5-2

Question:

(a) Show that the conclusions of Theorem 5.32 are not generally true without the assumption that

ϕ
∈
L
(
E
)

{\displaystyle \phi \in L(E)}

.

(b) Show that Theorem 5.33 fails without the assumption that

E

<
∞

{\displaystyle |E|<\infty}

.

Solution:

(a) For (i), Consider the function

f

# k

{

−
∞

,
x
>
k

0

,
x
≤
k

{\displaystyle f\_{k=}\left\{\begin{matrix}
 -\infty \ ,x>k\\
 0 \ \ ,x\le k
\end{matrix}\right. }

 and

lim

k
→
∞

f

k

=
0

{\displaystyle \lim\_{k \to \infty} f\_{k}= 0}

 and take

# E
R

{\displaystyle E=R }

.

In this case,

∫

E

lim

k
→
∞

f

k

=

∫

E

# 0
0

{\displaystyle \int \_{E} \lim\_{k \to \infty} f\_{k}= \int \_{E} 0=0}

.

Since

∫

E

f

k

=
−
∞

{\displaystyle \int \_{E} f\_{k} = -\infty }

 for all k, we have

lim

k
→
∞

∫

E

f

k

=
−
∞

{\displaystyle \lim\_{k \to \infty} \int \_{E} f\_{k}= -\infty }

. You could see that

lim

k
→
∞

∫

E

f

k

≠

∫

E

lim

k
→
∞

f

k

{\displaystyle \lim\_{k \to \infty} \int \_{E} f\_{k} \ne \int \_{E} \lim\_{k \to \infty} f\_{k} }

.

For (ii), i think the function

−

f

# k

{

∞

,
x
>
k

0

,
x
≤
k

{\displaystyle -f\_{k=}\left\{\begin{matrix}
 \infty \ ,x>k\\
 0 \ \ ,x\le k
\end{matrix}\right. }

 is a good counter—example, and this proof is similar to the proof above.

(b) Consider the density function of exponential distribution which comes from probability. It's

f

# k

{

1
k

e

−

1
k

x

,
x
>
0

0

,
x
≤
0

{\displaystyle f\_{k=}\left\{\begin{matrix}
 \frac{1}{k} e^{-\frac{1}{k}x} \ ,x>0\\
 0 \ \ ,x\le 0
\end{matrix}\right. }

 and

lim

k
→
∞

f

k

=
0

{\displaystyle \lim\_{k \to \infty} f\_{k}= 0}

. And we take

# E
R

{\displaystyle E=R }

.

We claim that

f
k

{\displaystyle f\_k}

 uniformly converges to

f
(
x
# )
0

{\displaystyle f(x) = 0}

 first.

We take

# N
⌊

1
ε

⌋
1

{\displaystyle N=\lfloor\frac{1}{ \varepsilon }\rfloor+1 }

, then for all

x
∈
R

{\displaystyle x \in R }

,

f

k

−
f

≤

1
k

e

−

1
k

x

≤

1
k

e

−

1
k

×
0

=

1
k

≤

1

⌊

1
ε

⌋
1

≤

1

1
ε

=
ε

{\displaystyle |f\_{k}-f| \le \frac{1}{k} e^{-\frac{1}{k}x} \le \frac{1}{k} e^{-\frac{1}{k} \times 0} = \frac{1}{k} \le \frac{1}{\lfloor\frac{1}{ \varepsilon }\rfloor+1} \le \frac{1}{\frac{1}{\varepsilon}}=\varepsilon }

 for all

k
≥
N

{\displaystyle k \ge N }

, which shows that

f
k

{\displaystyle f\_k}

 uniformly converges to

f

{\displaystyle f}

.

In this case,

∫

E

f

k

=
1

{\displaystyle \int \_{E} f\_{k} = 1 }

 for all

k

{\displaystyle k}

 and therefore

lim

k
→
∞

∫

E

f

k

=
1

{\displaystyle \lim\_{k \to \infty} \int \_{E} f\_{k} = 1 }

.

But

∫

E

# f

∫

E

# 0
0

{\displaystyle \int \_{E} f = \int \_{E} 0 =0 }

, and you could see that

lim

k
→
∞

∫

E

f

k

≠

∫

E

f

{\displaystyle \lim\_{k \to \infty} \int \_{E} f\_{k} \ne \int \_{E} f }

 in this case.

Note:

Theorem 5.32: (Monotone Convergence Theorem) Let

{

f

k

}

{\displaystyle \left\{f\_{k}\right\}}

 be a sequence of measurable functions on E:

(i) If

f
k

↗
f

{\displaystyle f\_k \nearrow f}

 a.e. on E and there exists φ ∈ L(E) such that

f
k

{\displaystyle f\_k}

 ≥ φ a.e. on E for all k, then

∫

E

f

k

→

∫

E

f

{\displaystyle \int\_{E} f\_{k} \rightarrow \int\_{E} f }

.

(ii)If

f

k

↘
f

{\displaystyle f\_{k} \searrow f }

 a.e. on E and there exists φ ∈ L(E) such that

f
k

{\displaystyle f\_k}

 ≤ φ a.e. on E for all k, then

∫

E

f

k

→

∫

E

f

{\displaystyle \int\_{E} f\_{k} \rightarrow \int\_{E} f }

.

Theorem  5.33: (Uniform Convergence Theorem) Let

f

k

∈
L
(
E
)

{\displaystyle f\_{k} \in L(E) }

 for

# k
1
,
2
,
…

{\displaystyle k=1,2, \ldots }

 , and let

{

f

k

}

{\displaystyle \left\{f\_{k}\right\}}

 converge uniformly to f on E,

E

<
∞

{\displaystyle |E|<+\infty}

. Then

f
∈
L
(
E
)

{\displaystyle f \in L(E)}

 and

∫

E

f

k

→

∫

E

f

{\displaystyle \int\_{E} f\_{k} \rightarrow \int\_{E} f }

.