1. A Koml´os Theorem for general Banach lattices of
measurable functions
E. Jim´enez Fern´andez, A. Juan, E. A. S´anchez P´erez
Instituto Universitario de Matem´atica Pura y Aplicada
(I.U.M.P.A.),
Universidad Polit´ecnica de Valencia.
Valencia 2010

2. Consider a Banach function space X(µ) of (classes of) finitely integrable functions over
a σ-finite measure space (Ω,Σ,µ) with the Fatou property. In 2010, Day and Lennard
proved that the theorem of Koml´os on convergence of Ces`aro sums holds also in these
spaces; i.e. for every bounded sequence (fn)n in X(µ), there exists a subsequence
(fnk
)k and a function f ∈ X(µ) such that for any further subsequence (hj )j of (fnk
)k , the
series 1
n ∑n
j=1 hj converges µ-a.e. to f. In this paper we generalize this result to a more
general class of Banach spaces of classes of measurable functions —spaces L1(ν) of
integrable functions with respect to a vector measure on a δ-ring— and explore to
which point the Fatou property and the Koml´os property are equivalent. In particular we
prove that this always holds for sublattices of spaces L1(ν) with the Fatou property, and
provide an example of a Banach space of measurable functions that is Fatou but do not
satisfy the Koml´os Theorem.

3. Definitions

4. (Ω,Σ,µ) will be a σ-finite measure space, X(µ) a Banach space over µ and BX(µ) the
unit ball of X(µ). Let L0(µ) be the space of all measurable real functions on Ω.
We will call Banach (or K¨othe) function space to any Banach space X ⊆ L0(µ) with
norm · X satisfying that
1 if f ∈ L0(µ), g ∈ X(µ) with |f| ≤ |g| µ-a.e. then f ∈ X and f X ≤ g X(µ).
2 For every A ∈ Σ of finite measure, the characteristic function χA belongs to X.

5. (Ω,Σ,µ) will be a σ-finite measure space, X(µ) a Banach space over µ and BX(µ) the
unit ball of X(µ). Let L0(µ) be the space of all measurable real functions on Ω.
We will call Banach (or K¨othe) function space to any Banach space X ⊆ L0(µ) with
norm · X satisfying that
1 if f ∈ L0(µ), g ∈ X(µ) with |f| ≤ |g| µ-a.e. then f ∈ X and f X ≤ g X(µ).
2 For every A ∈ Σ of finite measure, the characteristic function χA belongs to X.

6. 1 A Banach function space is order continuous if order bounded increasing
sequences are convergent in norm.
2 If a K¨othe function space X(µ) is order continuous, then X(µ) = X(µ)×.
3 Consequently, in this case (X(µ) , · X(µ)× ) is a K¨othe function space and for
every f ∈ X(µ),
f X(µ) = sup
g∈BX(µ)× Ω
fgdµ .
4 If Y(µ) is other Banach function space, we denote by M(X,Y) the space of
multiplication operators from X to Y.
5 X(µ) has the weak Fatou property if for every increasing sequence in X(µ) fn ↑ f,
if sup fn < ∞ then f ∈ X(µ).

7. In 1967 Koml´os proved that for every bounded sequence (fn)n in L1(µ), for a finite
measure µ, there exists a subsequence (fnk
)k and a function f ∈ L1(µ) such that for
any further subsequence (hj )j of (fnk
)k , the Ces`aro sums
1
n
n
∑
j=1
hj
converge µ-a.e. to f.

8. Theorem
Let R be a δ-ring and consider a vector measure ν : R → X, where X is a Banach
space. Let E be an ideal of L1(ν). Then E has the Fatou property if and only if it has
the Koml´os property.
Theorem
Let ν : R → X be a locally σ−finite vector measure and consider an ideal E of L1
w (ν).
Then E has the Koml´os property if and only if E has the weak σ-Fatou. In particular,
L1
w (ν) has the Koml´os property.

