2
INTRODUCTION TO LEBESGUE INTEGRATION
5
Example.
Consider the function
f
n
that connects the points (0
,
0)
,
(1
/
2
n, n
)
,
(1
/n,
0)
,
(1
,
0).
These are Riemann
integrable with Riemann integral
1
2
, but they converge to the zero function.
Lebesgue’s Idea.
Consider
f
: [
a, b
]
!
R
. Let the range of
f
be [
A, B
]. Let
A
=
y
0
< y
1
<
· · ·
< y
n
=
B
be a
partition of the range. For each subinterval, look at
E
i
=
f

1
[
y
i

1
, y
i
] =
{
x
:
f
(
x
)
2
[
y
i

1
, y
i
]
}
. Then
f
⇠
P
i
y
i
χ
E
i
.
So take
Z
B
A
= “ lim
X
y
i
length(
E
i
)”
Lecture 3: May 8
We want to define a function
m
which measures the size of subsets of
R
and generalizes the notion of the length of
an interval. Ideally, we want
m
:
P
(
R
)
!
[0
,
1
] =
R
+
[
{
1
}
to satisfy:
(1) The measure of an interval should be the length of the interval.
(2)
m
(
S
1
1
E
j
) =
P
1
1
m
(
E
j
) if the
E
j
are disjoint. (This is called
σ
additivity.)
(3) For
t
2
R
, let
E
+
t
=
{
x
+
t
:
x
2
E
}
. We want
m
(
E
+
t
) =
m
(
E
)
8
t
2
R
8
E
✓
R
. (This is called translation
invariance.)
Notice that
σ
additivity implies that
m
is monotonic: If
A
✓
B
then
m
(
A
)
m
(
B
).
Unfortunately, this is impossible, because of the Axiom of Choice.
Theorem.
Such a measure function,
m
, with the properties above, does not exist.
Proof.
Define a relation on
R
by
x
⇠
y
if
x

y
2
Q
.
This is an equivalence relation.
Partition
R
into disjoint
equivalence classes. Each class is of the form
x
0
+
Q
. Each of these equivalence classes are dense in
R
, since they are
just a translation of
Q
. In particular, each class intersects [0
,
1
2
]. By the axiom of choice, we can pick one element in
[0
,
1
2
] from each equivalence class. Let
E
be the collection of these elements (one from each equivalence class).
Now we claim that the sets
E
+
x, x
2
Q
are disjoint. Say (
E
+
x
1
)
\
(
E
+
x
2
)
6
=
;
. Then
e
1
+
x
1
=
e
2
+
x
2
for some
e
1
, e
2
2
E
. Then
e
1

e
2
=
x
1

x
2
2
Q
, so
e
1
⇠
e
2
. By the construction of
E
,
e
1
=
e
2
, and hence
x
1
=
x
2
.
By
σ
additivity,
m
⇣
S
x
2
Q
E
+
x
⌘
=
P
x
2
Q
m
(
E
+
x
), since
{
E
+
x
:
x
2
Q
}
is a countable collection of disjoint sets.
By translation invariance, we have
m
0
@
[
x
2
Q
E
+
x
1
A
=
X
x
2
Q
m
(
E
+
x
) =
X
x
2
Q
m
(
E
)
which equals 0 (if
m
(
E
) = 0) or
1
(if
m
(
E
)
>
0). Now
S
x
2
Q
E
+
x
=
R
, so
m
⇣
S
x
2
Q
E
+
x
⌘
=
m
(
R
) =
1
and
hence,
m
(
E
)
>
0.