Ex4-6

Question: Let f and g be measurable functions on

E

{\displaystyle E}

.

(a) If f and g are finite a.e. in

E

{\displaystyle E}

, show that f + g is measurable no matter how we define it at the points when it has the form

∞
(
−
∞
)

{\displaystyle +\infty +(-\infty )}

 or

−
∞

(

∞

)

{\displaystyle -\infty +\left(+\infty \right)}

.

(b)Show that fg is measurable without restriction on the finiteness of f and g. Show that f + g is measurable if it is defined to have the same value at every point where it has the form

∞
(
−
∞
)

{\displaystyle +\infty +(-\infty )}

 or

−
∞

(

∞

)

{\displaystyle -\infty +\left(+\infty \right)}

.

(Note that a function h defined on E is measurable if and only if both

{

# h
∞

}

{\displaystyle \left\{h=+\infty \right\}}

 and

{

# h
−
∞

}

{\displaystyle \left\{h=-\infty \right\}}

 are measurable and the restriction of h to the subset of E where h is finite is measurable.)

Let

f
:
E
→

R

∪

{

∞
,
−
∞

}

{\displaystyle f:E\to \mathbb {R} \cup \left\{\infty ,-\infty \right\}}

 and

g
:
E
→

R

∪

{

∞
,
−
∞

}

{\displaystyle g:E\to \mathbb {R} \cup \left\{\infty ,-\infty \right\}}

 be two functions satisfying that f is measurable and

g
(
x
# )
f
(
x
)

{\displaystyle g(x)=f(x)}

 for almost every

x
∈
E

{\displaystyle x \in E}

. Then,

g
(
x
)

{\displaystyle g(x)}

 is measurable.

We define a new function

h
(
x
)

{\displaystyle h(x)}

 such that

h
(
x
# )

{

g
(
x
)

 if

g
(
x
)
∈

R

0

 if

g
(
x
)
∈
{
±
∞
}

{\displaystyle h(x)=\left\{{\begin{array}{ll}g(x)&{\text{ if }}g(x)\in \mathbb {R} \\0&{\text{ if }}g(x)\in \{\pm \infty \}\end{array}}\right.}

.

Since g is measurable and g(x)=h(x) almost everywhere, h(x) is measurable. Now, it remains to show f+h is measurable.

Now, we have

f
(
x
)
h
(
x
# )

{

f
(
x
)
g
(
x
)

 if

g
(
x
)
∈

R

f
(
x
)

 if

g
(
x
)
∈
{
±
∞
}

{\displaystyle f(x)+h(x)=\left\{{\begin{array}{ll}f(x)+g(x)&{\text{ if }}g(x)\in \mathbb {R} \\f(x)&{\text{ if }}g(x)\in \{\pm \infty \}\end{array}}\right.}

.

Consider the set

{

f
h
>
a

}

=
(

{

f
>
a

}

∩

{

# g
±
∞

}

)
∪
(

{

f
g
>
a

}

∩

{

g
≠
±
∞

}

)

{\displaystyle \left\{f+h>a\right\}=(\left\{f>a\right\}\cap \left\{g=\pm \infty \right\})\cup (\left\{f+g>a\right\}\cap \left\{g\neq \pm \infty \right\})}

.

{

f
>
a

}

{\displaystyle \left\{f>a\right\}}

 and

{

# g
±
∞

}

{\displaystyle \left\{g=\pm \infty \right\}}

 is measurable obviously. And

{

g
≠
±
∞

}

{\displaystyle \left\{g\neq \pm \infty \right\}}

 is measurable since it is complement of A. Next, we will claim that

{

f
g
>
a

}

{\displaystyle \left\{f+g>a\right\}}

 is measurable as well.

Let us write

Q

{\displaystyle Q}

 as

# Q

{

r

1

,

r

2

,

r

3

.
.
.

}

{\displaystyle Q=\left\{r\_{1},r\_{2},r\_{3}...\right\}}

, since

Q

{\displaystyle Q}

 is countable. Then we could write

{

f
g
>
a

}

{\displaystyle \left\{f+g>a\right\}}

 as

{

f
g
>
a

}

=

⋃

# k
1

∞

{

f
>

r

k

>
−
g
a

}

=

⋃

# k
1

∞

(

{

f
>

r

k

}

)
∩
(

{

−
g
a
<

r

k

}

)

{\displaystyle {\begin{aligned}\left\{f+g>a\right\}&=\bigcup \_{k=1}^{\infty }\left\{f>r\_{k}>-g+a\right\}\\&=\bigcup \_{k=1}^{\infty }(\left\{f>r\_{k}\right\})\cap (\left\{-g+a<r\_{k}\right\})\end{aligned}}}

Since f and g are both measurable, both

{

f
>

r

k

}

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

 and

{

−
g
a
<

r

k

}

{\displaystyle \left\{-g+a<r\_{k}\right\}}

 are measurable, which implies that

{

f
g
>
a

}

{\displaystyle \left\{f+g>a\right\}}

 is measurable.

