So, in the last lecture we saw two major theorems,
one was the monotone convergence theorem and
the other one was Fatou’s lemma. So, let
us recall both of them.
So monotone convergence theorem. So, both
these results, help us in interchanging the
integral and the limit. So, monotone convergences
always will have a triple X, F, mu. X is a
space F is a sigma algebra and mu is countably
additive measure. We had FN measurable, non-negative
functions and FNs increasing okay. In that
case, of course, the limit will exist so limit
FN, X and going to infinity exits and we call
that F of X, okay. So this is true for all
X in X.
Then F is measurable, that is easy because
it is a limit of measurable functions. And
this is the conclusion of the theorem, limit
N going to infinity integral over X, FN D
mu is integral over X, F, D mu. So you can
interchange limit and integration right, FN
and D mu. So, this is what a monotone convergence
theorem is. And we already saw an example
of this when we studied measures, this was
if we had sets AN increasing to A, A. So that
would mean AN, A equal to the union of all
these ANs.
Then measure of AN increases to measure of
A that is precisely monotone convergence theorem,
applied to indicator function. So, you simply
take FN to be equal to Chi A. But one can
prove this simply by using the countable additivity
of the measure A. So this was one of the theorem
we saw.
The second one was Fatou’s lemma. In this
case also FNs were measurable, non-negative.
In this case, we are not assuming FNs to be
increasing, but we simply look at lim inf
of FNs. So, lim inf of the functions FN. So
how is this defined? So, let us we call that
this is simply lim inf of FN, X over N. Well
this is nothing but you take supremum of infimum
of FK of X, K greater than or equal to N and
supremum over N right. This is the definition
of lim inf.
Lim inf is also measurable. FNs are measurable,
implies lim inf FN measurable, FN measurable
and Fatou’s lemma tells me that integral
of the lim inf is less than or equal to the
lim inf of integrals, lim inf of the integrals
right, integral FN, D mu. D mu they are positive
numbers and you take the lim inf. So, this
is what Fatou’s lemma is.
So, these two theorems allow us to interchange
the limits and integrals in many cases, but
both of them deal with positive functions.
Well, sometimes the positivity is not necessary,
but this depends on the context depending
on what kind of functions you have. For example,
you may you be able to subtract some things
or add some and make functions positive. And
then I, then apply Fatou’s lemma, we will
see instances of such things when we go ahead
okay.
So we have one more theorem of this kind,
which will allow us to interchange the limits
and integrals. But before we state that, let
us prove a simple result which we already
know in the case of summation, but will prove
this in the general context of integration.
So let me write it as a proposition. So as
usual we have the triple X, F mu and let us
take FN, L1 mu. So recall that this is same
as X, F, mu.
So whenever we write a measure, the space
and the sigma algebra is understood, it takes
it. So X and F will exist. So, recall that
this is simply all complex valued measurable
functions, such that integral over X mod F,
D mu is finite. So, in the class of integrable
function, so we call this class of integrable
functions. Now, so if I take an FN, L1 mu.
Then the conclusion is modulus of integral
of X, F, D mu is less than or equal to integral
over X mod F, D mu.
So, everything makes sense here So, first
of all let us make sure that we understand
this. So, I am taking a, so call this as remarks
if you like. We are starting with a function
in L1 of mu. What does that mean? That means,
integral over X mod F, D mu is finite okay.
We call that F is a complex valued function.
So, this is a complex valued measurable function,
measurable function okay and so I can write
F as mu plus I times V, where U and V are
both the functions U and V are real valued
measurable functions, real valued measurable
functions.
So, we are writing F equal to. So, let us
recall that, so, we are writing F equal U
plus I, V and if I look at mod F, mod F is
of course bigger than mod U and mod V. So,
we have mod U less than to mod F, mod V less
than or equal to mod F. Now, U and V are measurable,
so, mod U and mod V are also measurable. So,
both of these gives us by monotonicity of
the integral. We have integral over X mod
U, D mu is less than or equal to integral
over X, mod F, D mu which is finite.
Similarly, integral over X mod V, D mu is
less than or equal to integral over X mod
F, D mu both are finite. So, both these functions
are, so both U and V are actually in L1 of
mu as a result, if I look at integral over
X, F, D mu. So, whatever discussion we just
had implies that integral over X, F, D mu
is nothing, but, well I know that this is
by definition, because of the linearity of
the integral I have this quantity and each
of them is finite right. Because of these
two, these two properties. So, let us elaborate
on that.
