Welcome to lecture 30 on measure and integration.
In the previous lecture, we had
started defining what is called the notion
of a function to be an integrable function
and
then we started looking at some of the properties
of integrable functions.
.
Let us just recall what is an integrable function
and then we will start looking at various
properties of these integrable functions.
We will prove one important theorem called
dominated convergence theorem.
.
If you recall, we said a measurable function
f on X to extended real . measurable
function f is said to be integrable with respect
to mu and written as mu integrable if both
the integral of the positive part of the function
and the negative part of the function are
finite. We say f is mu integrable if integral
f plus into d mu and integral f minus into
d
mu both are finite numbers. In that case,
we say the integral of f is equal to integral
of f
plus minus integral of f minus.
Let us just once again emphasize saying that
function is mu integrable if and only if both
f plus and f minus are having finite integrals
and the integral of f is written as integral
of
f plus minus integral of f minus. The class
of all integrable functions on the space X,
S,
mu is normally denoted by L 1 – capital
L lower 1 – of X, S, mu or sometimes we
drop S
and mu if we are clear from the context what
are the sigma algebras or what is the
measure; sometimes we just emphasize mu because
we know what is X and what is S.
These are various notations used for denoting
integrable functions – L 1 of X, S, mu or
L 1
bracket X or L 1 bracket mu; this is the space
of all mu-integrable functions.
.
We started looking at the properties of these
functions. The first important thing we
observed was a function f which is measurable
is integrable if and only if mod f which is
a nonnegative function is integrable. That
means to check whether a measurable function
is integrable or not, it is enough to look
at the integral of the function mod f and
see
whether that is finite or not and this is
always true; for the integrable function,
integral of
mod f integral mod of the integral of f into
d mu is less than or equal to integral of
mod f
into d mu.
This is an important criteria; this is an
equivalent way of defining integrability of
a
measurable function, namely mod f is measurable.
This is not equal; this is wrong here; it
should be less than or equal to; mod of integral
f into d mu is less than or equal to; this
is
a typing mistake here; this should have been
less than or equal to integral of mod f into
d
mu.
Let us recall some of the other properties
that we had proved. We said if f and g are
measurable functions and mod f is less than
or equal to g of x for almost all x with
respect to mu and g is integrable, then f
is also integrable; that means if a function
f of x
is dominated by an integrable function, then
that measurable function automatically
becomes integrable.
.
We also proved the following property: if
two functions f and g are equal almost
everywhere and one of them is integrable,
say f is integrable, then the function g is
also
integrable and integral of f is equal to integral
of g. That essentially says that the
integrable of the function does not change
if the function is changed, if the values
of the
function are changed almost everywhere.
So, f is equal to g almost everywhere; f and
g are measurable functions; one of them, say,
f is integrable implies g is integrable and
the integrable of the two are equal. We also
proved the following property: if f is an
integrable function and alpha is any real
number,
then alpha times f is also integrable and
the integral of alpha f is equal to alpha
times the
integral of f. We continue this study of properties
of integrable functions.
.
Next we want to check the integrability property;
if f and g are integrable functions, then
we want to show that f plus g is also integrable
and integral of f plus g is equal to integral
f plus integral g.
.
To prove this property, let us look at what
we are given. We are given that f and g are
integrable functions; that is, integral mod
f into d mu is finite and integral of g into
d mu
is also… absolute value of g with respect
to mu is also finite. To check whether the
function f plus g is integrable or not, we
have to look at the integral of f plus g,
the
.bsolute value, and show that integral of
absolute value of f plus g is also finite.
That
follows easily because absolute value of f
plus g is always less than or equal to absolute
value of f plus absolute value of g.
All are nonnegative measurable functions.
So, using the property of the integral for
nonnegative measurable functions, this implies
that integral of mod f plus g into d mu is
less than or equal to integral of mod f plus
mod g into d mu and that by linearity is the
same as integral mod f into d mu plus integral
mod g into d mu. We are given that both
of them are finite; so, this is finite .. It
implies that f plus g is
integrable.
