Chapter 3 Lebesgue Integration
3.2 Convergence of Integrals
3.2.2. Convergence in Measure
Proof: Let Eb be the set of x ∈ E for which f(x) = limn→∞fn(x) and supn≥1|fn(x)| ≤g(x). ThenEb is measurable andµ(Eb{) = 0, and so integrals overEbare the same as those overE. Thus, without loss of generality, assume that all the assumptions hold for every x∈ E. But then, f = limn→∞fn,
|f| ≤gand|f −fn| ≤2g. Hence, by the second part of Fatou’s Lemma,
n→∞lim Z
f dµ− Z
fndµ
≤ lim
n→∞
Z
|f−fn|dµ≤ Z
n→∞lim |f −fn|dµ= 0.
It is important to understand the role played by theLebesgue dominant g.
Namely, it acts as anumbrellato keep everything under control. To see that some such control is needed, consider Lebesgue measureλ[0,1]on [0,1],B[0,1]
and the functions fn =n1(0,n−1). Obviously, fn −→0 everywhere on [0,1], but kfnk1= 1 for alln∈Z+.
Unfortunately, in many circumstances it is difficult to find an appropriate Lebesgue dominant, and, for this reason, results like the following variation on Fatou’s Lemma are interesting and often helpful. See Exercises 3.2.21 and 3.2.24 for other variations.
Theorem 3.2.5 (Lieb’s Version of Fatou’s Lemma). Let (E,B, µ)be a measure space, {fn : n≥1} ∪ {f} ⊆L1(µ;R), and assume thatfn −→f (a.e.,µ). Then
(3.2.6)
n→∞lim
kfnkL1(µ;R)− kfkL1(µ;R)− kfn−fkL1(µ;R)
= lim
n→∞
Z
|fn| − |f| − |fn−f|
dµ= 0.
In particular, if kfnkL1(µ;R)−→ kfkL1(µ;R), thenkf −fnkL1(µ;R)−→0.
Proof: Since
kfnkL1(µ;R)−kfkL1(µ)−kfn−fkL1(µ;R)
≤
Z
|fn|−|f|−|fn−f|
dµ, n≥1, we need check only the second equality in (3.2.6). But, because
|fn| − |f| − |fn−f|
−→0 (a.e., µ) and
|fn| − |f| − |fn−f| ≤
|fn| − |fn−f|
+|f| ≤2|f|, (3.2.6) follows from Lebesgue’s Dominated Convergence Theorem.
§3.2.2. Convergence in Measure: We now have a great deal of evi-
3 Lebesgue Integration
be said about the opposite implication. To begin with, I point out that kfnkL1(µ;R) −→0 does not imply that fn −→ 0 (a.e.,µ). Indeed, define the functions {fn: n≥1}on [0,1] so that, form≥0 and 0≤` <2m,
f2m+`=1[2−m`,2−m(`+1)].
It is then clear that thesefn’s are non-negative and measurable on ([0,1],B[0,1]) and that limn→∞fn(x) = 1 for everyx∈[0,1]. On the other hand,
Z
[0,1]
fndλR= 2−m if 2m≤n <2m+1,
and thereforeR
[0,1]fndλR−→0 asn→ ∞.
As the preceding discussion makes clear, it may be useful to consider other notions of convergence. Keeping in mind that we are looking for a type of convergence that can be tested using integrals, we should take a hint from Markov’s inequality and say that the sequence {fn : n ≥1} of measurable functions on the measure space (E,B, µ) converges in µ-measure to the measurable function f if µ(|fn−f| ≥ )−→ 0 as n → ∞ for every > 0, in which case I will write fn −→f in µ-measure. Note that, by Markov’s inequality (3.1.7), ifkfn−fkL1(µ;R)−→0 thenfn−→f inµ-measure. Hence, this sort of convergence can be easily tested with integrals, and, as such, must be quite different (cf. Exercise 3.2.23) fromµ-almost everywhere convergence.
In fact, it takes a moment to see in what sense the limit is even uniquely determined by convergence inµ-measure. For this reason, suppose that{fn: n≥1}converges to bothf andgin µ-measure. Then, for >0,
µ(|f−g| ≥)≤µ |f −fn| ≥ 2
+µ |fn−g| ≥ 2
−→0 asn→ ∞.
