Ex4-13

**Chapter 4, Question 13**

Question: One difficulty encountered in trying to extend the proof of Egorov’s theorem to the continuous parameter case

f
y

(
x
)
→
f
(
x
)

{\displaystyle f\_y(x)\to f(x)}

 as

y
→

y
0

{\displaystyle y\to y\_0}

 is showing that the analogues of the sets

E

m

{\displaystyle E\_{m}}

 in Lemma 4.18 are measurable. This difficulty can often be overcome in individual cases. Suppose, for example, that

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]}

. Show that if

ε

{\displaystyle \varepsilon}

 and

δ

{\displaystyle \delta}

 satisfy

0
<
ε
,
δ
<
1

{\displaystyle 0<\varepsilon,\delta<1}

, the set

E

ε
,
δ

=
{
x
∈
E
:

f
(
x
,
y
)
−
f
(
x
)

≤
ε
 for all
y
<
δ
}

{\displaystyle E\_{\varepsilon,\delta}=\{x\in E: |f(x,y)-f(x)|\leq \varepsilon \text{ for all }y<\delta\}}

 is measurable.

Proof: Fix

ε
,
δ
∈
(
0
,
1
)

{\displaystyle \varepsilon,\delta\in(0,1)}

, let

{

y
k

}

# k
1

∞

{\displaystyle \{y\_k\}\_{k=1}^\infty}

 be a dense subset of

(
0
,
δ
)

{\displaystyle (0, \delta)}

 (for example, take

{

y
k

}

# k
1

∞

=

Q

∩
(
0
,
δ
)

{\displaystyle \{y\_k\}\_{k=1}^\infty = \mathbb Q\cap (0,\delta)}

). Define

E
k

=
{
x
∈
E
:

f
(
x
,

y
k

)
−
f
(
x
)

≤
ε
}

{\displaystyle E\_k=\{x\in E: |f(x,y\_k)-f(x)|\leq \varepsilon\}}

. We claim that each

E
k

{\displaystyle E\_k}

 is measurable and the following identity

E

ε
,
δ

=

⋂

# k
1

∞

{
x
∈
E
:

f
(
x
,

y
k

)
−
f
(
x
)

≤
ε
}

{\displaystyle E\_{\varepsilon,\delta}=\bigcap\_{k=1}^\infty \{x\in E: |f(x,y\_k)-f(x)|\leq \varepsilon\}}

holds. Then the result immediately shows

E

ε
,
δ

{\displaystyle E\_{\varepsilon,\delta}}

 is measurable.
First we have to show

f
(
x
)

{\displaystyle f(x)}

 is a measurable function on

E

{\displaystyle E}

. Consider the sequence

{
1

/

n

}

# n
1

∞

{\displaystyle \{1/n\}\_{n=1}^\infty}

 , we see

f
(
x
,
1

/

n
)

{\displaystyle f(x,1/n)}

 is a continuous function of

x

{\displaystyle x}

 relative to

E

{\displaystyle E}

, hence measurable. Moreover,

f
(
x
# )

lim

n
→
∞

f
(
x
,
1

/

n
)

{\displaystyle f(x)=\lim\_{n\to\infty} f(x,1/n)}

 for all

x
∈
E

{\displaystyle x \in E}

, by Theorem 4.12

f

{\displaystyle f}

 is measurable on

E

{\displaystyle E}

 as a limit of sequence of measurable functions. We also know that for every

k

{\displaystyle k}

,

f
(
x
,

y
k

)

{\displaystyle f(x,y\_k)}

 is a measurable function of

x

{\displaystyle x}

, hence

E
k

=
{

f
(
⋅
,

y
k

)
−
f
(
⋅
)

≤
ε
}

{\displaystyle E\_k=\{|f(\cdot, y\_k)-f(\cdot)|\leq\varepsilon\}}

 is measurable.

If

x
∈

E

ε
,
δ

{\displaystyle x\in E\_{\varepsilon,\delta}}

, then

f
(
x
,
y
)
−
f
(
x
)

≤
ε

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

 for all

y
∈
(
0
,
δ
)

{\displaystyle y\in(0,\delta)}

, which in turn is particularly true for

# y

y
k

.

{\displaystyle y=y\_k.}

 This shows

E

ε
,
δ

⊆

E
k

∀
k
∈

N

,

{\displaystyle E\_{\varepsilon,\delta}\subseteq E\_k\; \forall k\in\mathbb N,}

 and thus

E

ε
,
δ

⊆

⋂
k

E
k

{\displaystyle E\_{\varepsilon,\delta}\subseteq \bigcap\_k E\_k}

. On the other hand if

x
∈

⋂
k

E
k

{\displaystyle x\in \bigcap\_k E\_k}

 and

y
∈
(
0
,
δ
)

{\displaystyle y\in(0,\delta)}

, we may choose a sequence

{

z
n

}

# n
1

∞

{\displaystyle \{ z\_n \}\_{n=1}^\infty}

 from

{

y
k

}

# k
1

∞

{\displaystyle \{y\_k\}\_{k=1}^\infty}

 (treating

{

y
k

}

# k
1

∞

{\displaystyle \{y\_k\}\_{k=1}^\infty}

 as an unordered set) such that

z
n

→
y

{\displaystyle z\_n\to y}

. Then since for a fixed

x

{\displaystyle x}

,

f
(
x
,
y
)

{\displaystyle f(x, y)}

 is continuous on

y
∈
(
0
,
δ
)

{\displaystyle y\in(0,\delta)}

, we find that

f
(
x
,
y
)
−
f
(
x
)

=

lim

n
→
∞

f
(
x
,

z
n

)
−
f
(
x
)

≤
ε
.

This shows

x
∈

E

ε
,
δ

{\displaystyle x\in E\_{\varepsilon,\delta}}

. Finally, we conclude

E

ε
,
δ

{\displaystyle E\_{\varepsilon,\delta}}

 is measurable because it is a countable intersection of measurable sets.