Ex4-14

**Solution to Chapter 4, Exercise 14**

*Question: Let

f
(
x
,
y
)

{\displaystyle f(x, y)}

be as in Exercise 13. Show that given*

ϵ
>
0

{\displaystyle \epsilon > 0}

, there exists a closed

F
⊂
E

{\displaystyle F\subset E}

with

E
−
F

<
ϵ

{\displaystyle |E-F|<\epsilon}

such that

f
(
x
,
y
)

{\displaystyle f(x, y)}

converges uniformly for

x
∈
F

{\displaystyle x\in F}

to

f
(
x
)

{\displaystyle f(x)}

as

y
→
0

{\displaystyle y\to 0}

. (Follow the proof of Egorov's theorem, using the sets

E

ϵ
,
1

/

m

{\displaystyle E\_{\epsilon,1/m}}

in Exercise 13 in place of the sets

E

m

{\displaystyle E\_{m}}

in the proof of Lemma 4.18.)

Solution: Suppose *f
(
x
,
y
)

{\displaystyle f(x, y)}*is defined and continuous in the square

0
≤
x
≤
1
,
0
<
y
≤
1

{\displaystyle 0\leq x \leq 1, 0<y\leq 1}

and that

f
(
x
# )

lim

y
→
0

f
(
x
,
y
)

{\displaystyle f(x)=\lim\_{y\to 0}f(x,y)}

exists and is finite for

x

{\displaystyle x}

in a measurable subset

E

{\displaystyle E}

of

[
0
,
1
]

{\displaystyle [0,1]}

.

We first prove a lemma.

*Lemma: For any

ϵ
>
0
,
η
>
0

{\displaystyle \epsilon>0,\eta>0}

, there exists a closed subset

F

{\displaystyle F}

of

E

{\displaystyle E}

and integer

K
>
1

{\displaystyle K>1}

such that

E
−
F

<
η

{\displaystyle |E-F|<\eta}

 and*

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

{\displaystyle |f(x,y)-f(x)|\leq \epsilon}*

*for

x
∈
F

{\displaystyle x\in F}

 for all

y
<
1

/

K

{\displaystyle y<1/K}

.*

*Proof.* Given *ϵ
>
0
,
η
>
0

{\displaystyle \epsilon>0,\eta>0}*. Since

f
(
x
# )

lim

y
→
0

f
(
x
,
y
)

{\displaystyle f(x)=\lim\_{y\to 0}f(x,y)}

exists for

x

{\displaystyle x}

in *E

{\displaystyle E}*, for each *x
∈
E

{\displaystyle x \in E}*, there exists

δ
>
0

{\displaystyle \delta > 0}

such that if

y
<
δ

{\displaystyle y<\delta}

, then

f
(
x
,
y
)
−
f
(
x
)

<
ϵ

{\displaystyle |f(x,y)-f(x)|< \epsilon}*

By Question 13, for any integer

m
>
1

{\displaystyle m > 1}

, the set

E

ϵ
,
1

/

m

=

{

x
∈
E

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

for all

y
<

1
m

}

{\displaystyle E\_{\epsilon,1/m}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m}\right\}}

is measurable. We claim that

⋃

m

E

ϵ
,
1

/

m

=
E

{\displaystyle \bigcup\_{m}E\_{\epsilon,1/m}=E}

By definition, each

E

ϵ
,
1

/

m

{\displaystyle E\_{\epsilon,1/m}}

is a subset of

E

{\displaystyle E}

. Now for any *x
∈
E

{\displaystyle x \in E}*, there exists

δ
>
0

{\displaystyle \delta > 0}

such that

f
(
x
,
y
)
−
f
(
x
)

<
ϵ

{\displaystyle |f(x,y)-f(x)|< \epsilon}*

for

y
<
δ

{\displaystyle y<\delta}

.

Choose

m
′

>
1

{\displaystyle m'>1}

so that

1

/

m
′

<
δ

{\displaystyle 1/m'<\delta}

, then

x

{\displaystyle x}

is in the set

E

ϵ
,
1

/

m
′

=

{

x
∈
E

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

for all

y
<

1

m
′

}

{\displaystyle E\_{\epsilon,1/m'}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m'}\right\}}

so

x
∈

⋃

m

E

ϵ
,
1

/

m

{\displaystyle x\in \bigcup\_{m}E\_{\epsilon,1/m}}

and hence

E
⊆

⋃

m

E

ϵ
,
1

/

m

{\displaystyle E \subseteq \bigcup\_{m}E\_{\epsilon,1/m}}

.

Thus,

⋃

m

E

ϵ
,
1

/

m

=
E

{\displaystyle \bigcup\_{m}E\_{\epsilon,1/m}=E}

. We also have

E

ϵ
,
1

/

m

=

{

x
∈
E

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

for all

y
<

1
m

}

{\displaystyle E\_{\epsilon,1/m}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m}\right\}}

⊆

E

ϵ
,
1

/

(
m
1
)

=

{

x
∈
E

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

for all

y
<

1

m
1

}

{\displaystyle \subseteq E\_{\epsilon,1/(m+1)}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m+1}\right\}}

since if

x

{\displaystyle x}

already satisfy

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

for all

y
<
1

/

m

{\displaystyle |f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< 1/m}

, then

x

{\displaystyle x}

will also satisfy

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

for all

y
<
1

/

(
m
1
)

{\displaystyle |f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< 1/(m+1)}

.