Hence, µ(f 6=g) = lim&0µ |f−g| ≥
= 0, and so f =g (a.e.,µ). That is, convergence in µ-measure determines the limit function to precisely the same extent as does either µ-almost everywhere or k · kL1(µ;R)-convergence.
In particular, from the standpoint ofµ-integration theory, convergence in µ- measure has unique limits.
The following theorems are intended to further elucidate the notions of µ- almost everywhere convergence, convergence in µ-measure, and the relations between them.
Theorem 3.2.7. Let {fn : n ≥ 1} be a sequence of R-valued measur- able functions on the measure space (E,B, µ). Then there is an R-valued, measurable functionf for which
(3.2.8) lim
m→∞µ
sup
n≥m
|f−fn| ≥
= 0 for all >0
if and only if
(3.2.9) lim
m→∞µ
sup
n≥m
|fn−fm| ≥
= 0 for all >0.
Moreover, (3.2.8) implies that fn −→ f both (a.e., µ) and in µ-measure.
Finally, when µ(E) <∞, fn −→ f (a.e.,µ) if and only if (3.3.8) holds. In particular, on a finite measure space,µ-almost everywhere convergence implies convergence inµ-measure.
Proof: Set
∆ =n
x∈E: lim
n→∞fn(x) does not exist inR o
. For m ≥ 1 and > 0, define ∆m() =
supn≥m|fn−fm| ≥ . It is then easy to check (from Cauchy’s convergence criterion forR) that
∆ =
∞
[
`=1
∞
\
m=1
∆m 1` .
Since (3.2.9) implies that µ T∞
m=1∆m()
= 0 for every > 0, and, by the preceding,
µ(∆)≤
∞
X
`=1
µ
∞
\
m=1
∆m 1`
! ,
we see that (3.2.9) does indeed imply that {fn : n≥1} convergesµ-almost everywhere. In addition, if (cf. the second part of Lemma 3.2.1) f is an R-valued, measurable function to which {fn : n ≥ 1} converges µ-almost everywhere, then
sup
n≥m
|fn−f| ≤ sup
n≥m
|fn−fm|+|fm−f| ≤2 sup
n≥m
|fn−fm| (a.e.,µ);
and so (3.2.9) leads to the existence of anf for which (3.2.8) holds.
Next, suppose that (3.2.8) holds for somef. Then it is obvious thatfn−→
f both (a.e.,µ) and in µ-measure. In addition, (3.2.9) follows immediately from
µ
sup
n≥m
|fn−fm| ≥
≤µ
sup
n≥m
|fn−f| ≥ 2
+µ
sup
n≥m
|f−fm| ≥ 2
. Finally, suppose that µ(E) < ∞ and that fn −→ f (a.e,µ). Then, by (2.1.11),
m→∞lim µ
sup
n≥m
fn−f ≥
=µ
∞
\
m=1
sup
n≥m
fn−f ≥
!
= 0
3 Lebesgue Integration
for every > 0, and therefore (3.2.8) holds. In particular, this means that fn−→f in µ-measure.
Clearly, the first part of Theorem 3.2.7 provides a Cauchy criterion for µ- almost everywhere convergence. The following theorem gives a Cauchy crite- rion for convergence inµ-measure. In the process, it shows that, after passing to a subsequence, convergence in µ-measure leads to µ-almost everywhere convergence.
Theorem 3.2.10. Again let {fn : n ≥ 1} be a sequence of R-valued, measurable functions on the measure space (E,B, µ). Then there is an R- valued, measurable functionf to which{fn : n≥1} converges inµ-measure if and only if
(3.2.11) lim
m→∞sup
n≥m
µ |fn−fm| ≥
= 0 for all >0.
Furthermore, if fn −→ f in µ-measure, then one can extract a subsequence {fnj : j≥1} with the property that
i→∞limµ
sup
j≥i
|f−fnj| ≥
= 0 for all >0;
and thereforefni−→f (a.e., µ)as well as inµ-measure.
Proof: To see thatfn −→f inµ-measure implies (3.2.11), simply note that µ
fn−fm ≥
≤µ f−fn
≥2 +µ
f −fm ≥2
.
Conversely, assume that (3.2.11) holds, and choose 1≤n1<· · ·< ni <· · · for which
sup
n≥ni
µ
|fn−fni| ≥2−i−1
≤2−i−1, i≥1.
