3Uniform continuity and integration
IB Analysis II
3.3 Non-examinable fun*
Since there is time left in the lecture, we’ll write down a really remarkable result.
Theorem (Weierstrass Approximation Theorem*). If f : [0, 1] → R is continu-
ous, then there exists a sequence of polynomials (
p
n
) such that
p
n
→ f
uniformly.
In fact, the sequence can be given by
p
n
(x) =
n
X
k=0
f
k
n
n
k
x
k
(1 − x)
n−k
.
These are known as Bernstein polynomials.
Of course, there are many different sequences of polynomials converging
uniformly to
f
. Apart from the silly examples like adding
1
n
to each
p
n
, there
can also be vastly different ways of constructing such polynomial sequences.
Proof. For convenience, let
p
n,k
(x) =
n
k
x
k
(1 − x)
n−k
.
First we need a few facts about these functions. Clearly,
p
n,k
(
x
)
≥
0 for all
x ∈ [0, 1]. Also, by the binomial theorem,
n
X
k=0
n
k
x
k
y
n−k
= (x + y)
n
.
So we get
n
X
k=0
p
n,k
(x) = 1.
Differentiating the binomial theorem with respect to
x
and putting
y
= 1
− x
gives
n
X
k=0
n
k
kx
k−1
(1 − x)
n−k
= n.
We multiply by x to obtain
n
X
k=0
n
k
kx
k
(1 − x)
n−k
= nx.
In other words,
n
X
k=0
kp
n,k
(x) = nx.
Differentiating once more gives
n
X
k=0
k(k −1)p
n,k
(x) = n(n − 1)x
2
.
Adding these two results gives
n
X
k=0
k
2
p
n,k
(x) = n
2
x
2
+ nx(1 − x).
We will write our results in a rather weird way:
n
X
k=0
(nx − k)
2
p
n,k
(x) = n
2
x
2
− 2nx · nx + n
2
x
2
+ nx(1 − x) = nx(1 − x). (∗)
This is what we really need.
Now given
ε
, since
f
is continuous,
f
is uniformly continuous. So pick
δ
such
that |f(x) − f(y)| < ε whenever |x − y| < δ.
Since
P
p
n,k
(
x
) = 1,
f
(
x
) =
P
p
n,k
(
x
)
f
(
x
). Now for each fixed
x
, we can
write
|p
n
(x) − f(x)| =
n
X
k=0
f
k
n
− f(x)
p
n,k
(x)
≤
n
X
k=0
f
k
n
− f(x)
p
n,k
(x)
=
X
k:|x−k/n|<δ
f
k
n
− f(x)
p
n,k
(x)
+
X
k:|x−k/n|≥δ
f
k
n
− f(x)
p
n,k
(x)
≤ ε
n
X
k=0
p
n,k
(x) + 2 sup
[0,1]
|f|
X
k:|x−k/n|>δ
p
n,k
(x)
≤ ε + 2 sup
[0,1]
|f|·
1
δ
2
X
k:|x−k/n|>δ
x −
k
n
2
p
n,k
(x)
≤ ε + 2 sup
[0,1]
|f|·
1
δ
2
n
X
k=0
x −
k
n
2
p
n,k
(x)
= ε +
2 sup |f|
δ
2
n
2
nx(1 − x)
≤ ε +
2 sup |f|
δ
2
n
Hence given any
ε
and
δ
, we can pick
n
sufficiently large that that
|p
n
(
x
)
−f
(
x
)
| <
2ε. This is picked independently of x. So done.
Unrelatedly, we might be interested in the question — when is a function
Riemann integrable? A possible answer is if it satisfies the Riemann integrability
criterion, but this is not really helpful. We know that a function is integrable if
it is continuous. But it need not be. It could be discontinuous at finitely many
points and still be integrable. If it has countably many discontinuities, then we
still can integrate it. How many points of discontinuity can we accommodate if
we want to keep integrability?
To answer this questions, we have Lebesgue’s theorem. To state this theorem,
we need the following definition:
Definition (Lebesgue measure zero*). A subset
A ⊆ R
is said to have (Lebesgue)
measure zero if for any
ε >
0, there exists a countable (possibly finite) collection
of open intervals I
j
such that
A ⊆
∞
[
j=1
I
J
,
and
∞
X
j=1
|I
j
| < ε.
here
|I
j
|
is defined as the length of the interval, not the cardinality (obviously).
This is a way to characterize “small” sets, and in general a rather good way.
This will be studied in depth in the IID Probability and Measure course.
Example.
– The empty set has measure zero.
– Any finite set has measure zero.
– Any countable set has measure zero. If A = {a
0
, a
1
, ···}, take
I
j
=
a
j
−
ε
2
j+1
, a
j
+
ε
2
j+1
.
Then A is contained in the union and the sum of lengths is ε.
–
A countable union of sets of measure zero has measure zero, using a similar
proof strategy as above.
– Any (non-trivial) interval does not have measure zero.
–
The Cantor set, despite being uncountable, has measure zero. The Cantor
set is constructed as follows: start with
C
0
= [0
,
1]. Remove the middle
third
1
3
,
2
3
to obtain
C
1
=
0,
1
3
∪
2
3
, 1
. Removing the middle third of
each segment to obtain
C
2
=
0,
1
9
∪
2
9
,
3
9
∪
6
9
,
7
9
∪
8
9
, 1
.
Continue iteratively by removing the middle thirds of each part. Define
C =
∞
\
n=0
C
n
,
which is the Cantor set. Since each
C
n
consists of 2
n
disjoint closed
intervals of length 1
/
3
n
, the total length of the segments of
C
n
is
2
3
n
→
0.
So we can cover
C
by arbitrarily small union of intervals. Hence the Cantor
set has measure zero.
It is slightly trickier to show that
C
is uncountable, and to save time, we
are not doing it now.
Using this definition, we can have the following theorem:
Theorem (Lebesgue’s theorem on the Riemann integral*). Let
f
: [
a, b
]
→ R
be a bounded function, and let
D
f
be the set of points of discontinuities of
f
.
Then f is Riemann integrable if and only if D
f
has measure zero.
Using this result, a lot of our theorems follow easily of these. Apart from
the easy ones like the sum and product of integrable functions is integrable,
we can also easily show that the composition of a continuous function with an
integrable function is integrable, since composing with a continuous function
will not introduce more discontinuities.
Similarly, we can show that the uniform limit of integrable functions is
integrable, since the points of discontinuities of the uniform limit is at most the
(countable) union of all discontinuities of the functions in the sequence.
Proof is left as an exercise for the reader, in the example sheet.