Ex4-18

**Question.** If

f

{\displaystyle f}

 is measurable on

E

{\displaystyle E}

, define

ω
f

(
a
# )

{
f
>
a
}

=

{

x

∈
E
∣
f
(

x

)
>
a
}

 for

−
∞
<
a
<
∞

{\displaystyle -\infty < a < \infty}

.

(1) If

f
k

↗
f

{\displaystyle f\_k \nearrow f}

, show that

ω

f
k

↗

ω
f

{\displaystyle \omega\_{f\_k} \nearrow \omega\_f}

.

(2) If

f
k

⟶
m

f

{\displaystyle f\_k {\overset{m}{\longrightarrow}} f}

, show that

ω

f
k

→

ω
f

{\displaystyle \omega\_{f\_k} \to \omega\_f}

 at each point of continuity of

ω
f

{\displaystyle \omega\_f}

.

**Answer.**

**(1)** Since

f
k

↗
f

{\displaystyle f\_k \nearrow f}

, we have

ω

f
k

=

{

x

∈
E
∣

f
k

(

x

)
>
a
}

≤

{

x

∈
E
∣

f

k
1

(

x

)
>
a
}

=

ω

f

k
1

.

We claim that

⋃

k
∈

Z

{

x

∈
E
∣

f
k

(

x

)
>
a
# }
{

x

∈
E
∣
f
(

x

)
>
a
}
.

{\displaystyle \bigcup\_{k \in \Z^+} \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\} = \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}.}

First it is clear that

⋃

k
∈

Z

{

x

∈
E
∣

f
k

(

x

)
>
a
}
⊆
{

x

∈
E
∣
f
(

x

)
>
a
}

{\displaystyle \displaystyle \bigcup\_{k \in \Z^+} \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\} \sube \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}}

 since

f
k

↗
f

{\displaystyle f\_k \nearrow f}

. On the other hand, let

x

∈
{

x

∈
E
∣
f
(

x

)
>
a
}

{\displaystyle \mathbf{x} \in \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}}

. Since

f
k

↗
f

{\displaystyle f\_k \nearrow f}

, for all

ε
>
0

{\displaystyle \varepsilon > 0}

, there exists

K
∈

Z

{\displaystyle K\in\mathbb{Z}^+}

 such that for all

k
≥
K

{\displaystyle k \ge K}

f
(

x

)
−

f
k

(

x

)

<
ε
.

{\displaystyle |f(\mathbf{x}) - f\_k(\mathbf{x})| < \varepsilon.}

In particular, take

# ε
f
(

x

)
−
a

{\displaystyle \varepsilon = f(\mathbf{x}) - a}

. Then

ε
>
0

{\displaystyle \varepsilon > 0}

 and

f
(

x

)
−

f
K

(

x

)
<
f
(

x

)
−
a
,

{\displaystyle f(\mathbf{x}) - f\_K(\mathbf{x}) < f(\mathbf{x}) - a,}

which implies that

f
K

(

x

)
>
a

{\displaystyle f\_K(\mathbf{x}) > a}

. Thus,

x

∈
{

x

∈
E
∣

f
k

(

x

)
>
a
}

{\displaystyle \mathbf{x} \in \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\}}

. This shows that

{

x

∈
E
∣
f
(

x

)
>
a
}
⊆

⋃

k
∈

Z

{

x

∈
E
∣

f
k

(

x

)
>
a
}

{\displaystyle \displaystyle\{\mathbf{x} \in E \mid f(\mathbf{x}) > a\} \sube \bigcup\_{k \in \Z^+} \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\}}

, which then proves the claim.

Now we have

{

x

∈
E
∣

f
k

(

x

)
>
a
}
↗

⋃

k
∈

Z

{

x

∈
E
∣

f
k

(

x

)
>
a
# }
{

x

∈
E
∣
f
(

x

)
>
a
}
.

{\displaystyle \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\} \nearrow \bigcup\_{k \in \Z^+} \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\} = \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}.}

Therefore, by Theorem 3.26 (i), we conclude that

ω

f
k

↗

ω
f

{\displaystyle \omega\_{f\_k} \nearrow \omega\_f}

.

**(2)** Given

ε
>
0

{\displaystyle \varepsilon > 0}

. Let

E
k

=
{

x

∈
E
∣

f
k

(

x

)
−
f
(

x

)

>
ε
}
.

