Ex4-17

Question: Suppose that

f
k

→

m

f

{\displaystyle f\_k\xrightarrow{m}f}

 and

g
k

→

m

g

{\displaystyle g\_k\xrightarrow{m}g}

 on

E

{\displaystyle E}

. Show that

f
k

g
k

→

m

f
g

{\displaystyle f\_k+g\_k\xrightarrow{m}f+g}

 on

E

{\displaystyle E}

 and, if

E

<
∞

{\displaystyle |E|<+\infty}

, that

f
k

g
k

→

m

f
g

{\displaystyle f\_kg\_k\xrightarrow{m}fg}

 on

E

{\displaystyle E}

. If, in addition,

g

k

→
g

{\displaystyle g\_{k} \rightarrow g }

 on

E

{\displaystyle E}

,

g
≠
0

{\displaystyle g\neq0}

 a.e., and

E

<
∞

{\displaystyle |E|<+\infty}

, show that

f
k

/

g
k

→

m

f

/

g

{\displaystyle f\_k/g\_k\xrightarrow{m}f/g}

 on

E

{\displaystyle E}

. (For the product

f
k

g
k

{\displaystyle f\_kg\_k}

, write

f
k

g
k

−
f
# g
(

f
k

−
f
)
(

g
k

−
g
)
f
(

g
k

−
g
)
g
(

f
k

−
f
)

{\displaystyle f\_kg\_k-fg=(f\_k-f)(g\_k-g)+f(g\_k-g)+g(f\_k-f)}

. Consider each term separately, using the fact that a function that is finite on

E

{\displaystyle E}

,

E

<
∞

{\displaystyle |E|<+\infty}

 is bounded outside a subset of

E

{\displaystyle E}

 with small measure.)

Assume that

f
k

→

m

f

{\displaystyle f\_k\xrightarrow{m}f}

 and

g
k

→

m

g

{\displaystyle g\_k\xrightarrow{m}g}

 on

E

{\displaystyle E}

. Let

ϵ
>
0

{\displaystyle \epsilon > 0}

. There exists

K
1

{\displaystyle K\_1}

 such that

{

f
−

f
k

>
ϵ

/

2
}

<
ϵ

/

2

 if

k
>

K
1

{\displaystyle k>K\_1}

. Similarly, there exists

K
2

{\displaystyle K\_2}

 such that

{

g
−

g
k

>
ϵ

/

2
}

<
ϵ

/

2

 if

k
>

K
2

{\displaystyle k>K\_2}

.

Since

{

(
f
g
)
−
(

f
k

g
k

)

>
ϵ
}
⊆
{

f
−

f
k

>
ϵ

/

2
}
∪
{

g
−

g
k

<
ϵ

/

2
}

, thus

{

(
f
g
)
−
(

f
k

g
k

)

>
ϵ
}

<
ϵ

/

2
ϵ

/

# 2
ϵ

 if

k
>
max
{

K
1

,

K
2

}

{\displaystyle k>\max\{K\_1,K\_2\}}

.

This implies

f
k

g
k

→

m

f
g

{\displaystyle f\_k+g\_k\xrightarrow{m}f+g}

 on

E

{\displaystyle E}

.

**Multiplication:**

Further assume

E

<
∞

{\displaystyle |E|<\infty}

. Write

f
k

g
k

−
f
# g
(

f
k

−
f
)
(

g
k

−
g
)
f
(

g
k

−
g
)
g
(

f
k

−
f
)

{\displaystyle f\_kg\_k-fg=(f\_k-f)(g\_k-g)+f(g\_k-g)+g(f\_k-f)}

.

There exists

K
3

{\displaystyle K\_3}

 such that

{

f
−

f
k

>

ϵ

}

<
ϵ

/

2

 if

k
>

K
3

{\displaystyle k>K\_3}

. There exists

K
4

{\displaystyle K\_4}

 such that

{

g
−

g
k

>

ϵ

}

<
ϵ

/

2

 if

k
>

K
4

{\displaystyle k>K\_4}

.

