Real Analysis Notes
Thomas Goller
September 4, 2011
Contents
1 Abstract Measure Spaces
1.1 Basic Definitions . . . .
1.2 Measurable Functions .
1.3 Integration . . . . . . .
1.4 Counterexamples . . . .
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2 Banach Spaces
2.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Lp Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 `p Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Uniform Boundedness Principle
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10
Chapter 1
Abstract Measure Spaces
1.1
Basic Definitions
The discussion will be based on Stein’s Real Analysis.
Definition 1. A measure space consists of a set X equipped with:
1. A non-empty collection M of subsets of X closed under complements and
countable unions and intersections (a σ-algebra), which are the “measurable” sets.
µ
/ [0, ∞] with the property that if E1 , E2 , . . . is a
2. A measure M
countable family of disjoint sets in M, then
!
∞
∞
[
X
µ
En =
µ(En )
n=1
n=1
Now we state an assumption on a measure space that allows us to develop
integration as in the case of Rd .
Definition 2. The measure space (X, M, µ) is σ-finite if X can be written as
the union of countably many measurable sets with finite measure. (To see that
Rd is σ-finite, consider balls centered at the origin with radii going to infinity.)
From now on, we assume (X, M, µ) is a σ-finite measure space.
1.2
Measurable Functions
Definition 3. A function X
is measurable for all a ∈ R.
f
/ [−∞, ∞] is measurable if f −1 ([−∞, a))
Proposition 1. Some properties of measurable functions.
2
• If f, g are measurable and finite-valued a.e., then f + g and f g are measurable.
• If {fn }∞
n=1 is a sequence of measurable functions, then
lim fn (x),
n→∞
sup fn (x),
inf fn (x),
n
n
lim sup fn (x),
n→∞
lim inf fn (x)
n→∞
are measurable (lim only if it exists).
• If f is measurable and f (x) = g(x) for a.e. x, then g is measurable.
Definition 4. A simple function on X is of the form
N
X
ak χEk ,
k=1
where Ek are measurable sets of finite measure and ak are real numbers.
Proposition 2. If f is a non-negative measurable function, then there exists
a sequence of simple functions {ϕk }∞
k=1 that satisfies
ϕk (x) ≤ ϕk+1 (x)
and
lim ϕk (x) = f (x) for all x.
k→∞
In general, if f is only measurable, then we conclude that
|ϕk (x)| ≤ |ϕk+1 (x)|
and
lim ϕk (x) = f (x) for all x.
k→∞
Proposition 3 (Egorov). Suppose {fk }∞
k=1 is a sequence of measurable functions defined on a measurable set E ⊆ X with µ(E) < ∞, and fk → f a.e.
Then for each > 0, there is a set A ⊆ E, such that µ(E − A ) ≤ and fk → f
uniformly on A .
1.3
Integration
R
R
We define the Lebesgue integral X f (x) dµ(x), often written f , of a general
measurable function f by a four part process:
Definition 5. Let f be a measurable function.
PN
(i) If f (x) = k=1 ak χEk (x) is a simple function, then
Z
f :=
N
X
ak µ(Ek ).
k=1
(ii) If f is bounded and supported on a set of finite measure, then choose a
sequence of simple functions {ϕk }∞
k=1 such that |ϕk (x)| ≤ |ϕk+1 (x)| and
limk→∞ ϕk (x) = f (x) for all x. Then
Z
Z
f := lim
ϕn ,
n→∞
3
(iii) If f is non-negative, then
Z
Z
f := sup
g,
g
where the supremum is taken over all measurable functions g such that
0 ≤ g ≤ f and g is bounded and supported on a set of finite measure.
(iv) For general f , write f = f + − f − , a decomposition into non-negative
measurable functions. Then
Z
Z
Z
f := f + − f − .
Definition 6. Let f be measurable. Then f is Lebesgue integrable if
∞ in the sense of (iii) above.
R
|f | <
Proposition 4. Key properties of the Lebesgue integral are:
• Linearity. If f, g integrable and a, b ∈ R, then
Z
Z
Z
(af + bg) = a f + b g.
• Additivity. If E, F are disjoint measurable subsets of X, then
Z
Z
Z
f=
f+
f.
EtF
E
F
• Monotonicity. If f ≤ g, then
Z
Z
f≤
g.
• Triangle inequality.
Z Z
f ≤ |f |.
