So far we have seen simple functions, measurable
functions, and we have defined integrals for
positive simple functions and taking appropriate
supremum we defined integrals of positive
measurable functions. And we looked at some
properties like monotonicity and things like
that. And if you integrate over a set of measure
zero, you will get zero even if the function
is infinity at that set. And if the function
is zero you integrate you will get zero, even
if the set has infinite measure.
So, today will prove one or two important
theorems and integration, which will allow
us to interchange the integrals and limits
very easily. So this is one of the advantages
of Lebasque theory of integration over Riemann
integration. In Riemann integration we needed
a sequence of functions to converge uniformly
for us to interchange the limit and integrity.
That is not necessary in the case of Lebesgue
integral. So we will see that today.
So the first theorem we want to prove is Lebesgue,
Lebesgue monotone convergence theorem. So
we will call this MCT for monotone convergence
theorem. Let me write this in full detail.
So, as usual we have a triple X, F, mu. X
is any space F is a sigma algebra of subsets
of X. Mu is accountable additive measure okay.
So, let FN be the sequence of non-negative
measurable functions, measurable functions
defined on X such that FNs are increasing,
that means a F1 is less than to F2, less than
F3 etc.
So, FNs actually increase, So, FN of X at
each X would be an increasing sequence of
positive numbers so, they will have to converge,
they may go to infinity but they will, they
will converge to something. So, let us call
the limiting function F. So, let F of X to
be equal to limit and going to infinity FN
of X for every X in X. So this of course exists,
because if FNs are increasing, and then F
is of course measurable, then f is measurable,
it is a limit of measurable function.
So, that is not surprising, but what is interesting
is, is that we have a limit of N going to
infinity, integral of X, FN, D mu. So, remember,
these are all well-defined now, because FNs
are non-negative measurable functions, we
know how to define the integral of FN which
respect to a measure, this is actually equal
to integral of X, F, D mu.
So, to remark, we are interchanging limits
and integrations okay. So this is interchanging
limit and integration right, because we are
saying limit of N going to infinity integral
over X FN D mu, this is actually equal to
integral over X limit and going to infinity
FN. So, that is our F, right and d mu. So
we are saying this can be done, if you have
a sequence of increasing measurable functions
okay, non-negative increasing measurable functions.
So proof, so, this is one of the most useful
theorems you will see in measure theory, we
have three such theorems which will allow
us to interchange limits and integration we
will see one more today and then when we define
integration for real valued and complex valued
functions, the next the third term will come
okay. So, let us define again F of X to be
equal to limit and going to infinity F and
X right. So this is for every X and X. So
then F is measurable.
Then F is measurable, because it is the limit
of measurable functions, limit of measurable
functions, is measurable okay. We want to
compute the integral of F and see if it is
the limit of integral of FX, okay? Now what
do we know? We know that FNs are increasing,
so FN is less than or equal to FN plus 1 less
than or equal FN plus 2, etc.
Because of the monotonicity property of the
integral, we have integral over X, FN, D mu
is less than or equal to integral over X,
FN plus 1 D mu. So, if you look at the sequence
of positive numbers. So remember FNs are non-negative
measurable functions. And I am integrating
over X with respect to a positive measure.
So I am getting these are all positive numbers,
positive numbers.
Sometimes they can be infinity. But let us
not worry too much about it, we have a sequence
of positive numbers, let us call AN. What
we know is AN is less than to AN plus 1. So
AN is an increasing sequence of positive real
numbers. So it will converge, it may go to
infinity, or some finite number, let us say
A. Now, what do we know? We have, we also
have, we also have FN to be less than or equal
to F for every N, right? Because F is the
limit of the increase in sequence of FN.
So for each N, FN is less than or equal N,
and so integral of X FN, D mu will also be
less than or equal to integral of X, F D mu
for every N. But this is a sequence which
converges to A. So, A is less than or equal
to integral of X, F D mu. So, this is one
inequality right. So, what we have? We are
proving is that limit of N going to infinity
integral over X FN, D mu that is my A that
is less than or equal to integral over X,
F D mu.
Okay this much did not use anything too difficult
only, only thing we needed was the monotonicity
of the integral and the fact that these numbers
ANs were increasing and so, we got AN increasing
to A okay.
So, now, let us prove the other way. Suppose
zero less than to S less than to F and S is
a simple function. So recall that we defined
integral of F. So that is something, let us
recall immediately, they call integral of
non-negative function, F is the supremum over
all the simple functions less than or equal
to F integral of X, SD, right. So SD mu is
something which we know, we know how to define
this, using the measure and the expression
for S, and we take supremum over says things,
we get integral of F.