So, U I know that, mod U, D mu is finite,
this implies or. So, we write U as the positive
and the negative part, U plus minus U minus
and mod U is nothing but U plus, plus U minus
right. Because U is real valued right, U is
real valued and integral over X mod U, D mu
is simply integral over X the positive part
plus integral over X the negative part, negative
part is a positive function by the way. U
is U plus minus U minus right.
So, both of these are finite because this
is finite. So, this is a finite quantity,
this is a finite quantity, this will imply
that integral over X, U, D mu, which is the
difference of two finite positive numbers.
So, that is also a finite number. So, this
is a, this is an element in R, it is a finite
quantity okay. So, going back, so similarly
for V right because V is an L1, mu implies
both V plus and V minus will be L1 of mu.
And so, integral over X, V, D mu is equal
to integral over X, V plus D mu minus integral
over X, V minus D mu both our finite. So,
this is this is a real number.
So, all this would imply that if I look at
F, D mu. This is of course, integral over
X, U, D mu plus I times integral over X, V,
D mu and this is finite, this is a finite
quantity, this is another finite quantity.
So, this is like alpha plus I beta, so it
is a complex number, it makes sense. So, this
is a fixed complex number and we are trying
to prove that. So let us go back to the statement
of the result.
So we are trying to prove this inequality
okay? So the left hand side makes sense first
of all, it is an integral over X, F, D mu.
It is a complex number and I am taking the
modulus of that. So I will get a positive
number. I want to say that is less than or
equal to integral over X mod F, D mu, mod
F is a positive measurable function. So it
integrates, its integral makes sense, it may
be infinity or finite. But we are assuming
F to be in L1, because if it is already infinity,
there is nothing to prove, but we are assuming
F is in L1. So this is a RHS is a finite quantity.
All right. So let us try to prove this, so
prove this sort of one line. But it is something
which we use every now and then. So take alpha
in the complex plane, mod alpha equal to 1
such that, such that alpha times integral
over X, F, D mu. So, remember integral over
X, F, D mu is a complex number. I am taking
another complex number alpha, such that mod
alpha equal to 1.
So, that this is actually equal to the modulus
of the complex number we started with, right.
So, this we can do, so what we are doing is?
If I take a Z in the complex number, there
exists alpha such that mod alpha equal to
1 and alpha times Z is equal to mod Z right.
Because if for example, if Z is zero, any
alpha will do. Any alpha will do right, any
alpha says that, mod alpha will do one will
do.
If Z is not zero, then what is alpha? Alpha
is mod Z by is Z right? Because of this mod
alfa is 1, so that is the alpha I am taking
I know what is integral over X, F, D mu is?
It is a complex number. So, there is an alpha
like this okay. So, let us continue with the
proof. So, let us start with the right hand
side and X, F, D mu. This is equal to alpha
integral over X, F, D mu which by linearity,
now alpha is a complex number right.
So, when I integrate, it will go inside the
integral due to linearity, which is integral
over X alpha times F that is my function D
mu. This is by linearity of the integral the
alpha goes inside, which is equal to. So let
us look at this again. The right hand side,
so this is a complex function, complex valued
function, complex valued function. But the
left hand side, this is a positive function,
this is a positive number. Which means the
imaginary part of this would be zero right.
So, this I can write it as, so let me write
one more step. So, that this is clear, this
is real part of alpha times F, D mu plus I
times integral over X imaginary part of, imaginary
part of alpha times as F, D mu that is how
you write it because of linearity. But this
will have to be equal to zero because the
left hand side is a positive number. So, because
of that the imaginary part will have to be
zero which means, I can erase this part right,
they will be equal to the integral of the
real part.
But it is positive, right? Well, so, I can
write this to be less than or equal to integral
over X modulus of alpha F, D mu. Why is that?
Because if I look at real part of F, I know
that this is less than or equal to. Well,
real part of any function G, real part of
any function G is less than or equal to mod
G. So integral over X real part of G, D mu
will have to be less than or equal to integral,
integral of mod G, D mu right? So that is
all I am using here, right, this is from this
inequality. But alpha has modulus 1. So this
is simply X mod F, D mu. So that is the proof
that is one line proof. So integral the model
X of the integral is less than or equal to
the integral of the modulus.
So, let me let me tell you that we have seen
a case of this. So, let us recall that if
I take suppose ANs are complex numbers okay.
Then we know that if I take summation AN n
equal to one to infinity modulus of this,
this is less than or equal to summation in
equal to one to infinity, modulus of AN, this
is something which we know. And this is precisely
this inequality right because the summation
is an integral, summation is an integral.
