_{i=1}^{∞}be a countable collection of sets and let
Ω_{1}≡∪_{i=1}^{∞}E_{i}. Then there exists an algebra of sets, A, such that A⊇C and A iscountable.
Proof: Let C_{1} denote all finite unions of sets of C and also include Ω_{1} and
∅. Thus C_{1} is countable. Next let ℬ_{1} denote all sets of the form Ω_{1}∖ A such
that A ∈C_{1}. Next let C_{2} denote all finite unions of sets of ℬ_{1}∪C_{1}. Then let
ℬ_{2} denote all sets of the form Ω_{1}∖ A such that A ∈C_{2} and let C_{3} = ℬ_{2}∪C_{2}.
Continuing this way yields an increasing sequence,
{Cn}
each of which is countable.
Let
A ≡ ∪∞ C .
i=1 i
Then A is countable. Also A is an algebra. Here is why. Suppose A,B ∈A. Then there
exists n such that both A,B ∈C_{n−1}. It follows A ∪ B ∈C_{n}⊆A from the construction.
It only remains to show that A ∖ B ∈A. Taking complements with respect
to Ω_{1}, it follows from the construction that A^{C},B^{C} are both in ℬ_{n−1}⊆C_{n}.
Thus,
AC ∪ B ∈ Cn
and so
A ∖B = (AC ∪ B )C ∈ ℬ ⊆ C ⊆ A.
n n+1
This shows A is an algebra of sets of Ω_{1} which is also countable and contains
C.
Lemma 18.9.2Let
{fn}
be a sequence of functions in L^{1}
(Ω,S, μ)
. Then thereexists a σ finite set of S, Ω_{1}, and a σ algebra of subsets of Ω_{1},S_{1}, such that S_{1}⊆S,f_{n} = 0 off Ω_{1},f_{n}∈ L^{1}
(Ω1,S1,μ)
, and S_{1} = σ
(A)
, the σ algebra generated by A,for some A a countable algebra.
Proof: Let ℰ_{n} denote the sets which are of the form
{ ------}
f−n1(B (z,r)) : z ∈ ℚ + iℚ,r > 0,r ∈ ℚ, and 0 ∕∈ B (z,r)
Since each ℰ_{n} is countable, so is
ℰ ≡ ∪∞ ℰ
n=1 n
Now let Ω_{1}≡∪ℰ. I claim Ω_{1} is σ finite. To see this, let
{ }
Wn = ω ∈ Ω : |fk(ω)| > 1-for some k = 1,2,⋅⋅⋅,n
n
Thus if ω ∈ W_{n}, it follows that for some r ∈ ℚ, z ∈ ℚ + iℚ sufficiently close to
f_{k}
(ω )
ω ∈ f− 1(B(z,r)) ∈ ℰk
k
and so ω ∈∪_{k=1}^{n}ℰ_{k} and consequently, W_{n}∈∪_{k=1}^{n}ℰ_{k}. Also
∫ ∑n
μ(Wn ) 1-≤ |fk(ω)|dμ < ∞.
n Wn k=1
Now if ω ∈ Ω_{1}, then for some k,ω is contained in a set of ℰ_{k}. Therefore, for that
k,
f (ω ) ∈ B(z,r)
k
where r is a positive rational number and z ∈ ℚ + iℚ and B
(z,r)
does not contain 0.
Therefore, f_{k}
(ω)
is at a positive distance from 0 and so for large enough n,ω ∈ W_{n}.
Take n so large that 1∕n is less than the distance from B
(z,r)
to 0 and also larger than
k.
By Lemma 18.9.1 there exists a countable algebra of sets A which contains ℰ. Let
S_{1}≡ σ
(A )
. It remains to show f_{n}
(ω)
= 0 off Ω_{1} for all n. Let ω
∕∈
Ω_{1}. Then ω is not
contained in any set of ℰ_{k} and so f_{k}
(ω)
cannot be nonzero. Hence f_{k}
(ω)
= 0. This
proves the lemma.
The following Theorem is the main result on sequential compactness in L^{1}
(Ω,S,μ )
.
Theorem 18.9.3Let K ⊆ L^{1}
(Ω, S,μ)
be such that for some C > 0 and allf ∈ K,
||f|| 1 ≤ C (18.9.23)
L
(18.9.23)
and K also satisfies the property that if
{En}
is a decreasing sequence of measurable setssuch that ∩_{n=1}^{∞}E_{n} = ∅, then for all ε > 0 there exists n_{ε}such that if n ≥ n_{ε},then
|∫ |
|| ||
| En fdμ| < ε (18.9.24)
(18.9.24)
for all f ∈ K. Then every sequence of functions of K has an L^{1}
(Ω, μ)
weakly convergentsubsequence.
Proof: Take
{fn}
a sequence in K and let A,S_{1},Ω_{1} be as in Lemma 18.9.2. Thus A
is a countable algebra and by assumption, for each E ∈A,
{ ∫ }
fndμ
E
is a bounded sequence and so there exists a convergent subsequence. Therefore, from a
Cantor diagonalization argument, there exists a subsequence, denoted by
{gn}
such
that
{ ∫ }
gndμ
E
converges for every E ∈A.