So, I am going to use that now. So, take any
single function which is between zero and
F. Let us fix a number. C less than, strictly
less than one, strictly less than one, okay.
And define, define the following set EN, this
is equal to all those points in the space
X, such that FNX is greater than or equal
to C times S of X.
So, remember S is a simple function less than
or equal to F, C is something which is less
than or equal to one. So C times S is something
strictly less than F, and I am looking at
all those X such that FNX will be greater
than or equal to C times S of X. So, it is
easy to see, you see to see that EN increases
to the whole space X. Well, why is that? So
let us Let us justify this. If I take any
X in X, I know that FNX increases to F of
X, right for at each X, we know this happens.
But if I look at C times S of X, this I know
is strictly less than F of X. So, there X
is some N naught such that C times S of X
will be less than or equal to FN naught of
X less than or equal to F of X right. Because
the FNs converts to F. So, it has to be between
this number and this number as and then becomes
bigger and bigger and so, that particular
X will be in EN. And so well because FNs are
increasing ENs are increasing. So, this simply
tells me that UN of EN is the whole space
in equal to 1 to infinity. Moreover, it increases
right, so UN is smaller than E2, smaller than
E3 and so, okay.
So, well, how do we use this? We start with
integral over X, FN D mu number, these are
positive numbers increasing we know it converges.
We need to show that actually converges to
integral of F. Well, this is of course less
than or equal to, sorry greater than or equal
to integral over EN, FN D mu, we call that
integral over EN, FN D mu is integral over
the whole space X indicator of EN times F.
So, let us, let us recall that part right,
integral over EN, FN D mu. We know by definition
this is integral over X, indicator of EN,
FN D mu, which is of course, less than or
equal to integral over X, FN, D mu right.
Because Chi, Chi EN is less than or equal
to one. So, I can by monotonicity of the integral
I can do so, we get this right. This is of
course, greater than or equal to.
Well on EN, we have and inequality, right?
So we can use that using monotonicity of the
integral again, we get the constancy that
will come out of the integral, integral over
EN as D mu right. Because S, C times S is
smaller than FN on the set EN. So we use the
monotonicity of the integral again there okay.
So now we are in good shape because the right
hand side now. Well, what will happen to this
so let us, let us write down those again?
I have integral over X, FN D mu to be greater
than or equal to C times integral over EN,
S, D mu okay. So let us recall the result
we proved in the last class. So we proved
that if S is non-negative and simple. Then
if I define U of A to be equal to integral
in over A has D mu, then mu is a measure,
we prove that okay. So this is a measure and
ENs increased to the whole space, ENs increased
to whole space okay.
So mu of EN, we know by the property of the
measure will have to increase to mu of X.
So this increases as N goes to infinity to
C times integral over X, S, D. But for each
and I know this is bounded by integral over
X, FN D mu, right? So this is of course, less
than or equal to sorry, let us put this. So,
I know that this, this goes to our number
A right. This is what we call A. let us see
if it is here, yes. So, the A is simply the
limit of these integrals.
So, this goes to A. So, what we have is A
greater than or equal to C times. So, let
us, lets write that separately, so, that it
is clear. So, let us recall A. A is simply
limit of N going to infinity integral X, FN
D mu, what we have just proved is that A is
greater than or equal to C times integral
over X, S D mu. So, what was C? C was less
than one some fixed constant, what was S?
S was less than or equal to F.
So, this is true for all C and all S like
this right, any C between zero and one any
S between zero and F, we have this inequality.
Left hand side is independent of everything.
So, I can take supremum over C and supremum
over S less than or equal to F and I will
on the right hand side I will get the integral
of F. So, this implies that A which is 
the limit of FNs. FN D mu. I know is greater
than or equal to integral over X, F D mu right.
So, this is one inequality and the other way
inequality was the easy part which we did
right, which uses only the monotonicity. So,
we had this inequality. So, limit of FNs to
be less than or equal to integral of F, and
now we have limit of FNs to be greater than
or equal to limit. So, this tells me that
limit of N going to infinity X FN D mu is
actually equal to integral of X, F D mu okay.
So, this is the monotone convergence there,
if you have increasing sequence of measurable
functions. Then you have convergence in the
integral okay that is what monotone convergence
theorem tells you, but you have seen. So,
let me remark, you have already seen an instance
of this when we started with a measure, so,
we call. Suppose I have a measure mu and if
I have sequence of measurable sets A1 smaller
than A2 contained in A3 etc right.
