Welcome back, we were discussing we were building
up towards monotone convergence
theorem we said we defined point wise convergence.
And almost everywhere
convergence of a sequence of functions fn
of omega to f of omega we said that even if
fn
of omega convergence to f of omega for all
omega or for almost all omega. It does not
necessarily mean that integral f n d mu converges
to integral f d mu it is not true.
Generally we gave a counter example to show
that this limit of limit an integration
cannot be interchange, we gave an explicit
example where the integral of the limit and
limit of the integral where different. So,
the monotone convergence theorem gives once
sufficient condition under which you can in
fact interchange limit and integration. So,
the
condition is not just, convergence almost
everywhere, but monotone convergence almost
everywhere.
.
So, I will state the monotone convergence
theorem. So, you have, so what you have is
omega f mu sum measures space and let g n
be a sequence of measurable functions 
such
that g n increases to g mu almost everywhere
i.e., g n of omega is less than or equal to
g
.n plus 1 of omega 
mu almost everywhere 
and limit and infinity g n of omega equal
to g
of omega also mu almost everywhere 
then integral g n d mu increases to integral
g d mu.
So, this is the statement of the monotone
convergence theorem.
So, you have these are non negative function
they are non negative measurable functions
such that you have g n converging to g mu
almost everywhere, but you also it is not
just
convergence it is convergence monotonically.
So, you have g n of omega, so g n of
omega less than or equal to g n plus 1 of
omega, so this is true for all.
.
n is equal to 1 2 dot. So, this holds mu almost
everywhere which means that there may
be a set of mu measure 0 where may be monotone
it is the does not hold. So, g n of
omega is less than or equal to g n plus 1
of omega. So, for every omega you have a
sequence of a increasing numbers this will
be a sequence of increasing numbers except
it
may not hold in a mu measure 0 set. And similarly
you have convergence mu almost
everywhere g n of omega converges to g of
omega for almost all omega mu almost
everywhere in that case, you are guaranteed
that the limit and integral can be
interchange. So, here again I write an up
arrow, because the sequence of this integrals
will be bigger and bigger, because these functions
themselves getting bigger and bigger.
So, the limit of this integral another way
of writing this is limit ((refer time: 05:20))
into
.infinity integral is equal to integral of
limit. So, this is saying that therefore,
you can
interchange limit and integration. So, long
as you have monotonic convergence not just
convergence monotone convergence almost everywhere
this statement of this theorem.
Any questions on the statement?
So, this theorem is perhaps the most important
in integration theory it considered the
corner stone of integration theory and it
has lots of application and it also has some
important corollaries which we will do a little
later. So, the proof of the monotone
convergence theorem is little long, but not
difficult. So in fact, if you look at m i
t open
course where lecture number 11 there is a
full proof of the monotone convergence
theorem. And it runs to about 2 pages which
is not a very long proof come to think of
because it is a very fundamental theorem right
it is a very important found foundation
result, but it proofs only 2 pages long.
In fact, I suggest you look at it and how
do you suspect that you will prove it there
are
only 1 way to prove all results in integration
start with simple functions. So, what you
will assume? So, the where the proof builds
up is that you assume that you have you first
assume that the limit function is a simple
function. This g is some q some limit function
is simple and you have monotonic convergence
to a simple limit function then you have
then you can explicitly proof a that this
theorem holds. And the key ingredients in
that is
continuity of measures continuity of any general
measure not just probability measure.
You first prove it first simple functions
and then you use the supremum definition for
non negative functions. And you can use you
already proved m c t for simple functions
then you use the supremum definitions to prove
it for arbitrary non negative function. So,
there is all proves in integration go like
this. And, so actually I had half of it do
it in
class, but it will probably take all of today's
lecture and we I may lose you here in
between I guess. So, I decided again doing
it in class, but it is not a difficult proof
I mean
for a such a fundamental theorem in measure
theory. It is about 2 pages long and you can
actually follow it and if you have difficulty
following it a please do come to me it is
a
interesting proof. So, I will not spend class
time onward, because let us all be able to
today if I do it. So, in the example we gave,
so this.
