Welcome to lecture number 19, on Measure and
Integration. In the previous lecture, we had
started looking at the properties of integral
for non- negative, measurable functions. We
had looked at the linearity property of the
integral for non- negative, measurable functions
and then we said we will start looking at
the limiting properties of a functions, which
are non- negative measurable and integrals
of them.
Today we will prove some important theorems.
We will start with proving what is called
monotone convergence theorem. Then we will
prove Fatou’s lemma. Then go to define integral
for general functions.
Let us look at, what is called Monotone convergence
theorem. Monotone convergence theorem says:
Let f n be a sequence of functions in class
L plus. That means, f n is the sequence of
non- negative, measurable functions increasing
to a function f of x at a brief point. That
means, f of x for every x in X is limit n
going to infinity of f n of x. So, we are
given a sequence f n of non- negative measurable
functions, which is increasing and the limit
is f of x, Claim: The function f belongs to
L plus, this we have already observed and
the additional properties that the integral
of the limit f d mu is same as limit of the
integrals of f n d mu. That means, whenever
a sequence f n of non- negative of measurable
functions increases to f, then integral of
the limit is equal to limit of the integrals.
So, this is one of the first important theorem,
about convergence of sequences of non- negative
measurable functions and their integrals.
So, let us prove this property. We are given
f n, a sequence. Each f n belongs to L plus,
is a non- negative, measurable function, for
every n bigger than or equal to 1. So that
means, that implies there exist a sequence
will denote by s j n of functions, where n
bigger than or equal to 1, such that s j n
are non- negative measurable simple functions
for every n and for every j. And s n j increases
to f. Let us fix notion, which one we are
going to vary. Let us say that the upper one
n will be fixed. So, this s j n is going to
f n as j goes to infinity. So, for every n
fixed, s n j is a sequence or non- negative,
simple measurable functions increasing to
f n’s and f n’s increased to f. So, we
want to show We have already shown, but will
show it again that this implies f belongs
to L plus is a non- negative, measurable function
and integral f d mu is equal to limit n going
to infinity integral f n d mu.
So, to prove this we are going to use this
sequence s n and construct a new sequence
of non- negative, simple measurable functions
out of it. So, what we will do is the following,
Let us write that for n is equal to 1.
That s 1 1; s 1 2; …, s 1 j, say this s
1 j converges to f 1. So, the upper index
is going to give you, so s 2 1, s 2 2, …, s
2 j, this increases to f 2. In general, we
will have s n 1, s n 2, s n j will increase
to f n and so on and this increases to f.
Let us observe that as we go from left to
right, as we go from left to right, this is
increasing. So, everywhere left to right,
it is increasing and down to up there also
is increasing, so every sequence if you look
at them is increasing. So, this is the array
of non- negative, simple measurable function.
Each row is increasing to the function on
the right side and this is increasing upwards.
So, let us out of this I am going to define.
So, let us Look at the function.
So, let me define from this, a function define
g n to be the function, which is maximum of
s n j, where j is between 1 and n. So, look
at so in a sense what way I am doing is in
this picture. Look at the 1 say let us say
here is s 1 n and here is s 2 n and here is
s n n. So, I look at this column, s 1. We
are looking at the column s 1 n up to s n.
Let us look at this column say and call that
maximum of this to be g n. What is g; so,
g n? is the so, Let me write again. So g n
is the maximum. So, define g n equal to maximum
of s j n, where j equals 1 to n.
Let us observe that each g n is a maximum
of non- negative, simple measurable functions.
So, each g n is a non- negative, simple measurable
function for every n. And g n is increasing
because at the next stage n plus 1, all this
is going to be bigger at the next stage. If
you look at g n plus 1, that is going to be
s power 1 n plus 1, s power 2 n plus 1 and
so on and s n plus 1 n plus 1 n and s power
n n plus 1. So, all this one is going to be
bigger than everything on the left hand side.
And these are places where we are looking
at, the maximum. So, in the maximum, of this
is going to be bigger than or equal to maximum
of this because each function at the right
side is bigger than the left hand side function.
