Egorov’s theorem
1 Theorem (Egorov) Let (Ω‚A, µ) be a measure space, let E ⊆ Ω with E ∈ A and
µ(E ) < ∞, and let f n , f : E → R be measurable functions so that f n → f µ-a.e. on E .
Let ε > 0 Then there is some A ⊆ E with A ∈ A and µ(A) < ε so that f n → f uniformly
on E \ A.
Egorov’s theorem is also known as one of Littlewood’s principles:
Pointwise convergence is almost uniform.
– but note that this principle holds only on sets of finite measure.
Proof: For each η > 0, let
Aη =
∞ [
∞ ©
\
ª
x ∈ E : | f n (x) − f (x)| ≥ η .
N =1 n=N
|
{z
A η,N
}
Whenever f n (x) → f (x), x ∈ A η . Thus µ(A η = 0). Since E ⊃ A η,N & A η and µ(E ) < ∞,
µ(A η,N ) → 0 when N → ∞.
S
For each k ∈ N, pick Nk ∈ N with µ(A 1/k,Nk ) < 2−k ε, and let A = ∞
A
.
k=1 1/k,Nk
Clearly, µ(A) < ε.
I claim that f n → f uniformly on E \ A. To see this, let η > 0 and pick k > 1/η. If
x ∈ E \ A and n ≥ Nk , then x ∈ E \ A 1/k,Nk , so | f n (x) − f (x)| < 1/k < η.
Lusin’s theorem – another one of Littlewood’s principles
Littlewood stated is thus:
Every measurable function is almost continuous.
There are several ways to state this as a precise theorem, all (or mostly) bearing Lusin’s name.
We shall state and prove two of these.
But first, some lemmas.
A short series of lemmas
The following lemma is also one of Littlewood’s principles:
A Lebesgue measurable set of finite measure is almost a finite union of intervals.
2 Lemma Given a Lebesgue measurable set E ⊂ R with λ(E ) < ∞ and ε > 0, there
exists a finite union K of intervals with λ(E 4 K ) < ε.
Proof: By the definition of outer Lebesgue measure, there is a sequence of intervals
P
S
I k with E ⊂ ∞
I and ∞
λ(I k ) < λ(E ) + ε. Since the series converges, we can
k=1
k=1 k
P∞
S
S
find n with k=n+1 λ(I k ) < ε. Let K = nk=1 I k . Then E \ K ⊆ ∞
I , so λ(E \ K ) <
k=n+1 k
S∞
S∞
ε. Also, K \ E ⊆ k=1 I k \ E , so λ(K \ E ) ≤ λ( k=1 I k ) − λ(E ) < ε Thus λ(E 4 K ) =
λ(E \ K ) + λ(K \ E ) < 2ε.
Recall that a simple function is a measurable function which takes only a finite
number of values, and which vanishes outside a set of finite measure. In particular, it is a finite linear combination of characteristic functions of measurable sets
of finite measure.
By a step function we shall mean a finite linear combination of characteristic functions of bounded intervals.
3 Lemma Given a simple function s and ε > 0, there is a step function S so that
λ({x ∈ R : S(x) 6= s(x)}) < ε.
P
Proof: This is really a corollary to the previous lemma. Let s = nk=1 χ A k , let K k
P
be a finite union of intervals with λ(A k 4 K k ) < ε/n, and let S = nk=1 χK k . Then
Sn
S(x) = s(x) except for x ∈ k=1 (A k 4 K k ), and this set has measure < ε.
4 Lemma If f : R → C is Lebesgue measurable, there is a sequence of step functions converging to f almost everywhere.
Proof: Take a sequence (s n ) of simple functions so that s n → f pointwise. For each
n, let S n be a step function so that S n = s n except on a set D n , where λ(D n ) < 2−n .
Let
∞ [
∞
\
D=
Dn ,
N =1 n=N
and note that λ(D) = 0, because λ(
S∞
n=N
Dn ) ≤
P∞
n=N
λ(D n ) <
P∞
n=N
2−n = 21−N .
If x ∈ R\D then for some N , S n (x) = s n (x) for all n ≥ N , so that S n (x) → f (x) because
s n (x) → f (x).
Lusin’s theorem, take one
5 Theorem (Lusin) If f : R → C is Lebesgue measurable
¯ and ε > 0, there is some
Lebesgue measurable set E ⊂ R with λ(E ) < ε so that f ¯R\E is continuous.
