{\displaystyle x\in I-Z}

 then

f

{\displaystyle f}

 is continuous at

x

{\displaystyle x}

. Using

ε
−
δ

{\displaystyle \varepsilon-\delta}

 definition of continuity, choosing

# ε
1

/

2

{\displaystyle \varepsilon=1/2}

, there exist

δ
>
0

{\displaystyle \delta > 0}

 such that

f

{\displaystyle f}

 is constant on

(
x
−
2
δ
,
x
2
δ
)
∩
(
I
−
Z
)

{\displaystyle (x-2\delta,x+2\delta)\cap (I-Z)}

. Note that

(
[
x
−
δ
,
x
δ
]
∩
E
)
−
Z

{\displaystyle ([x-\delta,x+\delta]\cap E) -Z}

 and

(
[
x
−
δ
,
x
δ
]
−
E
)
−
Z

{\displaystyle ([x-\delta,x+\delta]-E)-Z}

 both has positive measure, they are nonempty, hence there are

p
,
q
∈
(
x
−
2
δ
,
x
2
δ
)
∩
(
I
−
Z
)

{\displaystyle p,q\in(x-2\delta,x+2\delta)\cap (I-Z) }

 such that

f
(
p
# )
1
,
f
(
q
# )
0

{\displaystyle f(p)=1, f(q)=0 }

, a contradiction.