TMA4225 Foundations of Analysis
2011
Littlewood’s three principles
Harald Hanche-Olsen
http://www.math.ntnu.no/~hanche/
Littlewood’s three principles for the Lebesgue measure on R are, roughly speaking:
1. Every measurable set of finite measure is nearly a finite union of intervals
2. Every measurable function is nearly continuous
3. Every convergent sequence of functions is nearly uniformly convergent
The second and third principles are also known as Lusin’s and Egorov’s theorems,
respectively. The first principle has no other name that I know of:
Theorem 1 (Littlewood’s first principle). Let E be a Lebesgue measurable subset of
R, and assume ε > 0 is given. Then there exists a finite union F of intervals such that
µ(E 4 F ) < ε.
Proof. We go back to the definition of Lebegue outer measure: There is a sequence
of intervals 〈I j 〉 j ∈N such that
X
[
E⊆
I j and
µI j < µE + ε.
j ∈N
j ∈N
In particular, the sum above converges, so we can fix some n with
X
µI j < ε.
j >n
Let
F=
[
Ij.
j ≤n
Then
³[
´ X
µ(F \ E ) ≤ µ
Ij \E ≤
µI j − µE < ε
j ∈N
and
j ∈N
³[
´
X
[
µ(E \ F ) ≤ µ
Ij \F ≤ µ
Ij ≤
µI j < ε,
j ∈N
j >n
j >n
so that µ(E 4 F ) ≤ µ(E \ F ) + µ(F \ E ) < 2ε (and hence the overly pedantic person
would replace ε by ε/2 throughout the proof ).
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Littlewood’s three principles
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Littlewood’s third principle (Egorov’s theorem) is different from the other two, in
that it holds in any measure space:
Theorem 2 (Egorov; Littlewood’s third principle). Assume that (X , Σ, µ) is a measure space with µX < ∞, and that 〈 f k 〉k∈N is a sequence of measurable, real-valued
functions with measurable domains converging pointwise a.e. in X to a real-valued
function f . Then, given any ε > 0, there is a measurable set E ⊆ X so that µ(X \E ) < ε
and f k → f uniformly on E .
Proof. To simplify the notation, consider h k = | f k − f |, so that h k ≥ 0 and h k → 0
a.e. on X . By assumption, for any natural number m
³\ [©
ª´
µ
x : x ∉ dom h k or h k (x) ≥ m −1 = 0,
n∈N k≥n
since pointwise convergence fails on this set. The union in the above expression
decreases with n, and by finiteness of the measure it follows that the measure of
the intersection is the limit of the measure of the union as n → ∞. Thus we can
pick n(m) so that
³ [ ©
ª´
µ
x : x ∉ dom h k or h k (x) ≥ m −1 < 2−m ε.
k≥n(m)
Put
F=
[
[ ©
ª
x : x ∉ dom h k or h k (x) ≥ m −1 .
m≥1 k≥n(m)
Then adding the above inequalities we get µF < ε. Also, h k → 0 uniformly on E :=
X \ F , for
\ \ ©
ª
E=
x ∈ dom h k : h k (x) < m −1 ,
m≥1 k≥n(m)
so for every m and every k ≥ n(m) it is true that E ⊆ dom h k and h k < m −1 on
E.
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Littlewood’s three principles
Theorem 3 (Lusin; Littlewood’s second principle). Let f be a measurable, realvalued function on an interval I ⊆ R. Then, for any ε > 0, there is a measurable set
E ⊆ I so that µ(I \ E ) < ε and f E is continuous.
Proof. First, assume that I is a bounded interval; say, I = (a, b), and that f is a
bounded function. We know that if we put
Z x
f
F (x) =
a
then F is continuous in I , and furthermore F 0 = f a.e. in I . Define
g h (x) =
1
2h
x+h
Z
x−h
f =
¢
1 ¡
F (x + h) − F (x − h) ,
2h
then it follows that g h is continuous, and g h → f a.e. when h ↓ 0. By Lusin’s theorem, there is a set E ⊆ I with µ(I \E ) < ε so that g 1/n → f uniformly on E as n → ∞.
But then f E is continuous.
Second, let I remain bounded, but allow f to be unbounded. Since
\©
ª
x ∈ I : | f (x)| > n = ;,
n∈N
we can pick n large enough so that
µE 0 < ε,
©
ª
where E 0 = x ∈ I : | f (x)| > n .
Apply the first part to the bounded function f χE 0 (details left to the reader.)
Finally, assume that I is an unbounded interval, which we may take to be open.
We can find a disjoint sequence of open, bounded subintervals I j ⊂ I so that
³ [ ´
µ I\
I j = 0.
j ∈N
(We can easily arrange for the set difference to be countable. For example, if I =
(0, ∞) we can let I j = ( j , j + 1).) Apply the second part to f I j for each j , finding
some measurable E j ⊆ I j with µ(I j \E j ) < 2− j ε so that f E j is continuous for each
j . Then
[
E=
Ej
j ∈N
has the properties we are looking for (again, details are left to the reader).
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Application. The following classical result shows that the Fourier transform of an
integrable function vanishes at infinity.
Lemma 4 (Riemann–Lebesgue). If f : R → R is Lebesgue integrable then
Z ∞
f (x)e i sx d x = 0.
lim
s→±∞ −∞
Proof. First, we prove the result when f = χI where I is a bounded interval, say
I = (a, b). This is trivial indeed:
Z
b
a
e i sx d x =
e i sa − e i sb
→ 0 when s → ±∞.
is
Just as trivially, the result extends to any finite union of bounded intervals.
Next, consider f = χE where E is Lebesgue measurable with finite measure. Let
ε > 0. There is a finite union J of bounded intervals so that µ(E 4 J ) < ε. By the first
part of the proof,
¯Z
¯
¯
¯
¯ e i sx d x ¯ < ε
J
when |s| is large enough, in which case
¯Z
¯ ¯Z
¯ ¯Z
¯
¯ ¯
¯ ¯
i sx
i sx
¯ e d x¯ ≤ ¯ e d x¯ + ¯
F
J
F 4J
¯
¯
e i sx d x ¯ < 2ε,
so the result is shown when f = χE . Next, it extends trivially to any simple function.
Finally, if f is integrable then given ε > 0 there is a simple integrable function g
so that
Z
| f − g | < ε.
R
By what we have just proved,
¯Z
¯
¯
¯
¯ g (x)e i sx d x ¯ < ε
R
when |s| is large enough, in which case
¯Z
¯ ¯Z ¡
¯ ¯Z
¯ Z
¢ i sx
¯
¯ ¯
¯ ¯
¯
i sx
i sx
f (x) − g (x) e d x ¯ + ¯ g (x)e d x ¯ < | f − g | + ε < 2ε,
¯ f (x)e d x ¯ ≤ ¯
R
R
R
and the proof is finished.
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