Note that this theorem does not claim that f is continuous at every x ∈ R \ E . It is the restriction of f that is continuous. To illustrate the difference, consider f = χQ , which is nowhere
continuous. However, its restriction to R \ Q is continuous: This restriction is constantly zero.
Proof: Take a sequence (S n ) of step functions converging a.e. to f . For each integer
N , Egorov’s theorem implies the existence of a measurable set A N ⊆ (N , N + 1) with
S
λ(A N ) < 2−|N | ε so that S n → f uniformly on (N , N + 1) \ A N . Let A = N ∈Z A n . Then
λ(A) < 3ε. Further, let D be the set of points where some S n is discontinuous. Since
each S n is discontinuous at only a finite number of points, D is countable. Let E =
Z ∪D ∪ A. Then λ(E ) = λ(A) < 3ε. Each S n is continuous on E , and the convergence
to f is uniform on (N , N +1)\E , so the restriction of f to (N , N +1)\E is continuous.
Therefore, so is the restriction of f to R \ E .
That last bit is subtle, so let us do it in more detail: If x ∈ R \ E , then x ∉ Z, so
x ∈ (N , N + 1) for some N ∈ Z. Given ε > 0 there is some δ > 0 so that, whenever
y ∈ (N , N + 1) \ E and |y − z| < δ, then | f (y) − f (x)| < ε. Here we can replace δ by a
smaller number, so that the δ-neighbourhood of x is contained in (N , N + 1). Then,
if y ∈ R \ E and |y − x| < δ, we also have y ∈ (N , N + 1) \ E , so that | f (y) − f (x)| < ε.
Lusin’s theorem, take two
Theorem 5 is good as far as it goes, but we can improve on it:
6 Theorem (Lusin) If f : R → C is Lebesgue measurable and ε > 0, there is some
Lebesgue measurable set G ⊂ R with λ(G) < ε and a continuous function g : R → C
so that f (x) = g (x) for all x ∈ R \ G.
¯
Proof: Begin by finding a set E as in Theorem 5, so that λ(E ) < ε, and f ¯R\E is continuous.
¯
Pick an open set G ⊇ E with λ(G) < 2ε. Then also f ¯R\G is continuous. We shall
expand this function to a function g : R → C. So we begin by defining g (x) = f (x)
when x ∈ R \ G.
The set G, being open, is a pairwise disjoint union of open intervals – the components of G. Say,
∞
G
G=
(a k , b k ).
k=1
Expand g by making it linear on [a k , b k ]:
g (x) =
(b k − x) f (a k ) + (x − a k ) f (b k )
bk − ak
for x ∈ (a k , b k ).
Then g is indeed continuous on R (exercise!), and g = f on R \ G.
Approximation in L1
We don’t need the full power of Lusin’s theorem to prove the following approximation result.
But it does seem like a nice application.
7 Theorem Let f ∈ L1 (λ) and ε > 0. Then there is a continuous function g ∈ L1 (λ)
with
Z
| f − g | d λ < ε.
R
Proof: We prove this only for f : R → R. If f is complex valued, we can apply the
real result to the real and imaginary parts of f separately.
Let B n = {x ∈ R : | f (x)| ≥ n}. Since χB n f → 0 as n → ∞, and |χB nRf | ≤ | f |, we can use
the dominated convergenceRtheorem (DCT) to conclude that B n | f | d λ → 0 when
n → ∞. Pick some n so that B n | f | d λ < ε.
Now define f n by


 f (x) if | f (x)| ≤ n,
f n (x) = n
if f (x) > n,


−n
if f (x) < −n.
Pick a continuous function g : R → R so that f (x) = g (x) for all x except for x in a set
A with λ(A) < ε/n. We may assume that |g | ≤ n everywhere, for otherwise, redefine
it by setting g (x) = n where before g (x) > n and g (x) = −n where before g (x) < −n.
Now
Z
Z
Z
| f − g | d λ ≤ | f − fn | d λ + | fn − g | d λ
R
ZR
ZR
=
| f − fn | d λ + | fn − g | d λ
B
A
Z n
≤
| f | d λ + 2nλ(A)
Bn
< ε + 2ε = 3ε.
Finally, g ∈ L1 (λ) because the triangle inequality yields
Z
Z
Z
|g | d λ ≤ |g − f | d λ + | f | d λ < ∞.
R
The proof is now complete.
R
R