So, I can write summation n equal to infinity
AN as integral over X, F, D mu for some suitable
X and mu and F and this tells me that the
summation, model X of the summation is less
than or equal to some of the modular. That
is precisely this inequalities okay, let us
continue.
Now we come to one of the most important theorems.
So this is called the dominated convergence
theorem. So, Lebesgue dominated convergence
theorem, dominated convergence theorem. Again,
this is one of those results, which allows
one of the most useful results in measure
theory, which allows to allows, allow us to
interchange limited and integral.
So, we will call this DCT domain convergence
theorem. So, let me state it. So, I have X,
F, mu as usual and I have a sequence of measurable
functions. So, let FN be complex valued. So,
now, you see it is complex valued. Earlier
two theorems, the monotone convergence theorem
and Fatous lemma required that the measurable
functions be non-negative.
Here, we are looking at much more general
class of complex valued measurable functions
for N equal to 1, 2, 3 etc. We have a sequence
of measurable functions such that, such that
the limit exist, So, limit N going to infinity
FN, X equal to F of X. So, F of X, then will
be automatically measurable right. So, then
as a conclusion, we know that F is measurable
okay.
Suppose there exist a function G in L1 of
mu. So, an integral function such that 
mod FN is. So mod FN at X is less than or
equal to G of X for every X in X okay. So
n for every N right. So, G is called the dominating
function. So, G dominates all the FX, in that
case. So, here is the strong part of the conclusion
in then F is in L1 and integral over X mod
FN minus F, D mu, goes to zero as N goes to
infinity, okay.
In particular, we also have limit N going
to infinity integral over X, FN, D mu equal
to integral over X, FN, D mu. So, again here
you are interchanging. So, this is interchanging,
interchanging limits and integration okay.
So, there are conditions one is FNs, we have
measurable functions, limit exist and more
importantly there is a G which dominate. So,
this is the dominating function right, the
dominating function G has to be integrable
okay.
So, there is a, there is a control on the
way FNs grow in some sense. In that case we
have this convergence this is a rather strong
convergence, we will see it later. In particular
we can interchange integrals and limits. So,
proof of this, well proof is not all that
difficult once we have Fatou’s lemma. But
it is an extremely useful result, we will
see some examples soon. So, let us prove the
stronger statement that FNs converts to F
in some sense okay.
So first of all F is measurable, that is trivial,
okay. Then, all FNs are dominated by G for
every X and for every N. But F is the limit
of, since F of X is given by the limit of
FNs, we immediately get that mod F of X is
less than or equal to G of X for every X okay.
Well, which also implies by monotonicity of
the integral mod F, D mu if I integrate. I
am going to get something less than or equal
to G, D mu, which is finite because I know
that G is in L1. So this implies that F is
in L1. So this much is sort of straightforward
from the monotonicity of the integral.
So now, what we do is? We look at, so consider
these functions. So look at 2G. So remember,
G is my dominating function minus model X
of FN minus F. So, for each N, so this is,
this is a sequence of measurable functions
right, sequence of measurable functions. So,
what does it mean? So, for each X we are looking
at two times G of X minus mod FN, X minus
F of X right.
But, recall that FNs are bounded by, a recall
that FNs and F are bounded by G, mod F of
X is also less than or equal to G of X right,
this is true for every X. So, this gives me
that these are positive functions. So, I have
a sequence of positive measurable functions
I can apply Fatou’s lemma. So, apply Fatou’s
lemma to the sequence of functions 2G minus
mod FN minus F. Remember these are positive
measurable functions. So, I can apply Fatou’s
lemma.
So, what does Fatou’s lemma say? Integral
lim inf is less than or equal to lim inf of
integrals of these functions. So, we get integral
lim inf of the functions we are looking at
that is 2G minus mod FN minus F, D mu is less
than or equal to lim inf of the integrals.
So lim inf of the integrals of the functions
we are looking, mod FN minus F, D mu.
Well, so, let us try to compute this? What
is the left hand side? Since FNs are converging
to F, since FN, X converges to F of X, for
every X lim inf of 2G minus mod FN minus F
will be equal to 2G. Because of this part
going to zero. Similarly on the right hand
side, on the right hand side this is two elements.
So we have two pieces. So I can look at 2G,
D mu as a separate thing. And I have plus
lim inf of integral over X minus mod FN minus
F, D.
So remember the minus is put inside the, I
will be taking the lim inf, right? So this
is so let me write it again integral over
X, 2G, D mu plus the minus will come out of
the integral but not out of the lim inf. So
let us write one more step, lim inf of minus
integral over X mod FN minus F, D mu. So I
am taking some positive numbers, looking at
its negative and taking the lim inf okay.