Proof. Prove these step by step, as in the definition of the integral, starting
with the simple functions. The triangle inequality follows from monotonicity
and linearity by considering f ≤ |f | and −f ≤ |f |.
Proposition 5. Key limit theorems.
• Fatou’s lemma. If {fn } is a sequence of non-negative measurable functions
on X, then
Z
Z
lim inf fn ≤ lim inf
n→∞
n→∞
4
fn .
• Monotone convergence. If {fn } is a sequence of non-negative measurable
functions with fn % f , then
Z
Z
fn = f.
lim
n→∞
• Dominated convergence. If {fn } is a sequence of measurable functions
with fn → f a.e., and such that |fn | ≤ g for some integrable g, then
Z
|fn − f | → 0 as n → ∞,
and consequently
Z
Z
fn →
as n → ∞.
f
Proof. Fatou’s lemma: Set f := lim inf n→∞ fn . Suppose 0 ≤ g ≤ f , where
g is bounded and supported on a set E of finite measure. Then gn (x) :=
min{g(x),
R
R fn (x)} is measurable, supported on E, and gn (x) → g(x), so that
gn → g by the
R bounded convergence theorem (combine boundedness with
Egorov
to
show
|gn − g| → 0 as n → ∞). Since gn ≤ fn , monotonicity gives
R
R
gn ≤ fn , whence
Z
Z
g ≤ lim inf
n→∞
R
fn .
R
Since f = supg g for such
R g, weRget the result.
R
Monotone
convergence:
fn ≤ f by monotonicity,
whence
R
R
R lim supn→∞ fn ≤
f . Conversely, Fatou’s lemma implies f ≤ lim inf n→∞ fn . Combining
these inequalities gives the result.
Dominated convergence: we will use the reverse Fatou’s lemma, which states
that if {fn } is a sequence of measurable functions, such that fn ≤ g for some
integrable function g, then
Z
Z
lim sup fn ≥ lim sup fn .
n→∞
n→∞
This is proved by applying Fatou’s lemma to the non-negative sequence {g−fn }.
Since |fn − f | < 2g, the reverse Fatou’s lemma implies
Z
Z
lim sup |fn − f | ≤ lim sup |fn − f | = 0,
n→∞
n→∞
whence we get the desired result using the linearity of the integral and the
triangle inequality.
1.4
Counterexamples
• A non-measurable subset of R (Stein 24). For x, y ∈ [0, 1], consider the
equivalence relation x ∼ y whenever x − y is rational. Using the axiom
5
of choice, let N be the set obtained by choosing one element from each
equivalence class. Then if {rk } is an enumeration of the rationals in [−1, 1],
we have
∞
[
[0, 1] ⊆
(N + rk ) ⊆ [−1, 2],
k=1
but the measure
P∞ of each N + rk is the same since they are translates of
N . But 1 ≤ k=1 m(N + rk ) ≤ 3, a contradiction.
f
/ R such that |f | is measurable. Let
• A non-measurable function R
N ⊆ R be non-measurable, e.g. as above. Then
(
1
x∈N
f (x) =
−1 x ∈
/N
is not measurable since f −1 ((−∞, 0)) = N c is not measurable, but |f | ≡ 1
is measurable.
• A sequence {ϕn } of non-negative bounded, measurable functions
on R
R
that converge
pointwise
to
an
integrable
function,
where
the
ϕ
do
not
n
R
converge to f . Let
(
1
−n ≤ x ≤ n
ϕn (x) = 2n
0
otherwise.
Then ϕn → 0 pointwise, but
R
ϕn = 1 for each n. See also Stein 61.
• A Cauchy sequence {fn } in the L1 norm that does not converge pointwise
a.e. Let f1 = χ[0,1] , f2 = χ[0, 12 ] , f3 = χ[ 12 ,1] , f4 = χ[0, 41 ] , etc. (Stein 92,
Ex. 12).
• A function f integrable but discontinuous at every x, even when corrected
on a set of measure 0. Set
(
x−1/2 if 0 < x < 1
f (x) =
0
otherwise.
Then if {rk } is an enumeration of the rationals,
F (x) =
∞
X
2−n f (x − rn )
n=1
is integrable, so the series converges a.e., but the series is unbounded on
every interval, and any function that agrees with F a.e. is unbounded in
any interval (Stein 92-93, Ex. 15).