..
So, the example we gave so we said the g n
of last class we said we took the functions
the sequence of function as n 1 over m your
sample space was 0 1 and you took I guess
you put solid dots here halo dots over there
and in this case the integral of g n of omega
was 1, but the limit function was 0 limit
of g n was 0 for every omega. So, it was not
possible in this case to interchange limit
an integration. In fact, you can see that
monotonicity does not hold after all if this
is your g n of omega g n plus 1 will look
like.
So, g n plus 1 would look like that will be
at n plus 1 and 1 over n plus 1.
So, while it is true, so g n has to get bigger.
So, it is true that g n plus 1 is bigger than
g n
over here, but in this legal interval g n
plus 1 is 0 where as g n is equal to n. So,
in this
set of positive measure this set has positive
measure namely the measure is 1 over n
minus 1 over n plus 1 in that little set you
have failure of monotonicity. So, when which
is 1 i mean if this were monotonic. So, if
you this 1 monotonic you will have a by the
by
m c t you will have convergence of integral
as well on the other hand if you were to take
g n of omega equal to 1 between 0 and 1 over
n. It is, so for what I am saying as if you
have if g n of omega where equal to 1 between
0 and 1 1 over n and. So, g n plus 1 will
be.
..
1 between a 0 and 1 over n plus 1. In this
case the sequence of function is so; in this
case
mono this is monotonically decreasing. So,
even in this case you will have convergence
of integrals. So, the reason it broke down
here was it was bigger it mean the monotone
is
in quite cold. So, there is 1 property I had
not proven, so there was some property of
linearity which I said is not easy to proof
using the supreme of definition. I said that
we
will get back to it after we do monotone convergence
theorem. So, let us get back to
linearity so.
..
So, this says that integral g plus h d mu
equal to integral g d mu plus integral h d
mu this
is the property we never proved. As we might
expect the proof goes in the usual
sequence you prove it for simple functions
you take g n h as both simple and prove it
and
then you have to generalize for this non negative
arbitrary non negative functions as well
and then you can go to g plus g minus that
is standard. So, now, so if first assume g
and h
are simple. So, you have a representation
of the g as sum over a i indicator of A i,
i is
equal to 1 through sum n finite.
And h as sum over, so I have what I mean is
g of omega is equal to sum over a i I A i
of
omega and similarly h is equal to j equal
1 through m bi indicator of some other Bi
some
other set. So, all is a i's and all is bi's
are measurable sets f measurable sets. So,
in this
case what will g plus h d g plus h d also
be simple. So, g plus h will have the
representation sum over i equals 1 through
n sum over j equals 1 through m a i i should
write B j here a I will have a i plus b j
indicator what A i intersection B j after
all the
functions g plus h will take the value a i
plus B j if your omega lies in A i intersection
B
j.
So, now, what you will assume? So, let us
assume that these 2 are canonical
representations. If I assume that these are
canonical representations then I will have
the
.this a i's are disjoint and b j's are also
disjoint. So, these guys will disjoint for
a different i
i's and j's. So, these intersection will be
disjointed. So, if i have, so if I am looking
at this
integral.
.
G plus h d mu that will simply be 
sum i equals 1 through n sum j equals 1 through
m a i
plus b j measure of that guy at by definition
measure A i intersection B j which correct
by definition. So, I can just rewrite this
as sum over i equals 1 through n a i, so this
a i
will come out of the j summation and then
I will write sum over j equals 1 through m
mu
A i intersection B j plus a sum over j equals
1 through m, so now, I am looking at that
summation. So, in which, so if I am looking
at that summation I can bring the i
summation in side write. So, that is this
algebra you will get b j sum over i equals
1
through n mu something A i intersection B
j.
Now, what is that equal to see these a i this
A i intersection B j if you just look at it
as
some sets index by j let us call this j now.
