Welcome to lecture 20 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?
Then we will start looking at various properties
of 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 valued measurable
function is said to be integrable with respect
to mu and is 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.
And in case, when integral f plus d mu and
integral f minus d mu, both are finite numbers,
then we say the integral of f is equal to
integral of f plus minus integral of f minus.
Let us once again emphasize in saying that
a function is mu- integrable, is if and only
if both f plus and f minus are having finite
integrals. The integral of f is written as
integral of f plus minus integral of f minus.
The class of all integrable functions on the
measure space X S mu is normally denoted by
L 1 capital L lower 1 X S mu or sometimes
we drop S and mu. If they the sigma- algebras
are clear from the context, that what are,
or what then, what is the measure? Or sometimes
we just emphasize mu, because we want to we
know what is X and what is S. So, 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. So, this is a space of
all mu- integrable functions.
We will start looking at the properties of
the functions.
The first important thing we observed was
a function, f is which is measurable, is integrable,
if and only if mod f which is a non- negative
function is integrable. That means, to check
whether a function a measurable function is
integrable or not, it is to enough to look
at the integral of the function, mod f and
see whether that is finite or not. And in
that case, and this is always true and that
for the integrable function, integral of a
mod integral mod of the integral of f d mu
is less than or equal to integral of a mod
f, d mu. This is an important criterion.
This is an equitant definition. Equitant way
of defining integral of a measurable function.
namely mod f is measurable, This is not equal,
this is wrong here. This is a typing mistake
here, It should be less than or equal to.
So, integral of mod of integral f d mu is
less than or equal to they should have been
less than or equal to integral of mod f, d
mu.
Let us recall some 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.
And we also prove the following property namely,
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. So, that essentially says that
the integral of the function does not change,
if the function is change, if the values of
the function the change almost everywhere.
So, f equal to g almost everywhere. f and
g measurable functions and one of them say
f is integrable implies g is integrable and
the integral of the two are equal. We also
proved the following property namely, f is
an integrable function and alpha is any real
number, then alpha time f is also integrable
and the integral of alpha f is equal to alpha
times the integral of f. So, we continue this
study of properties of integrable functions.
Next we want to check the integral property
namely, 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. So to prove
this property, let us look at what we are
given.
So, we are given that f and g are integrable
functions that is integral mod f d mu is finite
and integral of g d mu as absolute value of
g with respect to mu that is integral of mod
g d mu is also finite. So, to check whether
the function f plus g is integrable or not,
we have to look at the integral of absolute
value of f plus g and show that integral of
absolute value of f plus g is also finite.
But 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. And all are non- negative, measurable functions.
So, using the property of the integral for
non- negative measurable functions, this implies
that integral of mod f plus g d mu is less
than or equal to integral of mod f plus mod
g d mu. And that by linearity, it is same
as integral mod f d mu plus integral mod g
d mu. We are given that both of them are finite,
so this is finite, 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.
So, f and g integrable means f and g are in
L1 of mu, so that implies integral of f plus
d mu is finite or less than plus infinity.
And integral of f minus d mu is finite or
less than plus infinity. The integral of positive
part of g that is g plus is finite, integral
of g minus d mu is finite and we have to show
so to show integral f plus g plus d mu is
finite and integral f plus g minus d mu is
finite. So, these two properties we have to
show. this, somehow we have to Let us express
the positive part of f plus g with the positive
part of f and positive part of g. Similarly,
the negative part of f plus g, with the negative
part of f and negative part of g.
And that is done as follows: So what we do.
Look at f plus g. By definition, we can write
it as f plus g, positive part minus f plus
g, the negative part. So, that is by the definition
of the positive part and the negative part
of the function. Also, f plus g, we can also
write it as decompose f into positive part
and the negative part. So that is f plus minus
f minus. Similarly, write g as g plus minus
g minus.
Now, from these two it follows that integral
of f sorry not the integral from this it follows
that, f plus g plus, the positive part minus
f plus g minus, the negative part is equal
to f plus minus f minus plus g plus minus
g minus. So, from these two equations, it
follows this. is, so and
Now what we do is: we shift all the negative
terms 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.