.
To compute the integral of f plus g, we have
to go back to the definition of the integral.
f
and g are integrable; that implies integral
of f plus d mu is finite, integral of f minus
d mu
is finite, integral of g plus – the positive
part of g – is finite, and integral of g
minus d mu
is finite. We have to show that integral (f
plus g) plus into d mu is finite and integral
(f
plus g) minus into d mu is finite; these two
properties we have to show. To show this,
somehow we have to relate the positive part
of f plus g with the positive part of f and
positive part of g and similarly, the negative
part of f plus g with the negative part of
f
and negative part of g; that is done as follows.
.
What we do is look at f plus g. By definition,
we can write it as f plus g positive part
minus f plus g the negative part; that is
by the definition of the positive part and
the
negative part of a function. Also, f plus
g we can also write it as shown; decompose
f
into positive part and into the negative part
– that is, f plus minus f minus – and
similarly
write g as g plus minus g minus.
From these two, it follows that integral of
f, sorry not the integral; from this, it follows
that f plus g positive part minus f plus g
the negative part is equal to f plus minus
f minus
plus g plus minus g minus; from these two
equations, it follows this is so. Now, what
we
do is all the negative terms we shift on the
other side of the equation; this implies that
(f
plus g) plus plus f minus plus g minus is
equal to f plus plus g plus from here and
this
term on the other side will give me plus (f
plus g) minus.
We also rearrange the terms; now, you observe
that the left-hand side is a nonnegative
function and the right-hand side is a nonnegative
function. By the properties of integrals
for nonnegative functions, this implies that
integral of (f plus g) plus into d mu plus
integral of f minus into d mu plus integral
of g minus into d mu so, that is the integral
of
the left-hand side, is equal to integral of
f plus into d mu plus integral g plus into
d mu
plus integral f plus g minus into d mu.
From this equation by using the properties
of integral for nonnegative functions, the
linearity property, the integral of the left-hand
side is equal to integral of the right-hand
.ide .. Integral of the left-hand side consists
of integral of (f
plus g) plus plus integral of f minus plus
integral of g minus and that is equal to integral
of f plus plus integral of g plus plus integral
of (f plus g) minus.
Now, we observe that in this equation all
the terms are finite quantities or real numbers;
that is because f plus g we have already shown
is integrable; this first integral of (f plus
g) plus is finite, integral f minus is finite
and similarly all the terms are actually
nonnegative real numbers. We can again manipulate
them and shift terms on the lefthand side
and right-hand side. What we will do is this
term (f plus g) minus on the righthand side
. we will bring it on the left-hand side and
the terms
integral f minus into d mu and integral g
minus into d mu we shift it on the right-hand
side; that gives us the property.
.
Shifting implies that integral of (f plus
g) plus into d mu minus this term will give
you
integral (f plus g) minus into d mu; this
term we have shifted ..
Shift these two terms on the other side. It
is equal to integral f plus into d mu; that
is this
term and bringing this integral of f minus
on this side will give you integral f minus
into
d mu plus integral of g plus into d mu which
is already there and integral of g minus
from the left-hand side will give you integral
of g minus into d mu.
. This rearrangement of the terms here once
again gives you
that integral of (f plus g) plus into d mu
and the integral of the negative part of f
plus g is
.qual to integral of f plus minus integral
f minus plus integral g plus. Now by the
definition, the left-hand side is nothing
but integral of f plus g into d mu and the
righthand side is integral f into d mu plus
integral g into d mu.
That proves the linearity property of the
integral that if f and g are integrable functions,
not only f plus g is integrable but integral
of f plus g is equal to integral of f plus
integral
of g into d mu; that is the linearity property
of the integral .. We
have proved the basic properties of the integrals,
namely the integral of a function which
is integrable, of course it is a finite quantity
and it is linear; if you take a function f
multiplied by a scalar alpha, then alpha times
f is integrable and the integral of alpha
f is
equal to alpha times integral of f. Similarly,
if f and g are integrable, then f plus g is
integrable and the integral of f plus g is
equal to integral of f plus integral of g.