So, this is going to give us, that is g n
an increasing sequence of functions. Let us
write let g be equal to limit n going to infinity
of g n, so g is. So, all these g n’s are
increasing and they are going to increase
to some function g. So, what we are going
to show is: g is equal to f . So, that is
what we are going to check. Let g be equal
to, so then Clearly, by definition g is a
non- negative, simple measurable function
because it is a limit of increasing sequence
of non- negative, simple measurable functions.
So, g belongs to L plus also.
Let us observe also 
each g n is less than or equal to f n for
every n. So, that is because say g n is the
maximum of this. So, the maximum of this each
one of them is less than f 1 is less than
f 2 is less than f n, so the maximum of these
g n’s is just going to be less than or equal
to this f n for every n and f n is increasing
to f. So, that will imply that g n is less
than or equal to f n, for every f and f n
is less than or equal to f.
So, implies that g n is less than or equal
to f n and f n less than or equal to f for
every n. Hence g n is increases to g. So,
that implies g is less than or equal to f.
So, that is one observation that the function
g is less than or equal to f. We claim that
the other way round is also true. Claim: f
is also less than or equal to g. Let us note
that for every j between 1 and n, if I look
at s j n, we see g n is the maximum of this.
So, this s j n is less than or equal to g
n for every n. So, this is less than or equal
to g n for every n and j, j between less than
this. If we fix n, then g n is less than or
equal to so, in and this is less than or equal
to g so, s j n is less than or equal to g
n, which is less than or equal to g, for every
j between 1 and n and for every n.
Let us now fix j and let n go to infinity.
So, as n goes to infinity what happens. This
converges f n. Note that as n goes to infinity,
this s j n goes to f j. So, from this and
this, these two observations, s n j is less
than or equal to g, for every n. So, if we
fix j and let n go to infinity, then n crossover
j and s n j and as n goes to infinity converges
to f j. So, this implies f j is less than
or equal to g for every j, this we; so we
get.
So, implies that f j is less than or equal
to g for every j and f n’s f j are increasing.
So, this implies that f is also less than
or equal to g. We have already shown g is
less than or equal to f. Now we are saying
f is less than or equal to g, this implies
that f is equal to g. Hence, second observation
from here. Hence g belongs to L plus. So,
f belongs to L plus. We have once again proved
that, if f n’s are increasing to a function
f and f n’s are non- negative, measurable,
then f is also non- negative measurable. Now
note that integral of f d mu is same as integral
of g of d mu because f is equal to g and this
integral of f d mu is equal to integral of
g of d mu is equal to limit n going to infinity
of integral g n d mu because g n’s are non-
negative, simple measurable increasing to
g.
So by definition, this is so, but each g n
is less than or equal to f. If you recall,
each g n is less than or equal to f, then
integral of g n will be less than or equal
to integral of f. So, limit of integral of
f n’s will be less than or equal to integral
f d mu.
So, this is less than or equal to integral
f d mu, or we can even introduce in between.
So, g n is less than or equal to f n. So,
it is less than or equal to limit n going
to infinity integral f d mu, which is less
than or equal to integral f d mu. So, what
does this imply integral f d mu is less than
or equal to limit f n integral of f n d mu
and that is less than or equal to f d mu.
So, that implies that integral of f d mu is
equal to limit n going to infinity integral
f n d mu. So, that proves the theorem completely
integral of f d mu is equal to limit n going
to infinity integral of f n d mu.
So, this is a construction which is quite
useful. This is the kind of analysis 1 must
carry out. Let us go through the proof again,
so we understand what we are doing? Each f
j of f n is a measurable function. So, I can
look at a sequence s 1 to 1, s 1 to 2, s 1
j, s 1 n which is going to increase to f 1.
Similarly, at the top n is fixed at 2. So,
s 2 to 1, s 2 to 2, s 2 to j, s 2 to n that
increases to f 2 and so on.
So, each row is increasing to the function
on the right side and the functions f 1, f
2,…,f n are increasing to the function f.
So, what we do? We look at the maximum of
this column. So, what is the maximum in this
column? This column has the maximum at the
functions s 1 to n, s 2 to n and so on s n
to n. So, call this as g n. This function
is called g n, so the observation is: g n
is a maximum of non- negative, simple measurable
function. So, it is nonnegative simple measurable
each g n is less than or equal to f n, because
you are going only going up to this corner
only.