So, if you take, so these are disjoint sets,
because you assume that g n hr in canonical
representation it is. So, these sets are disjoint
we are adding the measures of disjoint sets
which means I should get the measure of yes
exactly you will you will get the measure
of the unions of all these guys over all these
sets over j. So, this is simply become. So,
essentially, so these are this is by countable
of
.negativity measure finite negativity of measures
it will become i equals 1 through n a i
mu of A i. I am just you just have to establish
that union over all that union of all these
seta over j is simply A i. And similarly the
if lift the argument around if you take union
over i this will give you the measure of the
j b j's. So, you will get j i equals 1 through
m
b j 
mu a well let me union over i, so this should
be B j. So, that is exactly the this is
nothing but the integral of a g write this
is plus integral h d mu.
So, proving it for simple functions is quite
elementary. Now, you have to prove it for
arbitrary functions arbitrary non negative
functions first measurable functions
measurable non negative functions. So, there
you use the monotone convergence
theorem. Now, I say next.
.
Let g and h be non negative. Let g n and h
n be 
simple functions a sequence of simple
functions such that g n increases to g and
h n increases to h. So, what I am doing now,
is
I am given 2 non negative functions I am approximating
this non-negative functions
from below by sequence of simple functions
g n and h n. So, I have g n increasing to
g
and h n increasing to h. Now, the question
is given 2 functions hg can always find simple
function that increase to let say g sum non
negative function how am I guarantee that
I
can always find simple functions g n that
increase to g. It is not clear, but I will
give you
.an explicit construction, so this is possible.
So, if you hold on a little bit i will show
you
that this is possible. For now, just take
this is possible we will show that this is
possible
little later then we will invoke monotone
convergence here. So, in this case, so then
you
will see then you can show that g n plus h
n will increase to g plus h. If g n increases
to g
h n increases to h then g n plus h n will
increase to v plus h value elementary actually
all
these can be almost everywhere it does not
have to be for all omega almost everywhere
is enough. So, you have, so if this is true
in that, so I have this, so I have we have
integral
g plus h d mu is equal to limit n tend to
infinity integral g n plus h n d mu why is
that
true? Convergence theorem this is, because
of m c t.
.
And this now, I have h n and g n are simple
functions. So, I have proved an linearity,
so
linearity among simple function is already
proven, so this will split. So, this integral
will,
so if you look at that that would.
..
Be limit n tending to infinity integral g
n d mu plus, so I can take the limit in side,
so
limit, so far I am, so I have just have skipped
1 step. So, I have written this integral has
integral g n d mu plus integral h n d mu and
I am putting limits separately that is that
is
always allowed does not it as long as the
limits are well defined. So, now, what happens?
So, now g n is again a monotonic sequence
of function. So, this will converge to this
will
give you what you want? This is again by m
c t monotone convergence theorem. So, is
that proof clear?
So, after the monotone convergence theorem
proving linearity of a going linearity of
integral integrals is very simple matter except
that we have to buy that this is possible
you can always for any non negative functions
g i can find a sequence of simple
functions which increases to g. It is possible
to this in many ways given any non negative
functions g i can always find a sequence of
simple functions that increase to g and I
can
do this in many ways. I will show you 1 explicit
way in which this can always be done 1
value of construction.
I will show you and this approximating a non
negative function from below using simple
functions is anyway very useful to compute
this see remember that, so far I only given
a
supreme of definition which I told you is
conceptually fine it helps to prove a number
of
.properties, but it does not tell you how
to evaluate the integral. So, far if I give
you some
you go on integrate g of x is equal to x square
over the 0 1 integral we actually do not
know how to do it except treated as a riemann
integral. Now I will tell you an explicit
way of computing. So, is this clear I am moving
to the next topic?
.
Approximating 
a non negative measurable function 
from below using simple function.
Let g be measurable. Define so; you are given
some non negative function g in some
measurable space. Now, I am going to define
sequence g n an explicit definition of g n
such that g n increases to g. This 1 particular
construction you may be able to do it in
other ways as well, but I will give you 1
construction which always work we define g
n
of omega equal to 
i over 2 rays to n if i over 2 raise to n
less than or equal to g of omega
less than or equal to well actual less than
i plus 1 over 2 power n and this is for i
equals 0
1 2 dot till n times 2 power n minus 1 and
if g of omega is bigger than or equal to n.
