So, we have two important results for non-
negative. So, we have got 2 important results
for sequences of non- negative measurable
functions. One of them is called the Monotone
convergence theorem which says: if f n is
an increasing sequence of functions, non-
negative, measurable functions increasing
to a function f, then integrals of f n will
converge to integral of f.
So, keep in mind monotone convergence theorem
is for a non negative sequence of non- negative,
measurable functions which is increasing to
f. And in case the sequence of non- negative,
measurable functions is not increasing, then
we have got Fatou’s lemma which says: that
that for any sequence f n of non- negative,
measurable functions, the integral of the
limit n going to infinity inferior of f n
d mu is going to be less than or equal to
is always less than or equal to limit n going
to infinity inferior of the integral of f
n d mu.
So, this is these are the two important theorems,
which help us to relate the limit of the integrals
with integral of the limits. We will see applications
of this in rest of our course. So with this,
we conclude the section on the definition
of integral for non- negative, simple non
negative measurable function. So, let us just
recall what we have done. We have started
with defining the integral of non- negative,
simple measurable functions. The functions
which look like linear combinations of indicator
functions: sigma A i, indicator function of
A i. For them, we defined the integral to
be nothing, but A i times summation of A i
times mu of A i. We showed it is independent
of the representation. We proved various properties
of the integral for non- negative, simple
measurable functions.
And then we looked at the class of non- negative,
measurable functions. Since the every non-
negative, measurable function is limit of
sum of sequences of non- negative, simple
measurable functions increasing to that function
f. So, we defined integral of the non- negative,
simple measurable function to be nothing,
but the limit of the integrals of that sequence
of non- negative, simple measurable functions
increasing to it. And we showed this integral
is independent of the limit of a sequence
s n. You choose or you select which increases
to f. Then we proved various properties including
Monotone convergence theorem and Fatou’s
lemma.
And now let us look at how can we define the
integral for a function, which is not necessarily
non- negative. So, for that we will do the
following: we will start with. So, let us
look at a function f.
So keep in mind, we have got a major space
(X, S, mu) which is complete. f is a function
which is defined on X, taking extended real
values and we want to define. So, to define
integral f d mu. This is what we want to know,
what it should look like and of course, we
would like this integral to have nice properties.
So, we would like to have it to be a fun diagonal
in a linear operation.
Now recall. So, let us recall, the function
f can be written as f plus minus f minus.
We can split into two parts: the positive
part and the negative part of the function,
where f plus and f minus, both are non- negative
functions. If f is measurable of course, with
respect to the sigma algebra S, that is true
if and only if both f plus and f minus are
measurable. So, f can be written. So, this
is the clue how we should go about.
So, f can be written as a difference of two
non- negative, measurable functions. If f
is measurable and integral of non- negative
functions is defined, then integral of f plus
is defined, integral of f minus is defined
and if our integration is going to be linear,
it is all but necessary that our integral.
we should define integral f d mu whatever
be the way.
So, we define it. It should have the property:
this is f plus d mu minus integral of f minus
d mu. So, this is what we would like to have.
This is defined, this is defined. Now the
question: is the difference defined?
So, the difference will be defined if both
of these quantities are finite numbers. So,
the left hand side is defined, if f plus d
mu is finite and integral f minus d mu is
finite. So, that says so; that means, whenever
f is a measurable function, that integral
f plus d mu is finite, integral f minus d
mu is finite. We can define its integral to
be equal to integral equal to integral of
f plus d mu minus integral f minus d mu. So,
with that we let us define what is the integral
what is called an integrable function.
So, a measurable function f defined on X,
taking extended real valued real values is
said to be mu- integrable. Of course, mu is
the measure on underlying space X, which is
fixed. So, it is said to be mu- integrable.
In both, integral of f plus d mu, f plus is
a non- negative measurable function and of
integral f minus d mu, f minus is a non- negative
measurable function. So by our earlier discussion,
both these numbers f plus d mu and f minus
d mu are defined.
So, if they are both finite in that case we
say that the function f is integrable and
its integral is defined as integral of f plus
minus integral f minus. So, integral of f
is defined as integral of the positive part
of the function minus the integral of the
negative part of the function. So, whenever
a function f is defined on X, we say f is
integrable. If both, the positive part and
the negative part have finite integrals. In
that case, we define the integral of f. So,
written as integral f d mu to be integral
of f plus minus integral of f minus.