This means that

E

ϵ
,
1

/

m

↗
E

{\displaystyle E\_{\epsilon,1/m}\nearrow E}

. Then since

E

ϵ
,
1

/

m

{\displaystyle E\_{\epsilon,1/m}}

are measurable, by Theorem 3.26,

lim

m
→
∞

E

ϵ
,
1

/

m

=

E

Since

E

ϵ
,
1

/

m

<

E

<
1
<
∞

, for any

η
>
0

{\displaystyle \eta>0}

, there exists

m
0

{\displaystyle m\_0}

such that

E
−

E

ϵ
,
1

/

m
0

=

E

−

E

ϵ
,
1

/

m
0

<

η
2

We know there exists a closed subset

F

{\displaystyle F}

of

E

ϵ
,
1

/

m
0

{\displaystyle E\_{\epsilon,1/m\_0}}

such that

E

ϵ
,
1

/

m
0

−
F

<

η
2

{\displaystyle |E\_{\epsilon,1/m\_0}-F|<\frac{\eta}{2}}

by Lemma 3.22 since

E

ϵ
,
1

/

m
0

{\displaystyle E\_{\epsilon,1/m\_0}}

is measurable. Then

E
−
# F
(
E
−

E

ϵ
,
1

/

m
0

)
∪
(

E

ϵ
,
1

/

m
0

−
F
)

{\displaystyle E-F=(E-E\_{\epsilon,1/m\_0})\cup(E\_{\epsilon,1/m\_0}-F)}

E
−
F

<

η
2

η
2

=
η

{\displaystyle |E-F|<\frac{\eta}{2}+\frac{\eta}{2}=\eta}

Also, in

F

{\displaystyle F}

,

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

for all

y
<

1

m
0

{\displaystyle |f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m\_0}}

Therefore, we have shown that for any

ϵ
>
0
,
η
>
0

{\displaystyle \epsilon>0,\eta>0}

, there exists a closed subset

F

{\displaystyle F}

of

E

{\displaystyle E}

and integer

K
>
1

{\displaystyle K>1}

such that

E
−
F

<
η

{\displaystyle |E-F|<\eta}

 and

f
(
x
,
y
)
−
f
(
x
)

≤
ϵ

{\displaystyle |f(x,y)-f(x)|\leq \epsilon}

for

x
∈
F

{\displaystyle x\in F}

for all

y
<
1

/

K

{\displaystyle y<1/K}

.

Now we proceed to prove the statement of the question.

Given

ϵ
>
0

{\displaystyle \epsilon > 0}

, choose closed sets

F
n

⊂
E

{\displaystyle F\_n\subset E}

, and integers

K
n

>
1

{\displaystyle K\_n>1}

for each

n
∈

N

{\displaystyle n \in \N}

so that

E
−

F
n

<

ϵ

2
n

{\displaystyle |E-F\_n|<\frac{\epsilon}{2^n}}

and

f
(
x
,
y
)
−
f
(
x
)

≤

1
n

{\displaystyle |f(x,y)-f(x)|\leq \frac{1}{n}}

for all

y
<
1

/

K
n

{\displaystyle y<1/K\_n}

and

x
∈

F
n

{\displaystyle x\in F\_n}

.

Define

# F

⋂
n

F
n

{\displaystyle F=\bigcap\_n F\_n}

. Then

F

{\displaystyle F}

is a closed set and

F
⊂

F
n

{\displaystyle F\subset F\_n}

for all

n

{\displaystyle n}

.

Now we want to show that

f
(
x
,
y
)

{\displaystyle f(x, y)}

converges uniformly to

f
(
x
)

{\displaystyle f(x)}

as

y
→
0

{\displaystyle y\to 0}

in

F

{\displaystyle F}

. For any

ϵ
′

>
0

{\displaystyle \epsilon'>0}

, choose

n
′

{\displaystyle n'}

so that

1

/

n
′

<

ϵ
′

{\displaystyle 1/n'<\epsilon'}

. Then take

# δ
1

/

K

n
′

{\displaystyle \delta=1/K\_{n'}}

. For all

x
∈
F

{\displaystyle x\in F}

,

x

{\displaystyle x}

is in

F

n
′

{\displaystyle F\_{n'}}

. Then if

y
<
δ

{\displaystyle y<\delta}

, we have

f
(
x
,
y
)
−
f
(
x
)

≤

1

n
′

<

ϵ
′

{\displaystyle |f(x,y)-f(x)|\leq \frac{1}{n'}<\epsilon'}

for all

x
∈
F

{\displaystyle x\in F}

.

This shows that

f
(
x
,
y
)

{\displaystyle f(x, y)}

converges uniformly to

f
(
x
)

{\displaystyle f(x)}

as

y
→
0

{\displaystyle y\to 0}

in

F

{\displaystyle F}

. Also,

E
−
# F
E
−

⋂

# n
1

∞

F
n

=

⋃

# n
1

∞

(
E
−

F
n

)

{\displaystyle E-F=E-\bigcap\_{n=1}^\infty F\_n=\bigcup\_{n=1}^\infty(E-F\_n)}

and so

E
−
F

≤

E
−

F
1

E
−

F
2

E
−

F
3

.
.
.

<

ϵ
2

ϵ
4

ϵ
8

.
.
.

=
ϵ

{\displaystyle \begin{align}
&<\frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{8}+...\\
&=\epsilon
\end{align}}