I
define it has simply n, so this is the sequence
of functions I define. If this is not no means
the only way you can do it, but this works
this is enough.
So, let me just ((refer time: 30:03)) tell
you what I am doing here off course if I have
to
draw picture I have to assume let say I have
to assume that my space is r alright let me
take omega equal to r.
..
And, so I have some non negative measurable
function. So, that is my g and this is my
omega and my space is real r I have sigma
algebra on it. Let say lebeg measure honor
just for the sake of concreteness. So, now,
what I am doing? Is I am I am at some level
quantizing that is really what I am doing.
So, I am looking at all those omegas, so let
me
draw may be little bit let me something like
that. So I am looking at some function, so
and I am considering let us say I fix some
i let say I fix an i and whenever my g of
omega is i over 2 to the n and i plus n over
2 to the n for those values of omega I define
it
as the i over to the n the lower end of the
interval.
So, I am looking at some value i over 2 to
the n let say that is my i over 2 power n.
So,
you fix an n am defining the nth function
now g n is what i am defining. So, you fix
an n
and let say you fix an i you look at i over
2 to the n and you also look at i plus 1 over
2 to
the n. So, i plus 1 over 2 to the n will be
somewhere here let say like that So, that
is i plus
1 over 2 to the n and you look at all those
omegas for which your function is in this
in
this range is the function takes value in
this range. So, in particular for us it will
be if we
look that as well, so it will be, so that
is it and here there will be that little set
and that
little set that guy there and perhaps 1 more
here that. So, you look at all the omegas
were
function takes values between i o over 2 to
the n and i plus 1 over 2 to the n. And in
all
this, so these are all omegas were the function
in that range and in all this ranges you
.define the function to be equal to over the
i over 2 power o mega. So, here again the
function will be here I do not have a different
color of chalk, but you see what I mean
right the function will be over here and similarly
here. Similarly if you look at i plus 1
over 2 to the n i plus 2 over 2 to the n you
will again approximate it from below you see
what I am doing here. So, in the sense I am
really just quantizing the function, but i
am
making sure that g n of omega is less than
or equal to gm of omega.
And you can very trivially see that g n of
omega is a simple function why is that true?
So, I can in fact, write, so let me write
in this way. So, can I not write g n of omega
as
sum over i equals 0 to n times 2 to the n
minus 1 see you get. So, you get n times 2
to the
n of these subdivisions write and if the function
is greater than equal to n you define the
function to be equal to n it is just that
is 1 way to do it. So, you have so many
summations the functional value is equal to
i over to power n 
indicator of omega for
which i over 2 power n less than or equal
to g of omega less than i plus 1 over 2 power
n
plus I will have just 1 more term that says
n times the indicator of omega for which g
of
omega is greater than or equal to n.
So, is that a correct representation of my
function agree with that? Now, this is already
in
a simple form, so only thing I have to verify
is that these sets are measurable or these
sets
measurable indeed they are measurable, because
g is a measurable function and after all
this set is simply the pre image of the interval
just a pre image of the interval i over 2
to
the n i plus over 2 to the n pre image of
that under the mapping g and this is an interval.
So, pre images of sets are measurable, so
that is a valid if measurable set reassembly
that
is valid f measurable set. So, you already
written g n in terms of a simple function
explicitly a simple function. So, now I have
the following 3 things are quite easy to
show.
.
.So, the 3, so the first is that for each
n g n is simple that we just argue. Number
2 we
have that g n is less than or equal to g n
plus 1 for all omega why is that true? Because
if
I take a finer quantization, so if I increase
n I am taking a finer quantization. So, in
that
case I will take, so this was split into 2
further subintervals and I will have 2, so
if I
spread this further I will lower approximate
in that in that little interval. And, so the
function I get the approximation I get will
be larger for a finite final quantization.
So,
that is true alright. So, this you can show
and 
g n, so g n increases to g.