Combining all these, we can conclude that

{

f
h
>
a

}

{\displaystyle \left\{f+h>a\right\}}

 is measurable, which means the function

f
h
:
E
→

R

n

∪

{

∞
,
−
∞

}

{\displaystyle f+h:E\to \mathbb {R} ^{n}\cup \left\{\infty ,-\infty \right\}}

 is measurable.

Since

f
h

{\displaystyle f+h}

 is measurable and

(
f
g
)
(
x
# )
(
f
h
)
(
x
)

{\displaystyle (f+g)(x)=(f+h)(x)}

almost everywhere, we have

f
g

{\displaystyle f+g}

 is measurable.

(b) We first prove two lemmas.

Lemma 1: ***Let

E

1

,

E

2

⊆

R

n

{\displaystyle E\_{1},E\_{2}\subseteq \mathbb {R} ^{n}}

 be measurable sets and

f
:

E

1

∪

E

2

→

R

{\displaystyle f:E\_{1}\cup E\_{2}\to \mathbb {R} }

 be a function defined on

E

1

∪

E

2

{\displaystyle E\_{1}\cup E\_{2}}

. Then

f

{\displaystyle f}

 is measurable if and only if its restrictions to

E
1

{\displaystyle E\_1 }

 (

f

E

1

=
f
:

E

1

→

R

{\displaystyle f|\_{E\_{1}}=f:E\_{1}\to \mathbb {R} }

) and

E
2

{\displaystyle E\_2}

 (

f

E

2

=
f
:

E

2

→

R

{\displaystyle f|\_{E\_{2}}=f:E\_{2}\to \mathbb {R} }

) are measurable.***

*Proof*: Suppose that

f

{\displaystyle f}

 is measurable. Then for any

a
∈

R

{\displaystyle a \in \R}

, the set

{

x

∈

E

1

∪

E

2

f
(

x

)
>
a
}

{\displaystyle \{\mathbf {x} \in E\_{1}\cup E\_{2}\ |\ f(\mathbf {x} )>a\}}

is measurable. Fix

a
∈

R

{\displaystyle a \in \R}

. Then

{

x

∈

E

1

∣

f

E

1

(

x

)
>
a
# }
{

x

∈

E

1

∪

E

2

∣
f
(

x

)
>
a
}
∩

E

1

{\displaystyle \{\mathbf {x} \in E\_{1}\mid f\_{E\_{1}}(\mathbf {x} )>a\}=\{\mathbf {x} \in E\_{1}\cup E\_{2}\mid f(\mathbf {x} )>a\}\cap E\_{1}}

which is measurable. Likewise, we show that

{

x

∈

E

2

∣

f

E

2

(

x

)
>
a
}

{\displaystyle \{\mathbf {x} \in E\_{2}\mid f\_{E\_{2}}(\mathbf {x} )>a\}}

 is measurable. Therefore, we conclude that

f

E

1

{\displaystyle f|\_{E\_{1}}}

 and

f

E

2

{\displaystyle f|\_{E\_{2}}}

 are measurable.

Conversely, suppose that

f

E

1

{\displaystyle f|\_{E\_{1}}}

 and

f

E

2

{\displaystyle f|\_{E\_{2}}}

 are measurable. Then by the formula

{

x

∈

E

1

∪

E

2

∣
f
(

x

)
>
a
}

=
{

x

∈

E

1

∣

f

E

1

(

x

)
>
a
}
∪
{

x

∈

E

2

∣

f

E

2

(

x

)
>
a
}
,

{\displaystyle {\begin{aligned}\{\mathbf {x} \in E\_{1}&\cup E\_{2}\mid f(\mathbf {x} )>a\}\\&=\{\mathbf {x} \in E\_{1}\mid f\_{E\_{1}}(\mathbf {x} )>a\}\cup \{\mathbf {x} \in E\_{2}\mid f\_{E\_{2}}(\mathbf {x} )>a\},\end{aligned}}}

we see immediately that

f

{\displaystyle f}

 is measurable.

