Uniformly Integrable Families of Functions
Michael Taylor
Let (X, F, µ) be a measure space, and assume µ(X) < ∞. Let δ : (0, 1] → (0, ∞)
be monotone increasing. Pick K ∈ (0, ∞). We define SδK to be the set of functions
f ∈ L1 (X, µ) with the property that kf kL1 ≤ K and, for all ε ∈ (0, 1],
Z
(1)
S ∈ F , µ(S) ≤ δ(ε) =⇒ |f | dµ ≤ ε.
S
A set S ⊂ L1 (X, µ) is said to be uniformly integrable if S ⊂ SδK for some such δ, K.
The following is a well known extension of the Lebesgue Dominated Convergence
Theorem, known as the Vitali Convergence Theorem.
Proposition 1. Assume {fk : k ∈ Z+ } is uniformly integrable, and fk → f in
measure. Then fk → f in L1 -norm.
This result has a converse, which we discuss below.
To prove Proposition 1, say {fk } ⊂ SδK . Take ε > 0, so δ(ε) > 0. Clearly
|f | < ∞, µ-a.e. By Egoroff’s Theorem, there exists S ∈ F such that
(2)
µ(S) ≤ δ(ε) and fk → f uniformly on X \ S.
Thus there exists N such that |fk (x) − f (x)| ≤ ε for all k ≥ N, x ∈ X \ S. Hence,
for all j, k ≥ N ,
Z
Z
Z
Z
|fj − fk | dµ ≤
|fj − fk | dµ + |fj | dµ + |fk | dµ
(3)
X
S
S
X\S
≤ 2εµ(X) + 2ε.
Thus (fk ) is Cauchy in L1 (X, µ). Hence there exists g ∈ L1 (X, µ) such that fk → g
in L1 -norm. Also fk → g in measure, so g = f, µ-a.e.
To see that Proposition 1 is actually an extension of the Dominated Convergence
Theorem (at least when µ(X) < ∞), suppose {fk } ⊂ L1 (X, µ) and there exists
F ∈ L1 (X, µ) such that |fk | ≤ F for each k. We claim the following.
Proposition 2. Given F ∈ L1 (X, µ), there exists monotone δ : (0, 1] → (0, ∞)
and K ∈ (0, ∞) such that F ∈ SδK .
Given this, we immediately have |fk | ≤ F ⇒ {fk } ⊂ SδK . The result stated
in Proposition 2 is an ingredient in a standard proof, via Egoroff’s Theorem, of
the Dominated Convergence Theorem. To prove this proposition, argue as follows.
Take K = kF kL1 . Given ε ∈ (0, 1], pick a positive simple function ϕε such that
Z
Z
ε
(4)
0 ≤ ϕε ≤ |F | and
ϕε dµ ≥ |f | dµ − .
2
1
2
Set Aε = sup ϕε < ∞, and then set δ1 (ε) = ε/2Aε . (Cf. [T], Chapter 3, Exercise 8.)
Then pick δ : (0, 1] → (0, ∞) monotone, such that δ ≤ δ1 , e.g., δ(ε) = inf {δ1 (s) :
ε ≤ s ≤ 1}.
Remark. Our proof of Proposition 1 does not work to establish the Dominated
Convergence Theorem, since we used the result that Cauchy sequences in L1 (X, µ)
converge. For the sake of brevity, we have avoided the slightly more elaborate
argument that would also give a self-contained proof of Dominated Convergence.
The following result extends Proposition 2. (See Proposition 6 for a further
extension.)
Proposition 3. If K ⊂ L1 (X, µ) is compact, then it is uniformly integrable.
Proof. Take K = supK kf kL1 . Pick {fk : k ∈ Z+ } dense in K, and for ε ∈ (0, 1],
let N (ε) denote the smallest integer such that
ε
∀ f ∈ K,
min kf − fk kL1 ≤ .
2
1≤k≤N (ε)
Apply Proposition 2 to Fε = sup1≤k≤N (ε) |fk |, to obtain δ̃ε such that Fε ∈ Sδ̃K . Set
ε
δ1 (ε) = δ̃ε (ε/2). Again take δ : (0, 1] → (0, ∞) monotone, such that δ ≤ δ1 . Then,
for each ε ∈ (0, 1], f ∈ K,
Z
S ∈ F , µ(S) ≤ δ(ε) ⇒
|f | dµ
S
(5)
≤
Z
min
1≤k≤N (ε)
ε ε
≤ + ,
2 2
kf − fk kL1 +
|Fε | dµ
S
so K ⊂ SδK .
The following converse to Proposition 1 is a special case of Proposition 3.
Corollary 4. If f, fk ∈ L1 (X, µ) and fk → f in L1 -norm, then {fk } is uniformly
integrable.
We mention another definition of uniform integrability. Take ϕ : [0, ∞) → [0, ∞),
monotonically decreasing, such that ϕ(λ) → 0 as λ → ∞. We define U ϕ to be the
set of elements f ∈ L1 (X, µ) such that
Z
(6)
|f | dµ ≤ ϕ(λ).
{x:|f (x)|≥λ}
If µ(X) < ∞, it is the case that a set S ⊂ L1 (X, µ) is uniformally integrable
(as defined before) if and only if there exists such ϕ so that S ⊂ U ϕ . See [R].
pp. 277–278, for a proof.