Let us
look at some more properties of this integral
which are going to be useful later on. Let
us
look at the next property.
.
For an integrable function f in L1 of mu,
let us look at we have already shown; if mod
f
is a nonnegative measurable function and if
you multiply it by the indicator function
of a
set E, then we already shown that this is
again a nonnegative measurable function; of
course, this function is less than or equal
to integral of mod f. So, nu of E is going
to be
always a finite quantity.
.he claim is nu is a measure, in fact a finite
measure, and it has the property that mu of
E
equal to 0 implies nu of E equal to 0. Whenever
a set E has got mu measure 0, the
measure of nu is also going to be equal to
0. This basically follows from the properties
of
the integral for nonnegative functions because
if f is integrable, then mod f is a
nonnegative measurable function and its integral
is a finite quantity.
So, nu of E is a finite measure for every
E, the function chi E times mod of f is less
than
or equal to mod f; this integral is going
to be less than or equal to integral of mod
f which
is finite. Obviously for nonnegative functions,
we have already proved this property: if
mu of E is 0, then the integral over E is
equal to 0; this property follows from our
earlier
discussions.
.
Let us look at the integral for the integrable
function – not the integral of mod f but
let us
look at integral of f times indicator function
of E. If you recall, we had already shown
that if f is measurable and E is a set in
the sigma algebra, then chi E times f is again
a
measurable function and just now we observed
that this number is going to be a finite
number because this is again an integrable
function.
.
Let us just observe this property once again
that if f belongs to L1 of mu and E is a set
in
the sigma algebra, then this implies that
chi E times f is a measurable function; that
we
have already seen; because f is a measurable
function and indicator function of E is a
measurable function, the product of measurable
function is measurable. We observed just
now that if you look at the absolute value
of chi of E times f, that is the same as indicator
function of E because that is negative into
absolute value of f.
This implies that the integral of chi E times
f absolute value into d mu is less than actually
is equal to integral chi E of mod f into d
mu which is less than or equal to integral
mod f
into d mu which is finite. What does that
imply? This implies that integral chi E into
d
mu, this we are denoting it by nu tilde of
E, is a real number – a finite real number.
That
is the observation and we want to claim that
mu of E is equal to 0 implies that nu tilde
of
E is also equal to 0 ..
.
The claim is that the claim is that this property.
Let us prove this property. Suppose mu
of E is equal to 0, then what is nu tilde
of E? nu tilde of E by definition is integral
chi E of
f into d mu which is same as the integral
of chi E f plus into d mu minus integral chi
E f
minus into d mu. Now, we observe mu of E equal
to 0, chi E of f plus is a nonnegative
function and properties of nonnegative functions
imply if the set has got measure 0, then
the integral of this is equal to 0.
The first integral is equal to 0 and the second
integral is equal to 0 by properties of
integrals of nonnegative measurable functions.
It implies nu tilde of E is equal to 0. What
we are saying is mu of E equal to 0 implies
nu tilde of E is also equal to 0 (Refer Slide
Time.; that is the property we are proving
here. Keep in mind that nu tilde of E is
defined as a real number for every E belonging
to S but it is not a nonnegative number
because f may not be a nonnegative function;
so, we cannot say nu tilde of E is a
measure. We will look at this property a bit
later; it may not be a measure but it has
some
properties similar to a measure.
.
Here is another important property. Let us
look at again the same value nu tilde of E
which is equal to integral of f over E d mu.
Suppose this is equal to 0 for every set E
in
the sigma algebra, then the claim is this
function f must be equal to 0 for almost all
x
belonging to mu.
.