So, each g n is less than or equal to f n
because s 1 n is less than f n, s 2 n is less
than f n, f 2 less than f n, f 1 is less than
f 2 and so on. So, this says g n will be less
than or equal to f n and each f n is less
than or equal to f. So, each g n is less than
or equal to f n is less than f. So, if you
write the limit of g, then the limit of this
g n to be equal to g, then g is less than
or equal to f by this simple construction.
Also for any fixed j, let us look at s n j.
So, let us look at s n j, where j is fixed
and n is going to vary, so as n varies, what
happens to these functions? So, for every
fixed j, this sequence of functions is going
to be s n j is less than or equal to g n and
g n is less than or equal to f. So, we get
g less than or equal to f. So, s n j is less
than or equal to g n for j between for between
1 and n. So, that will give us that f is also
less than or equal to j.
So, that will prove the theorem of limit of
increasing sequence of non- negative, measurable
functions. If f n is equal to sequence of
non- negative, measurable functions increasing
to f, then integral of f d mu is equal to
limit of n going to infinity of integral f
n d mu. So, this is called Monotone convergence
theorem. Monotone because we are looking at
monotonically increasing sequences, f n and
convergence because we are looking at the
convergence of the integrals of f, that is
integral f n d mu. So, this proves Monotone
convergence theorem.
Remark: We have proved the theorem, monotone
convergence for f n is an increasing sequence.
So, naturally the question arises will the
similar result hold, if I have a decreasing
sequence f n of non- negative measurable functions.
That results unfortunately is not true.
Here is an example which says that if f n
is a sequence of functions which are non-
negative, measurable and if that decreases
to a function f, then integral of f need not
be equal to integral of f n d mu. This example
is on the Lebesgue measurable space. Look
at X to be the real line R, the sigma- algebra
S to be the sigma- algebra of Lebesgue measurable
sets L and mu to be the Lebesgue measure lambda.
Look at the function f n, which is the indication
function of the interval n to infinity. Claim:
Each f n is actually a non- negative, simple
measurable function and is decreasing identically
to the identity function 0. That is quite
obvious to see. So what is f n? So, we are
looking at, so here is n and we are looking
at the interval n to infinity.
So, we are looking at this interval and we
are looking at the indicator function of n
to infinity. So, the function is 0 and it
is 1 here. So, the function in this side is
1, so, this is the function f n. It is 0 here,
up to here and then it starts and goes, so
that is the function f n. We take n plus 1
here, so this is n plus 1. So, n plus 1 will
be 0 here, but f n is equal to 1 here. So,
clearly f n of x is bigger than or equal to
f n plus 1 of x for every x.
So, f n is a sequence in L plus and f n is
decreasing. Claim: f n decrease to f of x
which is identically equal to 0 for every
x. and that because If I take any point x
on the real line, then I can find some integer
n, say n naught which is on the right side
of it then, for every x belonging to real
line R and fix it. I can find a point n, not
a positive integer n naught of course, it
will depend on x says that n naught of x is
bigger the x. So, that will imply that the
indicator function of n naught to infinity,
or n naught to infinity let us even n to infinity
at x is going to be equal to 0, for every
n bigger than or equal to n naught. That is
my f n of x. So f n of x is equal to 0, for
every n bigger than n naught. So that means,
f n of x converges to f of x which is equal
to zero.
So, f n is a sequence of non- negative measurable
functions which is decreasing to f identically
zero. But we if you look at the integral of
each f n. What is the integral of each f n?
So integral of f n, d lambda. So, there is
integral of the indicator function n to infinity
d lambda.
So, that is equal to lambda of n to plus infinity
and that is equal to plus infinity, for every
n. So, integral of f n is equal to plus infinity
for every n. And integral of f d lambda is
equal to 0 as f is 0. This implies that integral
f n d lambda does not converge to integral
f d lambda, whenever f n is a decreasing sequence
of function non- negative simple non- negative
even simple function. We are given example
here So, for decreasing sequences this result
does not hold. So, that gives an importance
to monotone convergence here. That means,
whenever a sequence f 1 of non- negative,
measurable function is increasing, then integral
f is equal to limit integral f n d mu. For
decreasing, this n does not hold. So, this
is what we have shown just now by an example.