So, this is what we call AN increasing an
increases to A. What is A? That is the union
AJ, J equal to infinity. In this case, you
know that mu of AN increases to mu of A. This
was a property, countable additivity property
of mu. Well now, this is part of the monotone
convergence theorem because what is mu AN,
we are looking at Chi A, AN. I know AN increases
to A. So Chi AN will increase to Chi A.
And by MCT, the monotone convergence theorem,
we know that integral of Chi AN, D mu will
increase to integral over Chi A, D mu right,
over the whole space X which is say I am saying
mu of AN increases to mu of A. So, you have
seen an instance of the monotone convergence
theorem for indicator functions okay. So,
this is a much more general theorem, which
is applicable to all positive measurable functions.
So, there are immediate corollaries to this,
which are very useful and you will also see
something which you have probably seen earlier.
So, corollary. So, again, we have this triple
and let us say we have a sequence of measurable
functions to infinity measurable okay. Define
F of X equal to summation FN of X in equal
to infinity. So I am adding positive numbers,
so it will exist it may be finite or it may
be infinite okay.
Then F is measurable and the more important
part is integral of F, D mu is the sum of
integrals of FNs okay. So let us corollary
one, let us prove this. So, positive part
is important otherwise, there are things that
can go wrong, one has to be careful about
convergence and so. Well, so how does one
prove this?
So let us start with the simpler case. Suppose
I have two functions, suppose F1, F2 non-negative
measurable functions, negative measurable
functions. Then I know that there are sequence
of, so there X is sequence of non-negative,
non-negative simple functions. So, we will
call them SJ1 and SJ2 okay. SJ1 will convert
F1, SJ2 converges to F2. So, we know this
for any non-negative measurable function,
I know there is a sequence of simple functions
increasing to that function.
But now by monotone convergence theorem, I
know that integral of SJ1 D mu will increase
to integral over F1 D mu and integral of SJ2
D mu, the second sequence will converge to
the second function. So, we know this by MCT
okay. But for simple functions we already
know linearity. So, remember that, so, we
proved that, so let us recall that if S and
T are simple. Then we know that integral of
S plus T, D mu equal to integral over X, S,
D mu plus integral over X, T, D mu, we did
this in the last class for simple functions.
So we use that.
So if I look at integral over X, SJ1 plus
SJ2 D mu. I know that this is equal to integral
over X SJ1 D mu plus integral over X SJ2 D
mu. So we can use monotone convergence theorem
again.
Right? Because SJ1 plus SJ2 will increase
to F1 plus F2. So by MCT, we have integral
over X SJ1 plus integral over sorry, SJ2 D
mu will increase to integral over X, F1 plus
F2, this is by MCT. But I know that this is
equal to because it is linear on simple functions.
This is simply the sum of two things plus
integral over X SJ2 D mu. And I know that
this increases to integral over X, F1, D mu
plus integral over X, F2 D mu.
So, remember we started this also converges
to this. So, we have what we have proved is
that linearity of integral over positive function.
So, whenever you have positive functions measurable,
you can add them up. So, of course, if you
have two, three functions of, you will get
three pieces and so on right. Finitely many
functions you can always do this.
So, now, let us get back to the proof of,
proof of this. So, we are looking at countably
many simple, measurable positive measurable
functions. We are, we are looking at F of
X equal to the sum of these things and we
want to say we want to prove this right, this
is what we want okay. So, we use monotone
convergence theorem again okay.
So, we use monotone convergence theorem again
okay? So recall we have F1, F2, F3, etc right?
This is what we call F. So let us take the
partial sums, so let us call that GN. GN of
X to be F1X plus F2X plus F3X plus etc, etc
plus FNX right. Then what we know is FJs are
all positive. So GNs are increasing right?
Where will they increase to? They will increase
to the whole sum which is F. So GNs increased
to F. So, by MCT, we will have integral of
GN D mu will increase to integral over X,
F, D mu.
But what is this GN is finite sum, we just
proved that if I have two functions, I know
how to the linearity of the integral or two
functions we know. So, similarly, for finite,
finite sum we have the same result. So, this
one would be simply summation J equal to one
to capital N integral over X, FJ, D mu right.
But as N goes to infinity what will happen
to this? So, as N goes to infinity this goes
to the sum J equal to one to infinity integral
over X, FJD right.
So, this and this will have to be same. So,
that is what precisely or statement of the
proof, statement of the theorem right. If
I have F to be the sum of FNs, then the integral
of F is the sum of the integral of FNs, right.