Let us prove this property. Given nu tilde
of E which is nothing but integral chi E times
f
into d mu is equal to 0 for every E belonging
to S; that is what is given to us. We want
to
show that if you take the set N which is x
belonging to X such that 
mod f of x bigger
.han 0 if we write the set N, then note that
this set N is a set in the sigma algebra and
we
want to show that mu of N… We want to show
f is 0 almost everywhere and this is the
set where f is not 0. So, we want to show
this is equal to 0 ..
This is the problem we want to show.
Let us look at consider the consider consider
the set say for example, let us look at. Let
us write, say, A n to be the set where x belongs
to X such that f of x is bigger than 1 over
n. Similarly, let us write B n to be the set
of x belonging X where f of x is less than
minus
1 by n. Now, the claim is that the set N is
nothing but union over A n union over B n
, n
equal to 1 to infinity union of union n equal
to 1 to infinity. That means all these sets
A n s
and B n s if you take their unions, that is
precise as a set N where N is… What is the
set
N? N is the set where f of x is not equal
to 0.
If f of x is not equal to 0, then either f
of x is positive or f of x is negative. If
it is
positive, then it is going to be bigger than
1 over n for some n. If x is positive and
bigger
than 1 over n, it is going to belong to 1
over n or if f of x is not 0 and it is negative,
that
means it is negative so it is going to be
less than minus 1 over n for some n; so, it
belongs to B n .
Every point x in N either belongs to A n or
belongs to B n and obviously if x belongs
to
A n or B n , then f of x is not equal to 0;
it belongs to N; so, N is equal to this. N
is written
as a countable union of sets and all of these
are sets in the sigma algebra S. We want to
show this union has got measure 0 .. In case
mu of N is not 0,
that will mean for some n, either A n has
got positive measure or B n has got positive
measure, because otherwise mu of N will be
less than or equal to sigma mu of A n s plus
sigma mu of B n s, all of them equal to 0.
.
What we are saying is the following: to show
that mu of n is equal to 0. Suppose mu of
n
is bigger than 0, then this condition implies
there exists some n 0 such that either mu
of A
A n0 naught is bigger than 0 or mu of B n
0 is bigger than 0, because if not then mu
of N
will be equal to 0. Let us look at these conditions.
The first one holds; if mu of A n0 is
bigger than 0, then look at the integral;
then integral of f over the set A n0 let us
look at;
let us look at integral of f over the set
An 0 .
That is equal to integral so which is same
as integral chi An0 times f into d mu. On
the set
A n0 , f is bigger than 1 over n. This is
bigger than obviously integral 1 over n 0
times mu
of A n0 . Let us observe that on the set A
n0 … Outside A n0 , this function is equal
to 0
.; the indicator function of A n0 times f
is 0; on A n0 , f is bigger
than 1 over n 0 . So, this function is bigger
than 1 over n 0 and outside A n0 it is 0.
This is
going to be bigger integral over A n0 of 1
over n 0 into d mu; that is what we are saying.
This is nothing but this integral and that
is bigger than 0. So, in case mu of A n0 is
bigger
than 0, integral of f over A n0 is going to
be bigger than 0 which is a contradiction;
it is
not true because we are given integral of
f over every set E is equal to 0, which is
not
true. If this holds ., then it is a contradiction.
Similarly, if this
holds, one can approve it is a contradiction;
then the integral of f over B n0 will be strictly
less than 0 and not equal to 0. In either
case, both of these are not possible; so,
our
assumption that mu of N is bigger than 0 must
be wrong and hence mu of N so implies.
.his implies that the measure of the set N
is equal to 0. N was the set where f of x
is
bigger than 0 .. This set has got measure
0. This is what we
wanted to prove. We have proved the property
that if integral of a function over f is an
integrable function . its integral over E
is equal to 0 for every E belonging to S,
then f
must be equal to 0 almost everywhere .. This
is a very nice
property and useful property.
.
Now, let us look at the additive property
of the integral over the sets; this just now
we
have proved .. Another property is that if
f is integrable, then f
must be a finite number for almost all x – a
similar argument as before; let us prove that
property also.