Rearrange the terms. Now observe that the
left hand side is a non- negative function
and the right hand side is a non- negative
function. So by the properties of integrals
for non- negative functions, implies that
integral of f plus g plus d mu plus integral
of f minus d mu plus integral of g minus d
mu. So that is the integral of the left hand
side equal to integral of f plus d mu and
plus integral g plus d mu plus integral f
plus g minus d mu.
So from this equation, by using the properties
of integral for non- negative functions, the
linearity property the integral of the left
hand side is equal to integral of the right
hand side. And 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, are real
numbers. That is because f plus g we have
already shown is integrable. So this first
integral of f plus g plus is finite, integral
f minus is finite. Similarly, or the all the
terms are non- negative real numbers.
We can again manipulate them. and treat Shift
comes on the left hand side and right hand
side. What we will do is: bring this term,
f plus g minus of the right hand side, on
the left hand side. The terms, integral f
minus d mu and integral g minus d mu, we shift
on the right hand side.
So, that gives us the property. So shifting
implies that integral of f plus g plus d mu
minus this term, integral of f plus g minus
d mu will give you integral f plus g minus
d mu. So this term we have shifted. Shift
these two term on other side, so it is equal
to integral f plus d mu, that is this term.
And this minus integral of f minus d mu bringing
on this side will give you integral f minus
d mu plus integral of g plus d mu, which is
already there. And integral of g minus from
the left hand side will give you integral
of g minus d mu.
So this rearrangement of the terms here. Once
again give you the integral of f plus g plus
d mu and the integral of the negative part
of f plus g is equal 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 d mu and
the right hand side is integral f d mu plus
integral g d mu. So, that proves the linearity
property of the integral. That is if f and
g are integrable functions, not only f plus
g is integrable, integral of f plus g d mu
is equal to integral of f d mu plus integral
of g d mu.
So, 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 say, f. Multiply it by
scalar alpha, then alpha times f is integrable.
Moreover, 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.
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, let f be in L1 of
mu. Let us look at We have already shown,
that if mod f is a non- negative measurable
function and if you multiply it by the indicator
function of asset E, that is nu of E, then
we already shown that this is again a non-
negative measurable function. And 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. Claim: nu is an measure
infinite measure. It has the property that
mu of E equal to 0 implies nu of E equal to
0. So, whenever asset E has got mu measure
0, the measure of nu also is going to be equal
to 0.
And this basically follows from the properties
of the integral for non- negative functions,
because if f is integrable, then mod f is
a non- negative 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. So this integral is going to be
less than or equal to integral of mod f which
is finite. Obviously, for non- negative functions,
we have already proved this property, that
mu of E is 0, then the integral over E is
equal to 0. So this property follows from
our earlier discussions.
Let us look at integral of for the integral
function not the integral of mod f, but let
us look at the integral of f time’s indicator
function of E. And 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 a is again a measurable function. Just
now, we have observed that this number is
going to be a finite number because, this
is again an integrable function.
So, let us just observe this property once
again that, if f belongs to L 1 of mu and
E is a set in the sigma algebra S, then this
implies that chi E times f is a measurable
function. That we have already seen. f is
a measurable function, indicator function
of E is a measurable function and it is a
product of measurable functions. measurable,
We observed just now, If you look at the absolute
value of chi of E times f, that is same as
indicator function of E because that is negative
into absolute value of f. So this implies
that the integral of chi E times the absolute
value f, d mu is less than or actually is
equal to integral chi E of mod f d mu, which
is less than or equal to integral mod f d
mu which is finite.
So, what does it imply? This implies that
integral chi E d mu is a real number. This
we are denoting by nu tilde of E, this is
a real number. Therefore, we can say, this
is a finite real number.
So, that is the observation. Claim: mu of
E is equal to 0 implies that tilde nu of E
is also equal to 0. So, the claim is that
this claim is that this property, so let us
prove this property.
Suppose mu of E is equal to 0, then what is
tilde nu of E? By definition, tilde nu of
E is integral chi E of f d mu, which is same
as the integral of chi E f plus d mu, minus
integral chi E of f minus d mu. Now we observe
mu of E equal to 0, chi E of f plus is a non-
negative function. Properties of non- negative
functions imply if the set has measure 0,
then the integral of this is equal to 0. So,
the first integral is equal to 0, second integral
is equal to 0. By properties of integrals
of non- negative measurable functions, nu
tilde of E is equal to 0.