Remember, they are all positive measurable
functions when you apply monotone convergence
theorem, you have to be careful, you apply
it to increasing sequence of functions.
Okay, so, let us look at another corollary.
This is something which you have probably
seen when you studied series of a positive
numbers and so on. Suppose I have numbers
AIJ positive I equal to 1, 2, 3 etc. similarly,
J equal to 1, 2, 3 etc. okay. Then summation
AIJ, J equal to one to infinity, summation
I equal to one to infinity, this is the same
as summation J equals one to infinity, summation
I equal to one to infinity AIJ.
So, remember we are changing the order of
the summation, changing the order of the summation
right. So, whenever AIJ are positive, you
can do that. So, that is an easy consequence
of the monotone convergence theorem. So, I
will leave it as an exercise to you. Well,
how do you see this? So, this is remember
summation is an integral right and we just
proved that if I add up functions I can change
summation and integrals, right that is what
we proved? So, use this use this result, use
this result and make sure that you are integral
over X is the summation.
So, remember the, if you look at natural numbers
with counting measure, then each of those
can be viewed as an integral. And so, you
have one summation and an integral and you
know how to interchange them using monotone
converges. So, use corollary one okay and
MCT.
Okay, so the first part we will stop with
the following lemma, which is another important
result, it is called Fatous lemma. So, as
usual, we have X, F, mu, and I have FN. Again,
positive measurable functions, measurable
okay. Then integral over X lim inf of FN,
D mu, this is less than or equal to lim inf
of integral FN, D mu okay.
So remember these are number and I am taking
the lim inf, here it is the function lim inf
FN. So let us recall that lim inf FN at X
is well, it is supremum over N infimum over
K greater than or equal to N, FKX okay. So,
the proof of this is two lines from using
MCT. So, all we do is, we look at this as
a collection of functions for each N. So,
define GN of X to be infimum over K greater
than or equal to N. So, for each N, you fix
N and look at all K greater than or equal
to N and look at FKX right, take the infimum
of that. These are positive ones. So, you
will have an infimum which may be zero does
not matter, but you will have an infimum.
Now you are looking at infimum of various
things. And as N increases the set you are
looking at us various smaller and smaller.
So the infimum will increase, so GN are measurable
of course, it goes supremum and infimum of
measurable functions are measurable and GNs
are increasing, GN are increasing. So it will
convert, where does it convert to the? GN
increases to supremum of GNs right, because
GNs are increasing, but supremum of GN is
precisely the lim inf of FN.
So GN increases to lim inf for FN. So, by
MCT, the monotone convergence theorem, integral
over X, GN D mu will converge to integral
over X lim inf of FN, okay.
So, let us use one inequality here. So, let
us recall them GN, so remember GN was GN of
X is infimum of K greater than or equal to
N FKX. And so this is of course less than
or equal to FNX right? Because I am looking
at, so what is this is infimum of FN, FNX,
FN plus 1X, FN plus 2X etc. So, FNX is one
such element and so infimum will be less than
or equal to that.
So this tells me again by monotonicity of
the integral, I have integral over X GN and
D mu to be less than or equal to integral
over X FN D mu, I know that this converges
to integral over X lim inf of FN D mu and
so of course, I can say this is less than
or equal to the lim inf of integral over X
FN D mu right. These are so what am I using?
If I have a sequence AN and so, this is a
simple exercise AN and BN positive, positive
sequences okay. I know that AN converges to
A or AN increases to A. AN increases to A
and AN is less than to BN okay. So this implies
A is less than to lim of BN okay that is all
we are using here.
So, this is called Fatous lemma. So Fatous
lemma is an inequality. It is called Fatous
in lemma, integral lim inf is less than or
equal to lim inf of integral, this is another
extremely useful result so. So you have just
seen two results were interchanging of the
integral and limits are in the changing of
the limits and integrals are involved. First
is the monotone convergence theorem, which
allows you to interchange the limit and integrals,
if you have a sequence of increasing measurable
functions, okay. If you have a sequence of
positive measurable functions, you have an
inequality which is given by Fatous lemma
okay.
So, so far we have seen monotone convergence
theorem and Fatous lemma, which allows us
to interchange integrals and limits in the.
In the next session, we will look at complex
valued functions and how to integrate them.
So, far we have integrated only positive functions,
starting from positive symbol functions, we
have gone to positive functions, we know the
integral is linear there. Now, we will extend
it to real valued functions, and then to complex
valued functions and then one more result
of this kind which will allow us to interact
limits and integrals will be put.