.
It says if f is integrable, then this implies
that mod f of x is finite almost everywhere
x.
Once again, the idea is let us write the set
N to be the set where mod f of x is equal
to
plus infinity; so, f of x is equal to infinity
or equal to minus infinity; we have put them
together in the set N. We have to show that
the set mu of N is equal to 0. Once again
if
not, let us write N as x belonging to X such
that mod f of x is bigger than say some
quantity; it is not equal to 0.
If this is not equal to 0 and N is a set where
f of x is equal to plus infinity, let us write
f
of x bigger than n. Sorry, let us write A
n to be the set where f of x is bigger than
n Refer
Slide Time.. Then, each set A n is in the
sigma algebra. Sorry, this is not required
.ecause we want to just want to show that
f is finite .. Now,
observe that integral of mod f into d mu I
can write as integral over N mod f into d
mu
plus integral over N complement mod f into
d mu because N and N complement together
make up the whole space and just now we observed
that integral of mod f over a set E is
a measure; so, integral of mod f over the
whole space can be written as integral of
mod f
over N plus integral of mod f over N complement.
On N, mu of N is equal to 0; so, the first
integral is 0. It is equal to 0 plus N sorry
yes no
let us observe this is so, if so, If mu of
N is bigger than 0, then what will happen?
Let us
assume mu of N is bigger than 0. Then this
integral is equal to this integral plus integral
over N plus integral over N complement.
.
That means integral of mod f into d mu is
always bigger than integral over N mod f into
d mu. Integral of mod f is integral over N
plus integral over N complement; let us just
drop the second term .. So, integral of mod
f over the whole
space is going to be bigger than integral
over N f into d mu. On N, the function takes
the
value plus infinity; this is going to be equal
to plus infinity multiplied with mu of N.
If mu of N is bigger than 0, then this will
be equal to plus infinity if mu of N is bigger
than 0. That is a contradiction; that is not
possible; not true as f belongs to L1 ; so,
this
integral must be a finite quantity. Here,
we are saying in that case it will be equal
to
.nfinity. That proves that if a function is
integrable, then it must be finite almost
everywhere.
. Let us come back to the question if f is
integrable and E
belongs to the set S, then the indicator function
of E times f is integrable; that we have
just now observed. Let us write nu tilde of
E as before: the integral of f over E. We
observe that this number may not be a nonnegative
number; however, it still has a
property something similar to that of countable
additive property for measures.
.
Let us state that property that if you take
sets E n s in the sigma algebra S which are
pairwise disjoint and E is the union of the
sets, then the claim is that the series which
is
integral of E i s f into d mu summation 1
to infinity – this series is absolutely
convergent.
If we write E as the union, then integral
of f over E is equal to summation of integral
of f
over E i s. Essentially, we want to say that
the integral of f of an integrable function
over a
set E can be written as summation integral
of E i s where E i s are pairwise disjoint.
We are saying it is always possible for f
to be an integrable function. Why are we saying
absolutely convergent? Here, one should note
that when E is equal to union E i , it does
not matter whether you write as E 1 union
E 2 union E 3 and so o or any other order,
say,
E 2 union E 1 ; the union does not depend
on the order in which you write the sequence
E i .
That means in this series, the summation should
not depend upon the order of the terms.
All are nonnegative; that means we should
prove that these are absolutely convergent.
.hat is what we want to prove: if E i s are
pairwise disjoint, then the series integral
over
E i of f into d mu summation 1 to infinity
is an absolutely convergent series and this
integral of f over E is summation of integral
over E i s.
.
Let us prove this property. We have got E
i is a sequence of sets in the sigma algebra;
they are pairwise disjoint and equal to empty
set for i not equal to j; E is equal to union
of E i s 1 to infinity. The first claim is
that the series summation i equal to 1 to
infinity of
integral over E i of f into d mu is absolutely
convergent. Let us observe what absolute
convergent means; that means absolute values
of these terms is a series of nonnegative
terms that must converge.