9. Lemma
Let R be a δ-ring and consider a sequence (An)n ⊂ R. Take U = ∪n≥1An. Then:
(1) The class of subsets R∩U is a δ-ring of subsets of U such that R∩U ⊆ R.
(2) (R∩U)loc = U ∩Rloc as δ-rings of subsets of U.
(3) If ν is a vector measure on R and νU is its restriction to R∩U, then
L1(ν)∩M ((R∩U)loc) = L1(νU ) and L1
w (ν)∩M ((R∩U)loc) = L1
w (νU ).

10. Lemma
Let (Ω,Σ,µ) be a finite measure space and let (Ω,Σ,η) be a σ-finite measure space
such that µ and η are equivalent measures. Let E(η) be a Banach function space over
η with the weak σ-Fatou property, and let Mg : E(η) → L1(µ) be a multiplication
operator given by a Σ-measurable a function g > 0. Then E(η) has the Koml´os
property.

11. Proof. Let (fn)n be a sequence of functions of E(η) such that fn ≤ M for all n ∈ N.
Consider the sequence (fng)n in L1(µ). We can apply Koml´os theorem in the space
L1(µ): there exists a subsequence (gfni
)i such that for any further subsequence (hnij
)j
of (fni
)i , 1
k ∑k
j=1 ghnij
converges µ −a.e. to a function h0 ∈ L1(µ). We need to show that
h :=
h0
g ∈ E, since η is equivalent to µ, and then 1
k ∑k
j=1 hnij
converges η-a.e. to h. For
each n ∈ N, consider the measurable function kn(w) := ´ınfk≥n | 1
k ∑k
i=1 fni
(w)|, w ∈ Ω. It
is clearly an increasing sequence kn ↑ |h| and kn E(η) ≤ 1
n ∑n
i=1 fni
≤ M.

12. The case of L1
w (ν)
Definition
Let R be a δ−ring of subsets of Ω, X a Banach space and ν : R → X a vector
measure. We say that ν is locally σ−finite with respect to a δ−ring R if given a set
B ∈ Rloc with ν (B) < ∞, B can be written as B = (∪n≥1An)∪N, with An ∈ R and
N ∈ Rloc a ν−null set.
Theorem
Let ν : R → X be a locally σ−finite vector measure and consider an ideal E of L1
w (ν).
Then E has the Koml´os property if and only if E has the weak σ-Fatou. In particular,
L1
w (ν) has the Koml´os property.

13. Example. A Fatou Banach lattice of measurable functions that is not Koml´os. Consider
the space ∞(I), where I = 2N. Take the sequence (fn) defined as follows. Let w ⊂ 2N.
Let us order w increasingly and write it as w = (nk )m
k=1, where m = |w|. Put fn(w) := 0
if n is not in w. In other case, put fn1
(w) = 1, fn2
(w) = −1, fn3
(w) = 1, fn4
(w) = 1,
fn5
(w) = −1, fn6
(w) = −1, fn7
(w) = 1, fn8
(w) = 1, fn9
(w) = 1, fn10
(w) = 1,fn11
(w) = 1,
fn12
(w) = 1, fn13
(w) = −1, fn14
(w) = −1... . This sequence is defined as follows. Take a
sequence defined as a1 = 3 and ar = 3ar−1 −2 for r ≥ 2. We define fn1
(w) = 1,
fnk
(w) = 1 if k ∈ {ar ,...,2ar −2} for any r ≥ 1, and fnk
= −1 in other case.
Clearly, this sequence is norm bounded by 1. Let us show that 1
m ∑m
k=1 fnk
do not
converge pointwise —and so n-almost everywhere for the counting measure n— for
any subsequence (fnk
) of (fn). If (fnk
) is such a subsequence, take the set
w0 := {nk : k ∈ N}. A direct computation shows that 1
2ar −2 ∑
2ar −2
k=1 fnk
(w0) = 1
2 and
1
ar −1 ∑
ar −1
k=1 fnk
(w0) = 0, for every r ∈ N; so the sequence ( 1
m ∑m
k=1 fnk
(w0))∞
k=1 do not
converge.