FUNCTIONAL ANALYSIS LECTURE NOTES:
EGOROFF AND LUSIN’S THEOREMS
CHRISTOPHER HEIL
1.
Egoroff’s Theorem
Egoroff’s Theorem is a useful fact that applies to general bounded positive measures.
Theorem 1
(Egoroff’s Theorem)
.
Suppose that
μ
is a finite measure on a measure space
X,
and
f
n
,
f
:
X
→
C
are measurable.
If
f
n
→
f
pointwise a.e., then for every
ε >
0 there
exists a measurable
E
⊆
X
such that
(a)
μ
(
E
)
< ε
, and
(b)
f
n
converges uniformly to
f
on
E
C
=
X
\
E
, i.e.,
lim
n
→∞
sup
x/
∈
E

f
(
x
)
−
f
n
(
x
)

= 0
.
Proof.
Let
Z
be the set of measure zero where
f
n
(
x
) does not converge to
f
(
x
). For
k
,
n
∈
N
,
define the measurable sets
E
n
(
k
) =
∞
uniontext
m
=
n
braceleftBig

f
−
f
m
 ≥
1
k
bracerightBig
and
Z
k
=
∞
intersectiontext
n
=1
E
n
(
k
)
.
Now, if
x
∈
Z
k
, then
x
∈
E
n
(
k
) for every
n
. Hence, for each
n
there must exist an
m
≥
n
such that

f
(
x
)
−
f
m
(
x
)

>
1
k
. Therefore
f
n
(
x
) does not converge to
f
(
x
), so
x
∈
Z
. Thus
Z
k
⊆
Z,
and therefore
μ
(
Z
k
) = 0 by monotonicity. Since
E
1
(
k
)
⊇
E
2
(
k
)
⊇ · · ·
, we therefore have by
continuity from above that
lim
n
→∞
μ
(
E
n
(
k
)) =
μ
(
Z
k
) = 0
.
Choose now any
ε >
0. Then for each
k
, we can find an
n
k
such that
μ
(
E
n
k
(
k
))
<
ε
2
k
.
Define
E
=
∞
uniontext
k
=1
E
n
k
(
k
)
,
c
circlecopyrt
2009 by Christopher Heil.
1
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EGOROFF’S AND LUSIN’S THEOREMS
then we have by subadditivity that
μ
(
E
)
≤
ε
. And if
x /
∈
E
, then
x /
∈
E
n
k
(
k
) for every
k
,
and therefore

f
(
x
)
−
f
m
(
x
)

<
1
k
for all
m
≥
n
k
. Thus, we have shown that for each
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 Fall '09
 HEIL
 Topology, Metric space, measure, Compact space, Lebesgue measure

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