So, we will denote by the symbol capital L
lower 1 of (X, S, mu) to be the class of all
mu-integrable functions. So in case, it is
not clear what is X, what is mu and what is
S, we can simple. So, sometimes simply write
it as L 1 of X or simply L 1 of mu. So, if
we understand what is underlying measure space.
So, the space of integrable functions, either
it will be a explicitly written as L 1 of
(X, S, mu) or sometimes simply as L 1 of X
or L 1 of mu. So, this is the class of all
integrable functions. That means, all functions
f such that integral f plus is finite, integral
f minus is finite and in that case integral
of f is defined as integral f plus minus integral
of f minus. So, for all integrable functions
f belonging to L 1 of X, we have integral
f d mu. We will now study the properties of
integral.
So, the first property is: let us fix functions
f and g which are integrable. and f and g
to be which are f and g to be real number
a n g to be real numbers. So, if f and g are
measurable functions at mod f of x is less
than g of x for almost all x and g belongs
to L 1 of mu, then the claim is f is in L1
of mu. So, this is a very simple property,
we want to check. namely that if.
Let f and g are measurable functions on X.
We are given that mod f of x is less than
or equal to g of x almost everywhere mu.
Claim: If g is in L 1 of X, then that implies
f is in L 1 of X. To prove that let us observe.
So, note what is given: f x is less than or
equal to g x almost everywhere X. Let us define
N to be the set of all points x belonging
to X, where mod f of x is not less than or
equal to g of x, where this property is not
true. We know that N belongs to the sigma-
algebra S and mu of N is equal to 0. n of
n is equal to 0.
Now, note because mod of f is a non- negative.
So, mod f belongs to L plus. g is a non- negative
measurable. So, g belongs to L plus and is
almost everywhere. So, mod f of x is less
than or equal to g of x on N c complement.
So, that is what is given to us. Thus, implies
integral mod f x, d mu x, which we can write
as integral over N mod f x d mu x plus integral
over N c complement mod f x d mu.
Now let us observe. So, let us now. So, observe
that the set N has measure 0. So, this part
is 0 and N c complement on N compliment, f
is mod f of x is less than or equal to g of
x.
So, this is equal to 0 plus integral over
N c complement of mod f x d mu x, and on N
c complement, f is less than or equal to g.
So, this is less than or equal to integral
over N c complement of g of x, d mu of x and
that is less than or equal to integral over
the whole space X of g of x, d mu x which
is finite.
So, what we have shown is: mod f of x is less
than or equal to g of x. Then we have shown
that the integral of mod f is finite, that
is integral of mod f is less than plus infinity.
So, this says. So, hence integral mod f d
mu is finite and Now let us note that f plus
is always less than or equal to mod f and
f minus is also less than or equal to mod
f. For any function f, the positive part is
less than or equal to mod f, the negative
part also is less than equal to mod f. So,
that implies that both integral f plus d mu
and integral f minus d mu are less than or
equal to integral mod f and is less than plus
infinity that is finite.
So, we have shown that the integral of f plus
and integral of f minus, both are finite and
whenever mod of f is less than or equal to.
So, whenever. So, what we have shown is whenever
mod. So, this property is true this implies
that integral of f plus and integral of f
minus both are finite. So, that implies. So,
implies f belongs to L 1 and further.
Let us calculate integral of mod f d mu. If
you recall mod f is nothing, but 
f plus plus f minus that means, integral of
mod f is equal to integral f plus d mu plus
integral of f minus d mu. So, integral of
mod f is less than and equal to integral of
g d mu.
.
And both of them are finite. So, that we have
already observed. I wanted to check that integral
of mod f is less than or equal to integral
of integral g d mu, which we have already
actually checked. We have also checked that
integral of f plus. So, which is less than
integral of mod f, which is less than. So,
mod f is less than integral g implies. So,
you do not have do this. So, is less than
or equal to integral of g d mu, that follows
from directly from that mod f. Now we have
shown f and is integrable and mod f is integral
which is finite. So, this integral mod f d
mu is less than or equal to integral g d mu.
So, this proves the first property namely,
if f and g are measurable functions and mod
f x is less than or equal to g x for almost
all x or x of mu and if g is integrable, then
f is integrable. So, what we are saying is:
if a function f of x which is measurable is
dominated by a function g which is integrable,
then the function f also becomes integrable.
Let us look at the next second property that
if f and g are equal almost everywhere and
f is integrable, then g is integrable and
the integrals of the two are equal and that
property is something similar to that we have
just now shown a similar analysis will work.
