THE FATOU LEMMA, DOMINATED AND MONOTONE CONVERGENCE
THEOREMS ARE EQUIVALENT TO EACH OTHER
Let
fn → f
almost everywhere on a measure space
X
of
nite measure.
Then the following
three facts are equivalent to each other.
Theorem 1 (the Fatou lemma).
fn ≥ 0 for all n then
ˆ
ˆ
f ≤ lim inf fn .
If
n→∞
Theorem 2 (Dominated Convergence Theorem).
|fn | ≤ g
ˆ
ˆ
fn
f = lim
If
for some
g ∈ L1
then
n→∞
Theorem 3 (Monotone Convergence Theorem).
ˆ
fn ≤ fn+1
ˆ
f = lim
fn
If
for all
n
then
n→∞
The proof of the equivalence (FL) ⇔ (DCT) ⇔ (MCT) if |X| < ∞
(FL) ⇒ (DCT).
Let
hn := g + |f | − |fn − f |.
Note
hn ≥ 0
|fn | ≤ g .
ˆ
lim inf hn ≤ lim inf hn .
by triangle inequality and the assumption
ˆ
n→∞
ˆ
ˆ
ˆ
|f | ≤ lim inf
|fn | ≤
n→∞
|f |
(1)
n→∞
Noting that Fatou lemma gives also
we see that
Fatou lemma gives
g<∞
is integrable and so we may cancel out the terms
ˆ
´
|f |,
´
g
in (1) to obtain
−|fn − f |,
0 ≤ lim inf
n→∞
ˆ
that is
lim sup
|fn − f | ≤ 0,
n→∞
from which the claim follows.
´
(DCT) ⇒ (MCT).
f =∞
since
f ≥ fn
Assume
for all
n.
fn ≥ 0
(in the general case consider
If not then the integrals are bounded, that is
ˆ
fn ≤ K.
Let
U := {x : fn (x) → ∞}.
Date :
fn − f0 ).
23 Feb 2017.
1
If
´
fn → ∞
then
FL, DCT AND MCT ARE EQUIVALENT TO EACH OTHER
2
Then
U=
\[
r
Unr ,
n
where
Unr := {x : fn (x) > r}
|Unr | ≤ K/r and so, since for each r the
[ r
Un = lim |Unr | ≤ K/r.
n→∞
n
S
r, U ⊂ n Unr we obtain
[ r
|U | ≤ Un ≤ K/r → 0.
n
By Chebyshev inequality
Moreover since for each
Hence
|U | = 0,
that is
f
is nite a.e. Now let
sets
Ar := {r − 1 ≤ f < r}
φ :=
∞
X
Unr
are increasing we obtain
and
r χ Ar .
r=1
φ ≥ f ≥ fn
Note
and so the claim follows from DCT if
Bs :=
s
[
φ ∈ L1 .
For this let
Ar
r=1
φ≤f +1
and note that
s
X
ˆ
r|Ar | =
P
r
X
n→∞
Bs
Bs
r=1
(Recall
f < s on Bs , an application of DCT gives
ˆ
ˆ
fn + |X| ≤ K + |X|.
f + |X| = lim
φ≤
and so, since
Bs
is the space.) Hence the partial sums of the series
r|Ar | =
´
φ
converge, that is
(MCT) ⇒ (FL).
1
φ∈L
P
r
r|Ar | are bounded, and so the series
, as required.
Let
φn (x) := inf{fk (x) : k ≥ n}.
Then
0 ≤ φn ≤ fn
ˆ
and so
ˆ
φn ≤
Noting that
φn ≤ φn+1
MCT to the sequence
and that
φn
fn
limn→∞ φn (x) = f (x)
gives
ˆ
ˆ
f = lim
n→∞
we observe that taking
ˆ
φn = lim inf
n→∞
lim inf
and applying
ˆ
φn ≤ lim inf
n→∞
fn ,
as required.
The case |X| = ∞
If
|X| = ∞
then the argument in (DCT)
by an implication (FL)
Since
fn ≤ f
we obtain
⇒ (MCT)
breaks down. However it can be replaced
⇒ (MCT) as follows:
´
´
fn ≤ f and so Fatou lemma gives
ˆ
ˆ
ˆ
ˆ
f ≤ lim inf fn ≤ lim sup fn ≤ f,
n→∞
n→∞
FL, DCT AND MCT ARE EQUIVALENT TO EACH OTHER
3
as required.
This however, makes
(DCT)
merely an implication of
fact. I am not aware of an argument that includes
(FL)
DCT
and
(MCT),
rather than an equivalent
as an equivalent fact. I would be grateful
for any comments.
Wojciech Ozanski, Mathematics Institute, University of Warwick, Coventry CV4 7AL, UK.
E-mail address :
[email protected]