If

k
>
max
{

K
3

,

K
4

}

{\displaystyle k>\max\{K\_3,K\_4\}}

, then

{

(

f
k

−
f
)
(

g
k

−
g
)

>
ϵ
}

≤

{

f
−

f
k

>

ϵ

}

{

g
−

g
k

>

ϵ

}

<
ϵ

.

Thus

(

f
k

−
f
)
(

g
k

−
g
)

→

m

0

{\displaystyle (f\_k-f)(g\_k-g)\xrightarrow{m}0}

.

Applying Question 15 to

{
f
}

{\displaystyle \{f\}}

, there is a closed

F
⊆
E

{\displaystyle F\subseteq E}

 and a finite

M

{\displaystyle M}

 such that

E
∖
F

<
ϵ

/

2

{\displaystyle |E\backslash F|<\epsilon/2}

 and

f
(
x
)

≤
M

{\displaystyle |f(x)|\leq M}

 for all

x
∈
F

{\displaystyle x\in F}

. We may assume

M
≠
0

{\displaystyle M\ne 0}

, as the case

# M
0

{\displaystyle M=0}

 is trivially true since

{

f
(

g
k

−
g
)

>
ϵ
}

=

{
x
∈
F
:

f
(

g
k

−
g
)

>
ϵ
}

{
x
∈
E
∖
F
:

f
(

g
k

−
g
)

>
ϵ
}

<
0
ϵ

/

2

.

Thus

{

f
(

g
k

−
g
)

>
ϵ
}

=

{
x
∈
F
:

f
(

g
k

−
g
)

>
ϵ
}

{
x
∈
E
∖
F
:

f
(

g
k

−
g
)

>
ϵ
}

≤

{
x
∈
F
:

g
k

−
g

>
ϵ

/

M
}

E
∖
F

<
ϵ

/

2
ϵ

/

2

\epsilon/2+\epsilon/2}

 for sufficiently large

k

{\displaystyle k}

. Thus

f
(

g
k

−
g
)

→

m

0

{\displaystyle f(g\_k-g)\xrightarrow{m}0}

. Similarly

g
(

f
k

−
f
)

→

m

0

{\displaystyle g(f\_k-f)\xrightarrow{m}0}

.

By our result of "Addition",

f
k

g
k

−
f
g

→

m

0

{\displaystyle f\_kg\_k-fg\xrightarrow{m}0}

, that is,

f
k

g
k

→

m

f
g

{\displaystyle f\_kg\_k\xrightarrow{m}fg}

 on

E

{\displaystyle E}

.

**Division:**

Assume in addition that

g

k

→
g

{\displaystyle g\_{k} \rightarrow g }

 on

E

{\displaystyle E}

,

g
≠
0

{\displaystyle g\neq0}

 a.e., and

E

<
∞

{\displaystyle |E|<\infty}

. Note that since we have proved "Multiplication", it suffices to show that

1

/

g
k

→

m

1

/

g

{\displaystyle 1/g\_k\xrightarrow{m}1/g}

 on

E

{\displaystyle E}

.

We use Theorem 4.21 (a.e. convergence and

E

<
∞

{\displaystyle |E|<\infty}

 implies convergence in measure). Note that since

g
≠
0

{\displaystyle g\neq0}

 a.e.,

1

/

g

{\displaystyle 1/g}

 is measurable and finite a.e. in

E

{\displaystyle E}

. Since

g

k

→
g

{\displaystyle g\_{k} \rightarrow g }

 on

E

{\displaystyle E}

, for sufficiently large

k

{\displaystyle k}

,

g
k

≠
0

{\displaystyle g\_k\neq0}

 a.e. so that

1

/

g
k

{\displaystyle 1/g\_k}

 is also measurable and finite a.e. in

E

{\displaystyle E}

.

By Theorem 4.21, since

1

/

g
k

→
1

/

g

{\displaystyle 1/g\_k\rightarrow1/g}

 a.e. on

E

{\displaystyle E}

 and

E

<
∞

{\displaystyle |E|<\infty}

, then

1

/

g
k

→

m

1

/

g

{\displaystyle 1/g\_k\xrightarrow{m}1/g}

 on

E

{\displaystyle E}

.