What we are saying is mu of E equal to 0 implies
tilde nu of E is also equal to 0. So that
is a property, we are proving here. But keep
in mind that tilde nu of E is defined as a
real number for every E belonging to S. It
is not a non- negative number because f may
not be non- negative function. So, we cannot
say tilde nu of E is a measure. We will look
at the property a bit later. It may not be
a measure, but it has some property similar
to a measure.
Here is another important property. Let us
look again at the value nu tilde of E. It
is equal to integral over E of f d mu or integral
chi over E d mu. Suppose this is equal to
0 for every set E in the sigma algebra S.
Claim: in this function f must be equal to
0 for almost all x belonging to mu.
Let us prove this property. Given tilde nu
of E, which is nothing but integral chi E
times f d mu is equal to 0 for every E belonging
to S. So, that is what is given to us, and
We want to show that If you take the set N,
which is x belonging to X such that 
mod f of x bigger than 0, if we write this
set N, then note that this set N is in the
sigma algebra S. We want to show that mu of
N is 0. we want to show of Almost everywhere,
f is 0 and this is the set where f is not
0, so we want to show this mu of N is equal
to 0. so this is the problem we want to show.
Now, let us look at 
consider the set say for example, let us look
at Let us write the set say A n, to be the
set, where x belongs to X such that f of x
is bigger than one over n. Similarly, let
us write the set B n, where x belonging to
X such that f of x is less than minus one
by n.
Claim: The set N is nothing but union n equal
to 1 to infinity of A n, union with union
n equal to 1 to infinity of B n. union of
Union n equal to 1 to infinity means, if you
take the union of all these sets, A n’s
and B n’s. That is precisely set N. where
n is What is that 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 one over n then 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 1 over minus
1 over n, for some n. So it belong to B n.
So, every set N every point x in N either
belongs to A n or belongs to B n. Obviously,
if x belongs to A n or B n, then f of x is
not equal to 0, so it belongs to N. so N is
equal to this.
So, N written as a countable union of sets.
All of these are sets in the sigma- algebra
S. We want to show this union has measure
zero. So 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. 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. So, let us write what
we are saying is the following: show that
mu of N is equal to 0.
Suppose, mu of N is bigger 0, then that implies,
then this condition implies there exists some
n naught, such that either mu of A n naught
is bigger than 0 or mu of B n naught is bigger
than 0. because If not, then mu of N will
be equal to 0.
Let us look at these conditions. Suppose the
first one if mu of A n naught is equal to
is bigger than 0, then look at the integral.
Then, integral of f ok then, integral of f
over the set A n naught. Let us look at integral
of f over the set A n naught, that is equal
to integral, so which is same as integral
chi A n naught times f d mu.
Now, on the set f, integral A n naught f d
mu is bigger than 1 over n. This is bigger
than this integral one by n naught times mu
of A n naught. Let us observe that outside
A n naught, this function is equal to 0, indicator
function of A n naught times is 0. And on
A n naught, f is bigger than 1 over n naught.
So, this function is bigger than one over
n naught d mu. and outside A n naught is zero
so this is going to be bigger than, so it
is bigger than integral over A n naught of
one over n naught d mu, So that is what we
are saying. Once that is true and this is
nothing but this integral and that is bigger
than 0.
So, in case mu of A n naught is bigger than
0, integral of f over A n naught is going
to be bigger than 0. This is a contradiction.
because which This is not true. We are given
integral of f over every set E equal to 0,
which is not true. So, if this holds, then
one can prove it is a contradiction. Similarly,
if this holds, one can prove it is a contradiction.
Then the integral of f over B n naught will
be less than strictly less than 0, not equal
to 0. So in either case, both of these are
not possible. So, our assumption that mu of
An naught is bigger than 0 must be wrong.
Hence mu of A n naught implies that the measure
of the set A n naught is equal to 0 and N
was the set where f of x is bigger than 0.
So, this set A n naught has got measure zero.
This is what we wanted to prove. So we have
proved the property that if integral of a
function, over f is an integrable function,
embedded integral over E is equal to 0, for
every E belonging to S, then f must be equal
to 0 almost everywhere. So, this is a very
nice property and useful property.
Thanks