.or that, let us note that absolute value
of integral E over E i of f into d mu is less
than or
equal to integral of mod f over E i into d
mu; that is a property of integrable functions
–
absolute value of the integral is less than
or equal to integral of the absolute value.
mod f
is a nonnegative function and E is a disjoint
union of sets; that implies that integral
over
E i of mod f into d mu if I sum it up i equal
to 1 to infinity, that is same as integral
over E
of mod f into d mu and f being integrable,
that is a finite member.
Here, we have used two things: one, for nonnegative
measurable functions, the integral
over a set is a measure; so, integral of mod
f into d mu over E i summation 1 to infinity
is
equal to integral of absolute value of f over
E and f being integrable, this is finite.
That
proves that the series integral f into d mu
over E i is absolutely convergent because
this
sum is less than or equal to this sum .. From
these two, it
implies that the series is absolutely convergent.
Once the series is absolutely convergent,
its sum is equal to sum of the partial sums.
Now, we can easily write this implies that
the
claim holds; so, the series is absolutely
convergent.
.
Hence, integral over E f into d mu is equal
to limit n going to infinity of the partial
sums,
i equal to 1 to n integral of f over E i d
mu and that is nothing but partial sums; that
is
same as saying that sigma i equal to 1 to
infinity integral over E i of f into d mu.
That
proves that though the integral of an integrable
function over a set (Refer Slide Time:
39:45) need not be a measure but we can say
this is a countable additivity property of
.his integral – that integral over E is
equal to summation of integrals over E i s
whenever
E is a union of pairwise disjoint sets E i
.; this is the property
that we have just now proved.
These were some of the properties that we
have proved about the integral of integrable
functions; now we want to prove an important
property; we want to analyze sequences of
integrable functions; we want to analyze the
property that if f n is a sequence of integrable
functions and it converges to a function f,
can we say that f is integrable and can we
say
integral of f n s will converge to integral
of f?
We have seen that this need not be true even
for nonnegative functions, but under some
suitable condition we can say that integral
of f n s will converge to integral of f and
that is
an important theorem called Lebesgue's dominated
convergence theorem. Let us prove
interchange of integral with the limits and
look at the theorem called Lebesgue's
dominated convergence theorem.
.
It says let f n be a sequence of measurable
functions such that there exists a function
g
which is integrable with the property that
integral of mod f n s are less than or equal
to g
for almost all x in mu for all n. Then the
claim is if f n s converge to f almost everywhere,
then the limit function is integrable – (i)
.. Secondly, integral of
the limit function is equal to limit of the
integrable functions.
.et us observe once again what is given and
what is true. We are saying here f n is a
sequence of measurable functions and all the
f n s are dominated by a single function g
which is integrable and this dominance could
be almost everywhere. All the f n s are
dominated by a single function g and the conclusion
is if f n s converge to f, then f is
integrable and integral of f is equal to limit
of integral of f n s. This is what is called
dominated convergence theorem; it is an important
theorem; let us prove this theorem.
.
.e are given that all the f n s are measurable
for n bigger than or equal to 1; mod f of
x is
less than or equal to g of x for almost all
x and for every n; we are given that f n of
x
converges to f of x almost everywhere. To
prove the required claim, for the time being
let us assume that this almost everywhere
is everywhere .. The
proof is not going to change much; we will
see that (( )). Let u assume for the time
being
that mod f n of x is less than or equal to
g of x where g is in L1 is an integrable function.
That implies that integral of mod f n of x
into d mu of x is less than or equal to integral
g
of x into d mu of x; g is integrable and so
that is finite.
That implies that each f n is an integrable
function; also, mod f n converges to mod f
because f n converges to f. Each mod f n is
less than or equal to g.
