Advanced Calculus
1
Convergence theorems
Definition 1.1. Given a sequence {an }∞
n=1 in R. Define
lim sup an := lim (sup{ak })
n→∞ k≥n
n→∞
lim inf an := lim (inf {ak })
n→∞
n→∞ k≥n
Remark 1.2. Note that there is a nonincreasing sequence
supan ≥ supan ≥ supan ≥ · · ·
n≥1
n≥2
n≥3
and a nondecreasing sequence
inf an ≤ inf an ≤ inf an ≤ · · ·
n≥1
n≥2
n≥3
and it is always true that
lim inf an ≤ lim sup an
n→∞
Proposition 1.3. A sequence
{an }∞
n=1
n→∞
in R converges to a if and only if
lim inf an = lim sup an = a
n→∞
n→∞
Proof. Suppose that {an }∞
n=1 converges a. Then for any > 0, there exists N ∈ N such that
if n ≥ N ,
|an − a| < Then for all n ≥ N ,
a − < an < a + and we have
sup an ≤ a + ,
inf an ≥ a − n≥N
n≥N
which imply
lim sup an ≤ a + ,
lim inf an ≥ a − n→∞
n→∞
1
Therefore,
a − ≤ lim inf an ≤ lim sup an ≤ a + n→∞
n→∞
Since is arbitrary, this implies
lim inf an = lim sup an = a = lim an
n→∞
n→∞
n→∞
For the converse, suppose that
lim inf an = lim sup an = a
n→∞
n→∞
Then for > 0, there exists N1 , N2 ∈ N such that
sup an < a + and
inf an > a − n≥N2
n≥N1
Take N = max{N1 , N2 }. Then for n ≥ N ,
a − < inf an ≤ an ≤ sup an < a + n≥N2
n≥N1
Therefore
|an − a| < which means
lim an = a
n→∞
Definition 1.4. Let X be a set. For n ∈ N, let fn : X → R be a function. We say that the
∞
sequence {fn }∞
n=1 of functions converges pointwise if for x ∈ X, {fn (x)}n=1 is a convergent
sequence in R. For n ∈ N, define
(supfk )(x) := sup{fk (x)}
k≥n
k≥n
(inf fk )(x) := inf {fk (x)}
k≥n
k≥n
(lim supfn )(x) := lim (sup{fk (x)})
n→∞ k≥n
n→∞
(lim inf fn )(x) := lim (inf {fk (x)})
n→∞
n→∞ k≥n
Throughout this section, E ⊂ R is a measurable set.
Proposition 1.5. Suppose that for each n ∈ N, fn : E → R is a measurable function.
Fix n ∈ N. If any of (supfk ), (inf fk ), (lim supfn ) and (lim inf fn ) is a function, then it is a
k≥n
k≥n
n→∞
n→∞
measurable function.
2
Proof. For a ∈ R, n ∈ N, note that
{x ∈ E|(supfk )(x) > a} =
k≥n
{x ∈ E|(inf fk )(x) ≥ a} =
k≥n
∞
[
{x ∈ E|fk (x) > a}
k=n
∞
\
{x ∈ E|fk (x) ≥ a}
k=n
which are measurable sets. Therefore (supfk ) and (inf fk ) are measurable. Furthermore, by
k≥n
k≥n
definition,
lim sup fn = inf {supfk },
lim inf fn = sup{inf fk }
n→∞
n≥1 k≥n
n→∞
n≥1 k≥n
and hence are also measurable.
Corollary 1.6. Let {fn }∞
n=1 be a sequence of measurable functions on E and f : E → R be
a function.
1. If {fn }∞
n=1 converges to f pointwise, then f is measurable.
2. Let
A := {x ∈ E|{fn (x)}∞
n=1 converges }
Then A is measurable.
Proof.
1. This follows from the equality
f = lim fn = lim sup fn
n→∞
n→∞
and the fact that lim supn→∞ fn is measurable.
2. Note that
A = (lim sup fn − lim inf fn )−1 ([0, 0])
n→∞
n→∞
which is measurable.
Theorem 1.7. (Fatou’s lemma) If {fn }∞
n=1 is a sequence of nonnegative measurable functions
on E and lim inf n→∞ fn converges almost everywhere, then
Z
Z
lim inf
fn dm ≥ (lim inf fn )dm
n→∞
E
E
3
n→∞
Proof. Let
A := {x ∈ E| lim inf fn (x) < ∞}
n→∞
and
f (x) :=
if x ∈ A
if x ∈ E − A
lim inf n→∞ fn (x),
0,
For any interval I ⊂ R,
f −1 (I) = (lim inf fn )−1 (I) or (lim inf fn )−1 (I) ∪ (E − A)
n→∞
n→∞
which is measurable. Hence f is a measurable function on E.