So, when because bigger and bigger, so y approximation
will get better and better,
because my g n of omega will be i over 2 to
the n, but g of omega will be very close it
has to lie between i over 2 to the n i plus
1 over 2 to the n as n becomes very large
g or g
n of omega will converts to g of omega for
every omega. So, I have a monotonically
increasing simple function sequence of simple
functions monotonically increasing to g.
Alright this sequence can in particular be
used in the previous group of linearity correct
with me?
This 1 let me draw a slightly bigger picture
for, so what I am trying to say is that yeah
no. So, what I am trying to say is that, so
let me draw slightly bigger picture so.
..
You have like that, so you approximated, so
earlier you approximated like that g was the
approximated like that. Now, you have a 1
more level of, so you are increasing n by
1 let
us say and, so you will have, so that will
split into that is further split into 2. So,
here i
may be different now, but that is the functional
approximation will now be, so here, so
just to zoom in, so I have let me zoom into
that a little bit. So, I had that write initially
I
had that. So, I said, so in that interval
I approximated the function as that guy now,
I am
going to increase n which means the width
of each of these intervals use to be 1 over
2 to
the n now, it is going to be 1 over 2 to the
n plus 1 it is mean it is half as big as the
earlier
width earlier.
So, it is going to be like that write and,
so if you look at those omegas for which g
is here
I will approximate the function with that
and here I will approximate it has that so
. Now,
therefore, g n plus 1 of omega is bigger than
or equal to g n of omega. Now, so the value
of i may be different in the 2 cases that
is, but you see the functional value is bigger,
because over equal to rather. So, g n plus
1 will be like that g n will be like that.
So, you
can just if you just draw big slightly bigger
diagram like that you will figure out why
this
is true? And this is converges this is simply
converges which is also reasonably easy to
show. So, I have now, by, so I since I have
all this I can definitely write, so by m c
t.
..
I have integral g d mu is equal t o integral
well limit of why, because I have monotonic
convergent, but integral g n d mu already
know why? It really have it in the simple
form,
so I can explicitly write down an expression
for in integral g t n d mu and by m c t. So,
this limit is equal to the integral of g d
mu. So, I am going to give you a explicit
formula
for integral g d mu in terms of a limit. So,
which is nice, because if I give you a function
to integrate you will go ahead and evaluate
the limit which is an explicit rather than
give
you a supreme of a set of simple functions
and, so on which will never be able to
compute in practice? So, this is a more practical
way to compute the integral, so integral
therefore.
..
Integral g d mu is equal to limit n tend to
infinity sum i equals 0 to n times 2 to the
n
minus 1 i over 2 to the n mu of omega for
which i over 2 to the n less than or equal
to g
less than i plus 1 over 2 to the n mu of that
plus there will be 1 more term is not it plus
n
times the measure of omega. So, for which
g of omega is bigger than or equal to n. So,
the if you evaluate this limit you already
have the integral of dg mu this is an explicit
formula. So, if your mu had to be lebesgue
measure you will simply take the, so
remember you are this set this set you will
just simply take the lebesgue measure which
is the length of those sets if it is some
other measure you have to take that measure.
So, in effect what you do in let us for again
continue assuming that mu is lebesgue
measure on our. So, in this case what you
are essentially doing is you are chopping
as up
the vertical axis for Riemann integral you
used to chop the horizontal axis it is. So,
that is
really the only essential difference. So,
for Riemann integral you used to chop of the
horizontal axis and look for the supremem
infimum within those intervals and you define
the upper Riemann integral lower Riemann integral
and complete the 2 see the 2 are
equal.
So, in this abstract integral this lebesgue
integral the mu is lebesgue measure what you
doing is you just chopping up the vertical
axis the y axis instead of chopping of the
x
.axis. And you looking at the measure of those
sets for which the g lies on certain range
that is all you doing, so somehow just chopping
up. So, chopping up other axis instead of
the, so chopping up the vertical axis instead
of chopping of the horizontal axis gives you
a much more general notion of an integral
alright. So, I will stop here.
.