This implies that mod f n of x is less than
or equal to g and that converges; that
implies that mod f of x is also less than
or equal to g of x for every x. That once
again
implies integral mod f into d mu is less than
integral g into d mu which is finite; this
again implies that f is in L 1 of mu.
Under the given conditions, we have shown
that if f n s are dominated by an integrable
function and f n s converge to f, then f is
an integrable function. To look at the limits,
let
us note f n converges to f and mod f n is
less than equal to mod g. This implies that
look at
the sequence f n minus g; look at this sequence.
This is a sequence of measurable
functions; of course, mod f n is less than
or equal to g. This will be negative; we want
.his is a sequence of nonnegative measurable
functions because mod f n is bigger than or
equal to g; that is given. So, g minus f n
is a sequence of measurable functions. Let
us
observe g minus f n is bigger than or equal
to 0 because g is bigger than or equal to
f n .
This is a sequence of nonnegative measurable
functions.
.
Let us write that g minus f n is a sequence
of nonnegative measurable functions and it
converges to g minus f because f n converges
to f. Now, we can apply Fatou's Lemma;
this implies by Fatou's Lemma that integral
limit inferior of g minus f n into d mu will
be
less than or equal to limit inferior of integral
of g minus f n into d mu; that is the
application of Fatou's Lemma. Recall we had
Fatou's Lemma which was applicable for a
sequence of functions which is not necessarily
increasing.
Look at this. Now, let us compute both sides.
What is the left-hand side? This is equal
to
integral limit infimum of g minus f n is g
minus limit inferior plus limit inferior of
minus
f n ; that is the left-hand side d mu. That
is equal to integral of g that is equal to
integral of
g into d mu. What can we say about this . f
n s all are integrable
function; so, everything is finite. This limit
inferior of minus f n is equal to minus limit
superior of f n s into d mu.
This is a property of limit superior and limit
inferior that limit inferior of minus f n
is
equal to minus of limit superior; so, this
is equal to integral of g into d mu minus
integral
limit superior of f n s; limit f n is convergent
and so limit superior is same as f of x into
d
.u .; that is the left hand side. Let us see
what the right-hand side is. Once again,
limit inferior of integral and so this is
equal to limit inferior of integral; integral
of g
minus f n is integral g and that does not
depend upon limit; so, it is integral g into
d mu
and then limit inferior of minus; that will
be minus limit superior of integral f n into
d mu.
From these two, this is less than or equal
to this. What does that imply?
.
That implies that integral g into d mu minus
integral f into d mu, I am just writing
integral this ., is less than or equal to
integral g into d mu minus
limit superior of integral f and d mu. Everything
is finite and so I can cancel out these;
negative sign gives you the other way inequality.
It implies integral f into d mu is bigger
than or equal to limit superior of integral
f n into d mu.
. Looking at the sequence g minus f n , we
got that g minus f n is
nonnegative converges to g minus f gives us
this .. Similarly, if
I look at the sequence g plus f n , that is
again a sequence of nonnegative measurable
functions and application of Fatou's Lemma
will give me that integral of f into d mu
is
less than or equal to limit inferior of integral
f n into d mu; a similar application of Fatou's
Lemma to this sequence will give me this ..
This is (1) and this is (2). (1) plus (2)
together imply that integral f into d mu is
bigger
than limit superior; that is always bigger
than limit inferior and that is bigger than
.ntegral f into d mu; it implies that limit
integral f n d mu exists, this limit exists
and this
is equal to integral f into d mu ..
That proves the dominated convergence theorem.
The proof of the dominated
convergence theorem is essentially very simple;
it is just a straightforward application of
Fatou's Lemma because mod f n is less than
or equal to g implies g minus f n and g plus
f n
both are sequences of nonnegative measurable
functions; apply Fatou's Lemma and you
have the conclusion that integral of f is
equal to limit of integral of f n s. We have
proved
this under the conditions that f n of x converges
to f of x and f n of x is dominated by g of
x for every x. The modification for this for
almost everywhere . is simple and we will
do it next time. Thank you very much.