Lemma 2: ***Let

E
⊆

R

n

{\displaystyle E \subseteq \R^n}

 be measurable and

f
:
E
→

R

¯

{\displaystyle f:E\to {\overline {\mathbb {R} }}}

 is a function defined on

E

{\displaystyle E}

, where

R

¯

=

R

∪
{
−
∞
,
∞
}

{\displaystyle {\overline {\mathbb {R} }}=\mathbb {R} \cup \{-\infty ,\infty \}}

 is the extended real line. If

E

1

=
{
f
(

x

# )
∞
}

{\displaystyle E\_{1}=\{f(\mathbf {x} )=\infty \}}

 and

E

2

=
{
f
(

x

# )
−
∞
}

{\displaystyle E\_{2}=\{f(\mathbf {x} )=-\infty \}}

, then

f

{\displaystyle f}

 is measurable if and only if

E
1

{\displaystyle E\_1 }

 and

E
2

{\displaystyle E\_2}

 are measurable and the restriction

f

E
∖
(

E

1

∪

E

2

)

{\displaystyle f|\_{E\setminus (E\_{1}\cup E\_{2})}}

 is measurable.***

*Proof:* Suppose that

f

{\displaystyle f}

 is measurable. Then

E
1

{\displaystyle E\_1 }

 and

E
2

{\displaystyle E\_2}

 are measurable by Corollary 4.2 (from the textbook). It follows that

E
∖
(

E

1

∪

E

2

)

{\displaystyle E\setminus (E\_{1}\cup E\_{2})}

 is measurable, and so is the restriction

f

E
∖
(

E

1

∪

E

2

)

{\displaystyle f|\_{E\setminus (E\_{1}\cup E\_{2})}}

 by Lemma 1.

Conversely, suppose that

E
1

{\displaystyle E\_1 }

 and

E
2

{\displaystyle E\_2}

 are measurable and the restriction

f

E
∖
(

E

1

∪

E

2

)

{\displaystyle f|\_{E\setminus (E\_{1}\cup E\_{2})}}

 is measurable. Then for any

a
∈

R

{\displaystyle a \in \R}

, we have

{

x

∈
E
∣
f
(

x

)
>
a
# }

E

1

∪
{

x

∈
E
∣
f

E
∖
(

E

1

∪

E

2

)

(

x

)
>
a
}

{\displaystyle \{\mathbf {x} \in E\mid f(\mathbf {x} )>a\}=E\_{1}\cup \{\mathbf {x} \in E\mid f|\_{E\setminus (E\_{1}\cup E\_{2})}(\mathbf {x} )>a\}}

which is measurable, and so is

f

{\displaystyle f}

.
We first show that

f
g

{\displaystyle f g}

 is measurable without the restriction on finiteness of

f

{\displaystyle f}

 and

g

{\displaystyle g}

. Let

E

1

{
f
# g
∞
}

and

E

2

=
{
f
# g
−
∞
}
.

{\displaystyle E\_{1}\{fg=\infty \}\qquad {\text{and}}\qquad E\_{2}=\{fg=-\infty \}.}

Then

E

1

=
{

x

∈
E
∣
f
(

x

# )
∞
,
g
(

x

)
>
0
}
∪
{

x

∈
E
∣
f
(

x

# )
−
∞
,
g
(

x

)
<
0
}

∪
{

x

∈
E
∣
f
(

x

)
>
0
,
g
(

x

# )
∞
}
∪
{

x

∈
E
∣
f
(

x

)
<
0
,
g
(

x

# )
−
∞
}
,

{\displaystyle {\begin{aligned}E\_{1}&=\{\mathbf {x} \in E\mid f(\mathbf {x} )=\infty ,g(\mathbf {x} )>0\}\cup \{\mathbf {x} \in E\mid f(\mathbf {x} )=-\infty ,g(\mathbf {x} )<0\}\\&\quad \cup \{\mathbf {x} \in E\mid f(\mathbf {x} )>0,g(\mathbf {x} )=\infty \}\cup \{\mathbf {x} \in E\mid f(\mathbf {x} )<0,g(\mathbf {x} )=-\infty \},\end{aligned}}}

which is measurable. Likewise, we show that

E
2

{\displaystyle E\_2}

 is measurable. By Lemma 2, it suffices to show that

# h
f
g

E
∖
(

E

1

∪

E

2

)

{\displaystyle h=fg|\_{E\setminus (E\_{1}\cup E\_{2})}}

 is measurable. Given

a
∈

R

{\displaystyle a \in \R}

. If

a
≥
0

{\displaystyle a \ge 0}

, then

{

x

∈
E
∖
(

E

1

∪

E

2

)
∣
h
(

x

)
>
a
}

=

{

x

∈
E
∖
(

E

1

∪

E

2

)

(

f

E
∖
{
f
±
∞
}

(

x

)

)

⋅

(

g

E
∖
{
g
±
∞
}

(

x

)

)

>
a

}

,

{\displaystyle {\begin{aligned}&\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\mid h(\mathbf {x} )>a\}\\&=\left\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\ {\big |}\ \left(f|\_{E\setminus \{f\pm \infty \}}(\mathbf {x} )\right)\cdot \left(g|\_{E\setminus \{g\pm \infty \}}(\mathbf {x} )\right)>a\right\},\end{aligned}}}

which is measurable by Theorem 4.10 (in the textbook).
If

a
<
0

{\displaystyle a<0}

, then

{

x

∈
E
∖
(

E

1

∪

E

2

)
∣
h
(

x

)
>
a
}

=
{

x

∈
E
∖
(

E

1

∪

E

2

)
∣
f
(

x

# )
0
}
∪
{

x

∈
E
∖
(

E

1

∪

E

2

)
∣
g
(

x

# )
0
}

∪

{

x

∈
E
∖
(

E

1

∪

E

2

)