We now specialize to the case (X, B, µ), where X is a compact metric space, B
the σ-algebra of Borel sets, and µ a finite Borel measure (necessarily regular if X
is metric). Let M(X) denote the space of finite signed Borel measures on X, the
dual of C(X), the space of continuous functions on X. We identify L1 (X, µ) with
a subspace of M(X) via f 7→ f µ.
3
Proposition 5. Let X be a compact metric space, with a finite Borel measure µ.
Assume δ : (0, 1] → (0, ∞) is strictly monotone increasing. Take K ∈ (0, ∞). Then
SδK is weak∗ compact in M(X).
Proof. Take fk ∈ SδK and set νk = fk µ, bounded in M(X). Passing to a subsequence (which we continue to denote νk ), we have νk → ν, weak∗ . We are claiming
that ν = gµ with g ∈ SδK . (Clearly the total variation of ν is ≤ K.)
We first establish this under the additional hypothesis that each fk ≥ 0. We aim
to show that, for each ε ∈ (0, 1],
(7)
S ∈ B, µ(S) ≤ δ(ε) =⇒ ν(S) ≤ ε.
The assertion that ν = gµ with g ∈ SδK follows from this, by the Radon-Nikodym
Theorem. If (7) fails, then there exist ε ∈ (0, 1], S ∈ B, and η > 0 such that
(8)
µ(S) ≤ δ(ε) and ν(S) ≥ ε + η.
Hence there exists a compact K ⊂ S such that
(9)
ν(K) ≥ ε + η/2.
Note that µ(K) ≤ δ(ε). Since δ is strictly monotonic, we deduce that there exists
an open set O ⊃ K such that µ(O) < δ(ε + η/4). Consequently,
Z
η
(10)
fk dµ ≤ ε + , ∀ k.
4
O
Now take ϕ ∈ C(X) such that 0 ≤ ϕ ≤ 1, ϕ = 1 on K, and supp ϕ ⊂ O. We have
Z
η
(11)
ϕfk dµ ≤ ε + , ∀ k,
4
i.e., hϕ, νk i ≤ ε + η/4. Hence
(12)
η
hϕ, νi = lim hϕ, νk i ≤ ε + .
k→∞
4
This implies
(13)
η
ν(K) ≤ ε + ,
4
which contradicts (9). This takes care of the case fk ≥ 0.
For general fk , write fk = fk+ − fk− , where for each k, fk+ and fk− are ≥ 0 and
are supported on disjoint sets, so |fk | = fk+ + fk− . If fk ∈ SδK , then also fk± ∈ SδK
and |fk | ∈ SδK . Passing to a subsequences (which we continue to denote fk± , etc.),
we have weak∗ limits fk± µ → ν ± , |fk |µ → ν̃. The argument just done gives ν ± =
g ± µ, ν̃ = g̃µ, with g ± , g̃ ∈ SδK . We also have fk µ = fk+ µ − fk− µ → ν + − ν − weak∗ ,
i.e., ν = ν + − ν − = (g + − g − )µ, while |fk |µ = fk+ µ + fk− µ → ν + + ν − weak∗ , so
ν̃ = ν + + ν − , and hence g + + g − ∈ SδK . Since g ± ≥ 0, it follows that |g + − g − | ≤
g + + g − , so g + − g − ∈ SδK , and Proposition 5 is proven.
We next produce further classes of uniformly integrable families of functions on
a finite measure space (X, F, µ). This result extends Proposition 3.
4
Proposition 6. Take p ∈ [1, ∞), with dual exponent p0 . Assume
(14)
0
K ⊂ Lp (X, µ) is compact,
B ⊂ Lp (X, µ) is bounded,
say v ∈ B ⇒ kvkLp0 (X) ≤ A. Then
(15)
{uv : u ∈ K, v ∈ B} is uniformly integrable on (X, µ).
Proof. Take ε > 0. Pick u1 , . . . , uK(ε) ∈ K such that
(16)
∀ u ∈ K,
min
1≤j≤K(ε)
ku − uj kLp (X) ≤
ε
.
2A
Say j = j(u) works. Then, making use of Hölder’s inequality, we have
Z
Z
Z
|uv| dµ ≤ |u − uj(u) | · |v| dµ + |uj(u) | · |v| dµ
S
S
S
Z
ε
· A + |uj(u) | · |v| dµ
2A
S
Z
´1/p
¡
ε
≤ +A
|uj(u) |p dµ
.
2
≤
(17)
S
p
Now, since the finite set |u1 | , . . . , |uK(ε) |p is uniformly integrable, we can take
δ = δ(ε) such that
Z
³ ε ´p
(18)
∀ j ∈ {1, . . . , K(ε)}, µ(S) ≤ δ(ε) =⇒ |uj |p dµ ≤
.
2A
S
Hence
(19)
Z
µ(S) ≤ δ(ε), u ∈ K, v ∈ B =⇒
|uv| dµ ≤ ε,
S
yielding the asserted uniform integrability.
Putting together Propositions 1 and 6, we have the following.
Proposition 7. In the setting of Proposition 6, assume uj ∈ Lp (X, µ), vj ∈
0
Lp (X, µ) satisfy
uj −→ u in Lp -norm,
(20)
and, for some A < ∞,
(21)
kvj kLp0 (X) ≤ A,
vj −→ v in measure.
Then
(22)
uj vj −→ uv in L1 -norm.
We leave the proof as an exercise for the reader.
References
[R] I. Rana, An Introduction to Measure and Integration, 2nd Ed., AMS, Providence, RI, 2002.
[T] M. Taylor, Measure Theory and Integration, AMS, Providence, RI, 2006.