So let us compute this again. So on the left
hand side, we have LHS, the simply integral.
So I know that this goes to zero. What remains
is simply 2G? So it is integral over X, 2G,
D mu. I know and this is less than or equal
to integral over X, 2G, D mu. Now, what I
have is lim inf of negative things. So, here
is a simple exercise lim inf of. So I take
sequence AN positive and I take lim inf of
minus ANs, okay this is minus of lim sup of
ANs.
So, apply that we will get the minus lim sup
of integral over X mod FN minus F, D mu. So,
this is the inequality we have. Now we call
that G is in L1, G was in L1. So, these are
finite quantities, right, these are positive
numbers, these are positive function. So,
I can cancel cancel this. So, let me, I can
cancel this. So what do I get? I get, here
I have a positive number and then I have a
negative sign. And here I have cancel, I will
get zero. So I am getting zero less than or
equal to minus lim sup integral over X mod
FN minus F, D mu.
But this is a positive quantity and minus
of that will be a negative quality and that
is positive. This implies that lim sup of
integral over X mod FN minus F, D mu is actually
equal to zero right? But lim sup of positive
things are zero meaning the limit itself is,
so limit of integral X mod FN minus F, D mu
is zero. So that was one part of the theorem.
So let us go back to the statement of the
theorem. So, what we have proved is this part
okay? This is an easy corollary or easy consequence
of what we have just proved. So, let me write
that as well.
So, from this conclusion, so that is a stronger
conclusion. So, integral, so, we have, we
already have integral over x mod FN minus
F, D mu goes to zero as N goes to infinity
that is what we just proved, okay. Now, if
I look at integral over X, FN, D mu. I know
this is a complex number minus integral over
X, F, D mu. I want to say that the left hand
side sequence goes to the right hand side
number, right this is what we want to prove.
So, to prove that I look at the modulus of
the difference of these two numbers. Well
integral is linear. So, you know that this
is FN minus F, D mu and then modulus okay.
Then we use the inequality we proved. So,
what was the inequality we proved at the beginning,
if I take a complex valued function G and
I integrate and take the modulus I know this
is less than or equal to modulus of the integral.
So, this is less than or equal to integral
over X mod FN minus F, D mu. I know this goes
to zero okay. So, this sequence converges
to this sequence, this number that is the.
So, integrals converge and you can interchange
limits and integration.
So, let us see how this is used? Okay, so
let us put this as a remark. How does this
get used in various places? So let us take
a special case okay, where mu X is finite
okay. Well what is the big deal? If the measure
of the whole space is finite the measure of
the space X is finite. Then constant functions
are integral, constant functions are integral
okay. What does that mean? So, if I take F
from X to C, F of X equal to K for every X.
K is some fixed complex number fixed complex
number. It could be 1, 2, I or any such fixed
complex number.
Then of course F is measurable that is trivial
because it is a constant function and if I
look at integral over X mod of D mu. Well,
this is equal to integral over X mod F will
be mod K, which is a constant and that comes
out. So, mod K comes out and what remains
is the total measure of the set that is X
and this is finite . So, such functions are
in element of mu. So, in such a case, suppose
we have measurable functions FN, complex valued,
measurable functions. And I know that mod
FN are less than or equal to let us say some
fixed constant K okay. K a positive constant.
Suppose, limit N going to infinity, FN X is
and let us say it is equal to F of X for every
X okay. Then I can conclude that integral
over X, FN minus F, D mu goes to zero. Well,
why is that? Apply DCT, what is the dominating
function? The constant function, the constant
function is the dominating function right.
Because it is an L1 and you know that if I
have sequence of measurable functions bounded
by a dominating function and you have convergence,
then this happens.
Now, you see this is much more powerful than
the remand aggression. In remand aggression
we needed uniform convergence for changing
the limit and the integration, here we have
only point wise convergence for every X, but
bounded by a dominating function. And if you
have that you can interchange limits and integration
okay.
So we will stop this session with just recalling
what we did. The main theorem we proved today
was the dominated convergence theorem, which
is extremely powerful combining it with Riemann
integration, where you require uniform convergence
to interchange limits and integrals. Here
we just need point wise convergence, but with
a dominating function, the dominating function
should be in L1. If the space has finite measure,
then it is enough to look at a dominating
function to be a constant. Of course, all
the time the constants may not work, but most
of the time that is what works you will see
examples as we go along okay.