Then µ
sup
j≥i
|fnj−fni|>2−i
≤µ
[
j≥i
|fnj+1−fnj| ≥2−j−1
≤
∞
X
j=i
µ |fnj+1−fnj| ≥2−j−1
≤2−i. From this it is clear that{fni : i≥1}satisfies (3.2.9) and therefore that there is an f for which (3.2.8) holds with {fn : n≥1} replaced by {fni : i≥1}.
Hence,fni−→f bothµ-almost everywhere and inµ-measure. In particular, when combined with (3.2.11), this means that
µ(|fm−f|> )≤ lim
i→∞µ |fm−fni| ≥ 2 + lim
i→∞µ |fni−f| ≥ 2
≤ sup
n≥m
µ |fn−fm| ≥2
−→0
as m→ ∞, and sofn−→f in µ-measure.
Notice that the preceding argument proves the final statement as well.
Namely, if fn −→ f in µ-measure, then (3.2.11) holds and therefore the ar- gument just given shows that there exists a subsequence {fni : i≥1} that satisfies (3.2.9). But this means that {fni : i ≥ 1} converges both (a.e.,µ) and in µ-measure, and, as a subsequence of a sequence that is already con- verging inµ-measure tof, we conclude thatf must be the function to which {fni: i≥1}is converging (a.e.,µ).
Because it is quite important to remember the relationships between the various sorts of convergence discussed in Theorems 3.2.7 and 3.2.10, I will summarize them as follows:
kfn−fkL1(µ;R)−→0 =⇒fn−→f in µ-measure
=⇒ lim
i→∞µ
sup
j≥i
|fnj −f| ≥
= 0, >0, for some subsequence{fni}
=⇒fni −→f (a.e.,µ) and
µ(E)<∞andfn−→f (a.e., µ) =⇒ fn −→f inµ-measure.
Notice that, when µ(E) = ∞, µ-almost everywhere convergence does not imply µ-convergence. For example, consider the functions1[n,∞) onRwith Lebesgue measure.
I next show that, at least as far as Theorems 3.2.3 through 3.2.5 are con- cerned, convergence inµ-measure is just as good asµ-almost everywhere con- vergence.
Theorem 3.2.12. Let f and {fn : n ≥ 1} all be measurable, R-valued functions on the measure space (E,B, µ), and assume that fn −→ f in µ- measure.
Fatou’s Lemma: If fn ≥0 (a.e.,µ) for each n ≥1, then f ≥ 0 (a.e.,µ)
and Z
f dµ≤ lim
n→∞
Z fndµ.
Lebesgue’s Dominated Convergence Theorem: If there is an inte- grable g on (E,B, µ) such that |fn| ≤ g (a.e.,µ) for each n ≥ 1, then f is integrable,limn→∞kfn−fkL1(µ;R)= 0,and soR
fndµ−→R
f dµasn→ ∞.
Lieb’s Version of Fatou’s Lemma: If supn≥1kfnkL1(µ;R) <∞, then f is integrable and
n→∞lim
kfnkL1(µ;R)− kfkL1(µ;R)− kfn−fkL1(µ)
= lim
n→∞
|fn| − |f| − |fn−f| L1(µ;
R)= 0.
In particular,kfn−fkL1(µ;R)−→0 if kfnkL1(µ;R)−→ kfkL1(µ;R)∈R.
3 Lebesgue Integration
Proof: Each of these results is obtained via the same trick from the corre- sponding result in§3.2.1. Thus I will prove the preceding statement of Fatou’s Lemma and will leave the proofs of the other assertions to the reader.
Choose a subsequence {fnm : m≥1} of {fn : n≥1} such thatR
fnmdµ tends to limn→∞R
fndµ. Next, choose a subsequence {fnmi : i ≥ 1} of {fnm : m ≥ 1} for which fnmi −→ f (a.e.,µ). Because each of the fnmi’s is non-negative (a.e.,µ), it is now clear thatf ≥0 (a.e.,µ). In addition, by restricting all integrals to the setEb on which thefnmi’s are non-negative and fnmi −→f, we can apply Theorem 3.2.3 to obtain
Z
f dµ= Z
Eˆ
f dµ≤ lim
i→∞
Z
Eˆ
fnmidµ= lim
m→∞
Z
fnmdµ= lim
n→∞
Z
fndµ.
§3.2.3. Elementary Properties of L1(µ;R): An important dividend of