2
INTRODUCTION TO LEBESGUE INTEGRATION
9
If
z
2
E
k
then
z
2
E
.
Indeed, if
f
was continuous at
z
, then for
"
=
1
2
k
,
9
δ
>
0 such that
8
x
2
[
z

δ
, z
+
δ
],

f
(
x
)

f
(
z
)

<
"
=
1
2
k
. Then if
x, y
2
[
z

δ
, z
+
δ
], then

f
(
x
)

f
(
y
)


f
(
x
)

f
(
z
)

+

f
(
z
)

f
(
y
)

<
2
1
2
k
=
1
k
, so
z /
2
E
k
.
Therefore
S
1
k
=1
E
k
✓
E
. In fact, if
z /
2
E
k
, there exists
δ
>
0 such that
8
x, y
2
[
z

δ
, z
+
δ
], we have

f
(
x
)

f
(
y
)

<
1
k
.
If we let
y
=
z
, then

f
(
x
)

f
(
z
)

<
1
k
. So if
z /
2
S
E
k
, for each
"
>
0 we can pick
k
such that
1
k
<
"
.
z /
2
E
k
, so if
δ
is chosen as above and
x
2
[
z

δ
, z
+
δ
], then

f
(
x
)

f
(
z
)

<
1
k
<
"
, so
f
is continuous at
z
, and thus
z /
2
E
. Hence
we can conclude that
E
=
S
1
k
=1
E
k
.
Side Note (Not part of proof)
: We claim now that
E
k
is closed, which will prove that
E
is
F
σ
, and
hence Borel. Let (
z
n
) be a sequence in
E
k
with
z
n
!
z
. Fix
δ
>
0, and get
z
n
2
[
z

δ
2
, z
+
δ
2
]. Since
z
n
2
E
k
, there exist points
x, y
2
[
z
n

δ
2
, z
n
+
δ
2
]
✓
[
z

δ
, z
+
δ
] such that

f
(
x
)

f
(
y
)

≥
1
k
. Thus
z
2
E
k
so
E
k
is closed.
Suppose
m
(
E
) = 0. Let
"
>
0 and find a partition
P
of [
a, b
] such that
U
(
f, P
)

L
(
f, P
)
<
"
. Denote Equivalently,
X
✓
sup
f
[
x
i

1
,x
i
]

inf
f
[
x
i

1
,x
i
]
◆
(
x
i

x
i

1
)
<
"
Denote
M
i
= sup
f
[
x
i

1
,x
i
]
and
m
i
= inf
f
[
x
i

1
,x
i
]
.
Since
m
(
E
) = 0, we can pick open intervals
{
I
n
}
1
n
=1
such that
S
1
n
=1
I
n
◆
E
and
P
1
n
=1
`
(
I
n
)
<
"
. Pick
k
such that
1
k
<
"
. Note that
S
1
n
=1
I
n
◆
E
◆
E
k
and
E
k
is closed and
E
k
✓
[
a, b
], so
E
k
is compact. By compactness, take a
finite subcover
I
1
, . . . , I
n
of
E
k
. Let
X
= [
a, b
]
\
S
n
j
=1
I
j
. If
z
2
Z
, then
z /
2
S
n
j
=1
I
j
◆
E
k
, so
z /
2
E
k
. Thus there
exists
δ
>
0 such that
8
x, y
2
[
z

δ
, z
+
δ
], we have

f
(
x
)

f
(
y
)

<
1
k
. Let
V
z
= (
z

δ
, z
+
δ
) and
V
z
= [
z

δ
, z
+
δ
].