So, let us look at we have got two functions,
f and g. And f of x is equal to g of x almost
everywhere that is x of mu. So, let us write
the set N, where x belonging to X and f x
is not equal to g x. Then by the given condition,
mu of N not equal to that is equal to 0 and
f x equal to g x for every x belonging to
N c complement. Now, let us look at. So, we
have given that the function f is f belongs
to L 1.
.
So, implies We want to show that g belongs
to L 1. And that is because if f x is equal
to g x almost everywhere, then that implies
mod f x is also less than or equal to mod
g x almost everywhere.
Right the sets where they are not equal. So,
Wherever the sets are equal, so that is mod
x is equal to g x, because on then on compliment
also that will happen. So, this is less than
or equal to this. That implies integral of
f x is equal to integral of g x. So, sorry.
So, we should say that f x is equal to g x
almost everywhere. So, f x is equal to g x
that implies mod f x is equal to mod g x.
And just now we have showed that whenever
f and g are equal almost everywhere, integral
mod f d mu is equal to integral mod g d mu.
So, either of them as finite implies other
is finite too. We are given this mod f d mu
is finite implies mod g d mu is finite.
Once again implies that g in L 1, and g is
L 1. so, integral of g d mu is equal to integral
g plus d mu minus integral g minus d mu. But
f is equal to g almost everywhere. We ask
the reader to verify this. So that means,
f of f plus must be equal to g plus and f
minus must be equal to g minus almost everywhere.
Once again, this integral is equal to minus
integral of f plus d mu minus integral of
f minus d mu which is nothing but equal to
integral of f d mu. So, our integral g is
equal to integral f, whenever f and g are
equal almost everywhere. So, these are simple
properties of our integrable functions that
we have looked at.
Note: if f is equal to g almost everywhere
and one of them is integrable, then the other
is integrable and the two integrals are equal.
Next let us check the property of linearity.
So, if f is in L 1, then we want to check
that alpha f is also in L 1. And alpha f of
d mu is equal to alpha times integral of f
d mu. So, to check that property, let us observe
one thing that saying that just now we looked
at. This kind of analysis namely, if f belongs
to L 1 of X, it is same as if and only if
mod f belongs to L 1 of X.
So, why is that? Once again let us do this
because this, we are going to use it again
and again. Saying that f belongs to L 1 implies
integral of f plus d mu is finite means less
than infinity and integral of f minus d mu
is finite means less than infinity. So, that
implies because What is mod f? Mod f is equal
to f plus, plus f minus, that implies integral
f plus d mu plus integral f minus d mu is
finite means less than infinity and this is
equal to integral of mod f d mu. So, f belonging
to L 1 implies integral of mod f is finite
conversely. Let us look at the converse part,
that if integral mod f is finite.
So, let us. So, conversely, let us this is
given to us that integral of mod f d mu is
finite. Once again let us observe that f plus
is less than or equal to mod f and f minus
is less than or equal to mod f. So, that implies
all are non- negative, measurable functions.
That implies integral of f plus d mu is less
than integral mod f d mu which is finite means
less than infinity and integral f minus d
mu is less than integral mod f d mu which
is finite means less than infinity. So, that
implies that f belongs to L 1.
So, saying that a function is integrable is
equivalent to saying that mod f which is a
non- negative, measurable function has got
finite integral. So, this property will be
used again and again. Let us see how that
the property is used in our proposition now.
So, a belongs to real line and f is L 1. So,
look at mod of a f. So, mod of a f is less
than or equal to mod a, mod f. All are non-
negative functions. So, integral of mod a
f is less than or equal to integral of this
which is mod a times integral mod f d mu which
is finite that is less than infinity. So,
that implies that a f is an integrable function.
Now, we can write not only it is integrable,
integral of a f d mu is equal to integral
of a f plus d mu minus integral of a f minus
d mu.
And now possibilities:
1) Either a is equal to 0, then in this case,
a f will be 0 and everything is 0. So no problem.
2) If a is positive, then this part is same
as a times f plus d mu minus a times integral
f minus d mu. If a is positive and so this
a comes out, because of the property for nonnegative
measurable for integral of nonnegative measurable
functions. So, this will be finite.
3) In case a is less than 0, this becomes
a of a minus, negative part and again that
thing is same. Similarly, for a minus. So,
that proves the property that if f is integrable
and a is a real number, then a f is integrable
and a comes out.
So, we will continue looking at the properties
of integrable functions in the next lecture.
We will show that in this, integral is a linear
operation on the space of integrable functions
and various other properties of this space
of integrable functions. So, we will continue
this study in the next lecture.
Thank you.