6
Chapter 2
Banach Spaces
2.1
Basic Definitions
Definition 7. A metric (distance measure) on a set S is a function S × S
such that
d
1. d(x, y) ≥ 0;
2. d(x, y) = 0 ⇐⇒ x = y;
3. d(x, y) = d(y, x);
4. d(x, z) ≤ d(x, y) + d(y, z).
Definition 8. A norm on a vector space X is a function X
k·k
/ R such that
1. kxk ≥ 0;
2. kxk = 0 ⇐⇒ x = 0;
3. kαxk = |α|kxk, α ∈ F;
4. kx + yk ≤ kxk + kyk.
One can use a norm to define a metric by setting
d(x, y) := kx − yk.
The metric triangle inequality holds since
d(x, z) = kx − zk = k(x − y) + (y − z)k ≤ kx − yk + ky − zk = d(x, y) + d(y, z).
In general, a metric is not sufficient to define a norm.
Definition 9. A Banach space is a complete normed vector space. A Hilbert
space is an inner-product space that is complete in the associated norm, hence
also a Banach space. A separable Hilbert space has a countable spanning
subset, hence has a (countable) orthonormal basis by Gram-Schmidt.
7
/R
2.2
Lp Spaces
Proposition 6. Let 1 ≤ p < ∞ and (X, µ) a measure space. The set of
/ C (that agree a.e.) such
equivalence classes of measurable functions X
that
Z
1/p
p
kf kp :=
|f |
<∞
f
X
p
is a normed vector space L (X, µ).
Proof. First we check that Lp is a vector space. It is clearly closed under (e.g.
complex) scalar multiplication. For vector addition, note that |f (x) + g(x)| ≤
2 max{|f (x)|, |g(x)|}, whence |f (x) + g(x)|p ≤ 2p (|f (x)|p + |g(x)|p ).
Now we check the properties of the norm. Clearly kf kp ≥ 0 for all f , and
kf kp = 0 if and only if f = 0. Likewise, kλf kp = |λ|kf kp is clear by the linearity
of the integral. The triangle inequality is true but difficult to prove – look up
Minkowski’s inequality.
For p = ∞, consider the equivalence classes of measurable functions that are
bounded except on a set of measure 0, with the norm
kf k∞ := inf{C ≥ 0 : |f (x)| ≤ C a.e. x}.
Then
kf k∞ = lim kf kp
p→∞
holds if f is in L (X, µ) for some p < ∞. (If f ∈ Lp and f ∈ L∞ , then f ∈ Lq
for each p ≤ q ≤ ∞. For the set E on which |f | ≥ 1 must
haveR finite measure, so
R
we can bound f on this set by some M , and obtain |f |q ≤ |f |p + µ(E) · M .)
The fact that Lp (X, µ) for 1 ≤ p ≤ ∞ is complete (and hence a Banach
space) is the Riesz-Fischer theorem. When p < ∞, Lp is separable (for
X = Rd , the countable collection is rχR (x), where r is a complex number with
rational real and imaginary parts and R is a rectangle with rational coordinates).
In the special case when p = 2, L2 is in fact a Hilbert space under the inner
product
Z
(f, g) = f (x)g(x)dx.
p
A useful tool in the study of Lp spaces is Hölder’s inequality, which states
that if f and g are measurable and 1 ≤ p ≤ q ≤ ∞ such that p1 + 1q = 1, then
kf gk1 ≤ kf kp kgkq ,
namely
Z
Z
|f g| ≤
|f |p
1/p Z
|g|q
In particular, if f ∈ Lp and g ∈ Lq , then f g ∈ L1 .
8
1/q
.
2.3
`p Spaces
Consider the vector space of sequences over R or C. For 0 < p < ∞, `p is the
subspace of all sequences x = {xk } such that
X
|xk |p < ∞.
k≥1
If p ≥ 1, then
1/p

kxkp := 
X
|xk |p 
k≥1
defines a norm on `p , and `p is complete, hence a Banach space (it’s the Lp
space on the integers with the counting measure). If 0 < p < 1, then `p does
not have a norm, but still has the metric
X
d(x, y) =
|xk − yk |p .
k≥1
(To see that the natural attempt to define a norm fails to satisfy the triangle
inequality, let p = 12 and set
1
y = (0, , 0, . . . ).
2
1
x = ( , 0, 0, . . . ),
2
Then
kx + yk =
while
1
1
√ +√
2
2
2
= 2,
1 1
+ = 1.)