{\displaystyle E\_k = \{\mathbf{x} \in E \mid |f\_k(\mathbf{x}) -f(\mathbf{x})| > \varepsilon\}.}

Since

f
k

⟶
m

f

{\displaystyle f\_k {\overset{m}{\longrightarrow}} f}

, for any

δ
>
0

{\displaystyle \delta > 0}

, there exists

K
∈

Z

{\displaystyle K\in\mathbb{Z}^+}

 such that

E
k

<
δ

{\displaystyle |E\_k| < \delta}

 for all

k
≥
K

{\displaystyle k \ge K}

 (by Exercise 16). Since

f
k

=
(

f
k

−
f
)
f

⟹

f
k

≤

f
k

−
f

f

,

we have

{

x

∈
E
∣

f
k

(

x

)
>
a
}
⊆

E
k

∪
{

x

∈
E
∣
f
(

x

)
>
a
−
ε
}
.

{\displaystyle \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\} \sube E\_k \cup \{\mathbf{x} \in E \mid f(\mathbf{x}) > a - \varepsilon\}.}

It follows by that

ω

f
k

(
a
)
≤

E
k

ω
f

(
a
−
ε
)
<
δ

ω
f

(
a
−
ε
)

{\displaystyle \omega\_{f\_k}(a) \le |E\_k| + \omega\_f(a - \varepsilon) < \delta + \omega\_f(a - \varepsilon)}

for all

k
≥
K

{\displaystyle k \ge K}

. This means that

ω
f

(
a
−
ε
)

{\displaystyle \omega\_f(a - \varepsilon)}

 is an eventual upper bound of

{

ω

f
k

(
a
)
}

{\displaystyle \{\omega\_{f\_k}(a)\}}

. Hence,

lim sup

k
→
∞

ω

f
k

(
a
)
≤

w
f

(
a
−
ε
)
.

{\displaystyle \limsup\_{k \to \infty} \omega\_{f\_k}(a) \le w\_f(a - \varepsilon).}

Analogously, since

# f
(
f
−

f
k

)

f
k

⟹

f

≤

f
−

f
k

f
k

,

we have

{

x

∈
E
∣
f
(

x

)
>
a
ε
}
⊆

E
k

∪
{

x

∈
E
∣

f
k

(

x

)
>
a
}
.

{\displaystyle \{\mathbf{x} \in E \mid f(\mathbf{x}) > a + \varepsilon\} \sube E\_k \cup \{\mathbf{x} \in E \mid f\_k(\mathbf{x}) > a\}.}

It follows that

ω
f

(
a
ε
)
≤

E
k

ω

f
k

(
a
)
<
δ

ω

f
k

(
a
)

{\displaystyle \omega\_f(a + \varepsilon) \le |E\_k|+ \omega\_{f\_k}(a) < \delta + \omega\_{f\_k}(a)}

for all

k
≥
K

{\displaystyle k \ge K}

. Thus,

lim inf

k
→
∞

ω

f
k

(
a
)
≥

ω
f

(
a
ε
)
.

{\displaystyle \liminf\_{k \to \infty} \omega\_{f\_k}(a) \ge \omega\_f(a + \varepsilon).}

Now, let

a

{\displaystyle a}

 be a point of continuity of

ω
f

{\displaystyle \omega\_f}

. Taking limits as

ε
→
0

{\displaystyle \varepsilon \to 0}

 in the inequalities above, we find that

lim sup

k
→
∞

ω

f
k

(
a
)
≤

ω
f

(
a
)
≤

lim inf

k
→
∞

ω

f
k

(
a
)
.

{\displaystyle \limsup\_{k \to \infty} \omega\_{f\_k}(a) \le \omega\_f(a) \le \liminf\_{k \to \infty} \omega\_{f\_k}(a).}

Since

lim inf

k
→
∞

ω

f
k

(
a
)
≤

lim sup

k
→
∞

ω

f
k

(
a
)

{\displaystyle \displaystyle \liminf\_{k \to \infty} \omega\_{f\_k}(a) \le \limsup\_{k \to \infty} \omega\_{f\_k}(a)}

 is always true, we conclude that

lim

k
→
∞

ω

f
k

(
a
# )

ω
f

(
a
)
,

{\displaystyle \lim\_{k \to \infty} \omega\_{f\_k}(a) = \omega\_f(a),}

which proves the assertion.