However, 1 can prove not n equality, but some
kind of inequality for a sequence of non-
negative measurable functions and that is
also an important result.
.
So, let us prove the result which is called
Fatou’s lemma. It says: let f n be a sequence
of non- negative, measurable functions. Then
the integral of limit inferior of f n d mu
is less than or equal to limit inferior of
the integrals f n d mu. So, this is only on
inequality and it need not be any equality.
So what we are saying is, if f n is a sequence
of non- negative, measurable functions, then
it is always true that the integral of the
limit inferior of f n is less than or equal
to limit inferior of the integral of f n.
So, let us give a proof of this theorem, so
to prove this theorem.
So, let us just once recall, what is so, f
n is the sequence of non- negative, measurable
functions. So, each f n is a non- negative,
measurable function. We want to look at limit
inferior of f n as n goes to infinity, this
is a function. Let us observe how this function
limit inferior of f n at a point x is defined.
You take the infimum from some stage on words.
So, m bigger than or equal to n of f n of
x. Look at the numbers, f n of f x and f m
of x, for m bigger than or equal to n. So,
I am looking at the tail of the sequence,
f n of x from m onwards. So, this number ‘infimum’
will depend on m. Let me take the supremum
of this overall m. so, first take the infimum
from some stage on words and then take the
supremum of these infimums.
So, let us observe that this infimum. Let
us put a bracket here. So, observe so, let
me call the infimum from the stage n onwards
as phi n. So, infimum of m bigger than or
equal to n of f n of x. phi m of x to be defined
as the infimum from the stage n onwards of
f m of x. So, then because it is an infimum
of a sequence of functions, which are non-
negative measurable. Clearly, note that each
phi n is also a non- negative, measurable
function.
So, it is a non- negative measurable function.
that is one. Secondly, we are taking the infimum
from some stage n onwards. So, if you increase,
the Claim is: this phi n is increasing this
is an increasing sequence. because phi. So,
phi n is the infimum from the stage n onwards,
is going to be less than or equal to the infimum
from the stage n plus 1 onwards because we
will have more numbers for which you have
taking infimum. So, infimum can the infimum
when you take infimum of more numbers, then
infimum will decrease. So, the infimum from
the stage n plus 1 n plus 1 onwards, so that
says that the infimum from the stage n plus
1 onwards, will be bigger than or equal to
the infimum from the stage n onwards. So increasing,
that is phi n plus 1 of x is bigger than or
equal to phi n of x, for every n. So, it is
an increasing sequence of non- negative, measurable
functions. And its limit is nothing, but the
limit inferior.
So, it is increasing and limit n going to
infinity of phi n is equal to limit inferior
of f n. n going to infinity, So the stage
is set perfectly for an application of Monotone
convergence theorem, phi n is a sequence of
non- negative measurable functions phi n s
are increasing.
Monotone convergence theorem, so, we can apply
implies by monotone convergence theorem, by
monotone convergence theorem that integral
of limit n going to infinity of phi n d mu,
is equal to limit n going to infinity of integral
phi n d mu n going to infinity. So, this is
nothing, but so, this side is nothing left
hand side is nothing, but integral of limit
inferior n going to infinity of f n, d mu.
So, that is equal to limit of integral phi
n of integral phi n’s. Now let us look at
what is phi n? This phi n is the infimum from
the stage n onwards. So each phi n is less
than or equal to f n. So, that is the observation
from here by the definition of phi n. We have
that each phi n is less than or equal to f
n, so, integral of phi n will be less than
or equal to integral of f n. So, it will be
less than or equal to limit inferior of n
going to infinity.
So, what we are observing here is, because
each phi n is less than or equal to f n. So,
this implies. So, This is what we are using
here. that is phi n is less than or equal
to f n, Then the n the limit n going to infinity
of integrals of phi n d mu are increasing.
so it limit exist However, integral of phi
n and integral of f n may not exists. So the
limit n going to infinity of integral phi
n d mu will be less than or equal to limit
inferior of integral f n d mu. So, this is
what is being used in this conclusion. And
that proves the theorem, what is called the
Fatou’s lemma.
Thanks