Let
gn := inf fk
k≥n
Then
g1 ≤ g2 ≤ g3 ≤ · · ·
and
lim gn = f on A
n→∞
Let ϕ be a simple function on E such that ϕ ≤ f , and let
En := {x ∈ E|gn (x) ≥ ϕ(x)}
Then
E1 ⊂ E2 ⊂ E3 ⊂ · · ·
and
∞
[
Ek = E
n=1
For k ≥ n, m ∈ N,
Z
ϕdm ≤
gn dm ≤
Therefore for any m ∈ N,
Z
fk dm ≤
Z
ϕdm ≤ lim inf
ϕdm = sup
and hence
n∈N
Z
k→∞
En ∩[−m,m]
Z
m∈N
fk dm
E
Z
ϕdm ≤ lim inf
ϕdm = sup
E
E∩[−m,m]
Z
E∩[−m,m]
fk dm
En ∩[−m,m]
En ∩[−m,m]
En ∩[−m,m]
Z
Z
Z
n→∞
E∩[−m,m]
fn dm
E
Since ϕ is arbitrary, we may take the supremum with respect to ϕ to get
Z
Z
f dm ≤ lim inf
fn dm
E
n→∞
4
E
Since m(E − A) = 0, this implies
Z
Z
Z
(lim inf fn )dm =
f dm ≤ lim inf
fn dm
n→∞
E
n→∞
E
E
Remark 1.8. In general,
Z
Z
fn dm 6=
lim inf
n→∞
E
lim inf fn dm
n→∞
E
For example, take fn = nχ[0, 1 ] . Then for x ∈ (0, 1],
n
lim inf fn (x) = lim fn (x) = 0
n→∞
n→∞
and
lim inf fn (0) = ∞
n→∞
This implies
Z
fn dm = 0
[0,1]
But for n ∈ N,
Z
Z
1
χ[0, 1 ] dm = n( ) = 1
n
n
[0,1]
fn dm = n
[0,1]
Therefore,
Z
fn dm = 1 6= 0
lim inf
n→∞
[0,1]
Theorem 1.9. (Lebesgue monotone convergence theorem) Given a sequence of measurable
functions {fn }∞
n=1 on E such that
0 ≤ f1 (x) ≤ f2 (x) ≤ f3 (x) ≤ · · ·
and {fn }∞
n=1 converges to f pointwise on E. Then
Z
Z
fn dm =
f dm
lim
n→∞
Proof. Since fn ≤ f ,
E
E
Z
Z
fn dm ≤
E
f dm
E
for all n ∈ N. By Fatou’s lemma,
Z
Z
Z
Z
Z
lim inf
fn dm ≤ lim sup fn dm ≤
f dm = (lim inf fn )dm ≤ lim inf
fn dm
n→∞
and hence
n→∞
E
E
Z
E
Z
f dm = lim sup
E
E
n→∞
n→∞
Z
fn dm = lim inf
E
n→∞
n→∞
5
Z
fn dm = lim
E
n→∞
fn dm
E
E
R
R
Definition 1.10. If f : E → R is measurable and both E f + dm and E f − dm are finite,
then f is said to be integrable and we define
Z
Z
Z
+
f dm :=
f dm −
f − dm
E
E
E
Theorem 1.11. (Lebesgue dominated convergence theorem) Given a sequence of measurable
functions {fn }∞
n=1 on E such that |fn | ≤ g where g : E → R is integrable. If f : E → R is a
function such that fn → f pointwise, then f is integrable over E and
Z
Z
lim
fn dm =
f dm
n→∞
E
E
Proof. Since |fn | ≤ g, we have −g ≤ fn ≤ g and hence 0 ≤ fn + g ≤ 2g. By Fatou’s lemma,
Z
Z
(f + g)dm ≤ lim inf (fn + g)dm
n→∞
E
E
Therefore,
Z
Z
f dm ≤ lim inf
fn dm
n→∞
E
E
Since g − fn ≥ 0, by Fatou’s lemma,
Z
Z
(g − f )dm ≤ lim inf (g − fn )dm
n→∞
E
So
E
Z
Z
f dm ≤ lim inf (−
−
and
fn dm)
n→∞
E
E
Z
Z
f dm ≥ lim sup
Therefore limn→∞
R
E
fn dm
n→∞
E
fn dm exists and equals to
E
R
E
f dm.
The following is a consequence of the Riemann-Lebesgue theorem which shows that
Lebesgue’s theory of integration coincides with Riemann integration for Riemann integrable
functions.
Theorem 1.12. Let f : [a, b] → R be a bounded function. If f is Riemann integrable, then
f is Lebesgue integrable on [a, b] and
Z
Z
f dm =
[a,b]
f (x)dx
a
6
b