Let
{ ∫ }
ℳ ≡ E ∈ S1 = σ (A) such that lim gndμ exists .
n→∞ E
Then it has been shown that A⊆ℳ. Suppose E_{k}↑ E where E_{k}∈ℳ. Then letting
ε > 0 be given, the assumption shows that for k large enough,
||∫ ||
|| g dμ||< ε
| E∖Ek n |
for all g_{n}. Therefore, picking such a k,
|∫ ∫ | |∫ ∫ |
|| g dμ − g dμ ||≤ 2ε+ || g dμ − g dμ||< 3ε
|E n E m | | Ek n Ek m |
provided m,n are large enough. Therefore,
{∫ }
Egndμ
is a Cauchy sequence and so it
converges.
In the case that E_{k}↓ E use the assumption to conclude there exists a k large enough
that
||∫ ||
|| g dμ||< ε
| Ek∖E n |
for all g_{n}. Then
|∫ ∫ | |∫ ∫ |
|| g dμ − g dμ|| = || g dμ− g dμ||
| E n E m | |Ek n Ek m |
||∫ || ||∫ ||
+ || gndμ||+ || gmdμ||
| |Ek∖E | |Ek∖|E |
||∫ ∫ ||
≤ |E gndμ− E gmdμ|+ 2ε < 3ε
k k
provided m,n large enough. Again
{∫ g dμ }
E n
is a Cauchy sequence. This shows ℳ is a
monotone class and so by the monotone class theorem, Theorem 10.10.5 on Page 892 it
follows ℳ = S_{1}≡ σ
(A )
.
Therefore, picking E ∈S_{1}, you can define a complex measure,
∫
λ (E ) ≡ nli→m∞ gndμ
E
Then λ ≪ μ and so by Corollary 18.2.9 on Page 1853 and the fact shown above that Ω_{1} is
σ finite there exists a unique S_{1} measurable g ∈ L^{1}
(Ω ,μ)
1
such that
∫ ∫
λ(E) = gdμ ≡ lim g dμ.
E n→ ∞ E n
Extend g to equal 0 outside Ω_{1}.
It remains to show
{gn}
converges weakly. It has just been shown that for every s a
simple function measurable with respect to S_{1}
∫ ∫ ∫ ∫
gnsdμ = gnsdμ → gsdμ = gsdμ
Ω Ω1 Ω1 Ω
Now let f ∈ L^{∞}
(Ω1,S1,μ)
and pick a uniformly bounded representative of this function.
Then by Theorem 9.3.9 on Page 644 there exists a sequence of simple functions
converging uniformly to f and so
^{′} is a space I don’t know much about due to a possible lack of σ
finiteness of Ω. However, it does follow that for i the inclusion map of L^{1}
(Ω1,S1,μ)
into
L^{1}
(Ω, S,μ)
which merely extends the function as 0 off Ω_{1} and f ∈
( )
L1 (Ω, S,μ)
^{′}, there
exists h ∈ L^{∞}
(Ω1)
such that for all g ∈ L^{1}
(Ω1,S1,μ)
∫
i∗f (g) = hgd μ.
Ω1
This is because i^{∗}f ∈
(L1(Ω1,S1,μ))
^{′} and Ω_{1} is σ finite and so the Riesz representation
theorem applies to get a unique such h ∈ L^{∞}
(Ω1)
. Then since all the g_{n} equal 0 off
Ω_{1},
∫
f (g ) = i∗f (g ) = hg dμ
n n Ω1 n
for a unique h ∈ L^{∞}
(Ω1,S1,μ)
due to the Riesz representation theorem which holds here
because Ω_{1} was shown to be σ finite. Therefore,
∫ ∫
lim f (g ) = lim hg dμ = hgdμ = i∗f (g) = f (g).
n→∞ n n→ ∞ Ω1 n Ω1
This proves the theorem.
For more on this theorem see [?]. I have only discussed the sufficiency of the
conditions to give sequential compactness. They also discuss the necessity of these
conditions.
There is another nice condition which implies the above results which is seen in books
on probability. It is the concept of equi integrability.
Definition 18.9.4Let
(Ω,S,μ)
be a measure space in which μ
(Ω)
< ∞. ThenK ⊆ L^{1}
(Ω,S,μ)
is said to be equi integrable if
∫
lim sup |f|dμ = 0
λ→ ∞f∈K [|f|≥λ]
Lemma 18.9.5Let K be an equi integrable set. Then there exists C > 0 such that forall f ∈ K,
||f||L1 ≤ C (18.9.25)
(18.9.25)
and K also satisfies the property that if
{En}
is a decreasing sequence of measurable setssuch that ∩_{n=1}^{∞}E_{n} = ∅, then for all ε > 0 there exists n_{ε}such that if n ≥ n_{ε},then