(

f

E
∖
{
f
±
∞
}

(

x

)

)

⋅

(

g

E
∖
{
g
±
∞
}

(

x

)

)

>
a

}

,

{\displaystyle {\begin{aligned}&\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\mid h(\mathbf {x} )>a\}\\&=\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\mid f(\mathbf {x} )=0\}\cup \{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\mid g(\mathbf {x} )=0\}\\&\quad \cup \left\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\ {\big |}\ \left(f|\_{E\setminus \{f\pm \infty \}}(\mathbf {x} )\right)\cdot \left(g|\_{E\setminus \{g\pm \infty \}}(\mathbf {x} )\right)>a\right\},\end{aligned}}}

which is also measurable. This proves that

h

{\displaystyle h}

 is measurable. Therefore,

f
g

{\displaystyle f g}

 is measurable.

Now, let

f
,
g
:
E
→

R

¯

{\displaystyle f,g:E\to {\overline {\mathbb {R} }}}

 be measurable. Given

a
∈

R

{\displaystyle a \in \R}

, define

f
# g
a

{\displaystyle f+g=a}

 whenever it is of the form

∞
(
−
∞
)

{\displaystyle +\infty +(-\infty )}

 or

−
∞
∞

{\displaystyle -\infty +\infty }

. Then for the set

E

1

=
{
f
# g
∞
}

{\displaystyle E\_{1}=\{f+g=\infty \}}

, we have

E

1

=
{
f
∈

R

,
# g
∞
}
∪
{
# g
∞
}
∪
{
# f
∞
,
g
∈

R

}

{\displaystyle E\_{1}=\{f\in \mathbb {R} ,g=\infty \}\cup \{g=g=\infty \}\cup \{f=\infty ,g\in \mathbb {R} \}}

which is measurable by Corollary 4.2 (from the textbook). Likewise, the set

E

2

=
{
f
# g
−
∞
}

{\displaystyle E\_{2}=\{f+g=-\infty \}}

 is measurable. By Lemma 2, it suffices to show that

# h
(
f
g
)

E
∖
(

E

1

∪

E

2

)

{\displaystyle h=(f+g)|{E\setminus (E\_{1}\cup E\_{2})}}

 is measurable, Let

b
∈

R

{\displaystyle b\in\mathbb{R}}

. If

b
≥
a

{\displaystyle b \geq a}

, then

{

x

∈
E
∖
(

E

1

∪

E

2

)
∣
h
(

x

)
>
a
}

=

{

x

∈
E
∖
(

E

1

∪

E

2

)

(

f

E
∖
{
f
±
∞
}

(

x

)

)

(

g

E
∖
{
g
±
∞
}

(

x

)

)

>
b

}

,

{\displaystyle {\begin{aligned}&\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\mid h(\mathbf {x} )>a\}\\&=\left\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\ {\big |}\ \left(f|\_{E\setminus \{f\pm \infty \}}(\mathbf {x} )\right)+\left(g|\_{E\setminus \{g\pm \infty \}}(\mathbf {x} )\right)>b\right\},\end{aligned}}}

which is measurable By Theorem 4.9 (in the textbook).

If

b
<
a

{\displaystyle b<a}

, then

{

x

∈
E
∖
(

E

1

∪

E

2

)
∣
h
(

x

)
>
b
}

=
{
# f
∞
,
# g
−
∞
}
∪
{
# f
−
∞
,
# g
∞
}

∪

{

x

∈
E
∖
(

E

1

∪

E

2

)

(

f

E
∖
{
f
±
∞
}

(

x

)

)

(

g

E
∖
{
g
±
∞
}

(

x

)

)

>
b

}

{\displaystyle {\begin{aligned}&\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\mid h(\mathbf {x} )>b\}\\&=\{f=\infty ,g=-\infty \}\cup \{f=-\infty ,g=\infty \}\\&\quad \cup \left\{\mathbf {x} \in E\setminus (E\_{1}\cup E\_{2})\ {\big |}\ \left(f|\_{E\setminus \{f\pm \infty \}}(\mathbf {x} )\right)+\left(g|\_{E\setminus \{g\pm \infty \}}(\mathbf {x} )\right)>b\right\}\end{aligned}}}

which is also measurable. This proves that

h

{\displaystyle h}

 is measurable. Therefore,

f
g

{\displaystyle f+g}

 is measurable.