2 2
is the space of all bounded sequences, which is a Banach
kxk + kyk =
If p = ∞, then `∞
space under the norm
kxk∞ := sup |xk |.
k
Note that
`p ⊂ `q ⊂ `∞
for 1 ≤ p < q < ∞ (since only finitely many elements of the sequence can have
absolute value ≥ 1). Also, `2 is the only `p space that is a Hilbert space.
9
Chapter 3
Uniform Boundedness
Principle
The discussion in this chapter, based on Chapter 8 of Munkres, is geared toward
proving the Uniform Boundedness Principle.
Definition 10. A topological space X is a Baire space if: Given any countable
collection
S {An } of closed sets of X each of which has empty interior in X, their
union An also has empty interior in X.
Example 1.
• Q in the standard topology is not a Baire space. Every point is closed,
and the union of countably many points makes up the whole space.
• Z+ is a Baire space. Every subset is open, so the only subset with empty
interior is the empty set.
• Every closed subspace of R is a complete metric space, hence a Baire space.
In fact, the irrationals in R also form a Baire space. (See Exercise 6.)
Lemma 1. X is a Baire space if and only if given any countable collection
T
{Un } of open sets in X, each of which is dense in X, their intersection Un is
also dense in X.
Proof. The result follows easily from the facts that complements of open sets
are closed sets (and vice versa) and A ⊆ X is dense if and only if X − A has
empty interior.
Before proving that complete metric spaces are Baire spaces, we will need a
simple result.
Lemma 2. If X is a complete metric space, then X is regular. Namely for each
pair of a point x and a closed set B disjoint from x, there exist disjoint open
sets containing x and B, respectively.
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Proof. Since x ∈ B c and B c is open, there is an open ball Bδ (x) of radius δ > 0,
centered at x, whose closure is contained in B c . Then the open sets Bδ (x) and
X − Bδ (x) give the result.
Now we come to the first big result.
Theorem 1 (Baire Category Theorem). If X is a complete metric space, then
X is a Baire space.
Proof. Let {An } be a countable collection of closed subsets of X, each with
empty interior. We need to show that any non-empty open set U0 contains a
point not in any of the An .
Since A1 does not contain U0 , choose y ∈ U0 not in A1 . Since X is regular
and A1 is closed, we can in fact choose a neighborhood U1 of y of diameter < 1
whose closure is contained in U0 and disjoint from A1 . Now repeat by induction,
where at the nth step we choose a point of Un−1 not in An , and a neighborhood
Un of that point with diameter < 1/n and with closure contained in Un−1 and
disjoint from An .
T
Then we claim that the intersection Un is non-empty, which will give the
result. This is a result of the following lemma.
Lemma 3. Let C1 ⊇ C2 ⊇ · · · be a nested sequence T
of nonempty closed sets in
the complete metric space X. If diam Cn → 0, then Cn 6= ∅.
Proof. Choose xn ∈ Cn for each n. Then {xn } is Cauchy, and eventually lies
inside each Ck , whence the limit x is in each Ck and thus in the intersection.
Now we come to a crucial theorem.
Theorem 2 (Uniform boundedness principle). Let X be a Banach space and
Y be a normed vector space. Suppose F is a collection of continuous (bounded)
linear operators from X to Y , such that for each x ∈ X, there is an Mx ≥ 0
satisfying
sup kT xk ≤ Mx .
T ∈F
Then
sup kT k < ∞.
T ∈F
The gist of the theorem is that we want a sufficient condition to ensure the
norms of a collection of operators are bounded. An obvious necessary condition
is that the norms are pointwise bounded in the assumed sense, which turns out
to be sufficient as well.
Proof. For n ∈ Z≥1 , define
An :=
x ∈ X : sup kT xk ≤ n .
T ∈F
11
Then An is closed (since X is complete and each T ∈ F is continuous), contains
0, and is symmetric about the origin. Thus if some An contains a non-empty
open set, then it contains a ball of radius δ > 0 centered at the origin, whence
for any x ∈ X,
x δ 2kxk
2kxk T
≤
kT xk =
n,
δ kxk 2 δ
whence kT k ≤ 2n/δ, giving the result. So we may assume that each An has
empty interior.
S
But X = An by the
S pointwise boundedness assumption, so by the Baire
category theorem, X = An has empty interior, contradiction.
Corollary 1. If a sequence of bounded linear operators on a Banach space
converges pointwise, then these pointwise limits define a bounded linear operator.
12