Welcome to lecture number 19 on measure and
integration. In the previous lecture, we
had started looking at the properties of integral
for nonnegative measurable functions.
We had looked at the linearity property of
the integral for nonnegative measurable
functions and then we said we will start looking
at the limiting properties of functions
which are nonnegative 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 and then go over to
define integral for general functions.
.
Let us look at what is called as monotone
convergence theorem. The monotone
convergence theorem says that let f n be a
sequence of functions in class L plus. That
means f n is a sequence of nonnegative measurable
functions increasing to a function f of
x at every point; that means f of x for every
x in X is limit n going to infinity of f n
of x.
We are given a sequence of f n of nonnegative
measurable functions which is increasing
and the limit is f of x.
Then the claim is the function f belongs to
L plus, this we have already observed, and
the
additional property is that the integral of
the limit f into d mu is same as limit of
the
integrals of f n into d mu. That means whenever
a sequence f n of nonnegative measurable
functions increases to f, then integral of
the limit is equal to limit of the integrals.
This is
the first important theorem about convergence
of sequences of nonnegative measurable
functions and their integrals. Let us prove
this property.
.
We are given f n is a sequence; each f n belongs
to L plus; it is a nonnegative measurable
function for every n bigger than or equal
to 1. That implies there exists a sequence,
we
will denote it by s j n, of functions n bigger
than or equal to 1 such that s j n are
nonnegative measurable simple functions for
every n and for every j and s j n increases
to
f of…
Let us fix the notation – which one we are
going to vary. Let us say that the upper one
will be fixed; so, this is going to f n as
j goes to infinity .. For
every n fixed, s j n is a sequence on nonnegative
simple measurable functions increasing
to f n s and f n s increase to f. We want
to show, we have already shown but we will
show
again, that this implies f belongs to L plus
is a nonnegative function and integral f into
d
mu is equal to limit n going to infinity integral
f n into d mu. To prove this, we are going
to use this sequence s n s and construct a
new sequence of nonnegative simple measurable
functions out of it.
.
What we will do is the following. Let us write
for n is equal to 1 that s 1 1, s 2 1 up to
s j 1
converges to f 1 . The upper index is going
to give you s 1 2, s 2 2 up to s j 2 increases
to f 2 .
In general, we will have s 1 n, s 2 n up to
s j n will increase to f n and so on; this
increases
to f ..
Let us observe that as we go from left to
right, this is increasing; everywhere left
to right,
it is increasing .; down to up, that also
is increasing. If you look
at every sequence, this is an array of nonnegative
simple measurable functions; each row
is increasing to the function on the right
side and this is increasing upwards.
.
Let us look at the function. Let me define
from this a function; let us define g n to
be the
function which is maximum of s j n, j between
1 and n.
.
In essence, what I am doing is in this picture,
let us say here is s n 1, here is s n 2 and
here
is s n n. I look at this column; we are looking
at the column s n 1; let us look at this
column and call the maximum of this to be
g n .
.
What is g n ? Let me write again. g n is the
maximum; so, define g n equal to maximum of
s j n, j equal to 1 to n. Let us observe that
each g n is a maximum of nonnegative simple
measurable functions; each g n is a nonnegative
simple measurable function for every n.
g n is increasing because at the next stage
n plus 1…
.
This is going to be bigger in the next stage.
If we look at g n plus 1 , that is going to
be s n plus
1
1, s n plus 1 2 and so on s n plus 1 n plus
1 n and s n plus 1 n. This one (Refer Slide
Time:
08:05) is going to be bigger than everything
on the left-hand side and we are looking at
.he maximum; the maximum of these is going
to be bigger than or equal to maximum of
these because at each the right-hand side
function is bigger than the left-hand side
function. So, this is going to give us that
at each g n is an increasing sequence of
functions ..
.
Let us write let g be equal to limit n going
to infinity of g n .
.
All these g n s are increasing and they are
going to increase to some function g. What
we
are going to show is g is equal to f; that
is what we are going to check.
.
Clearly by definition, g is a nonnegative
simple measurable function because it is a
limit
of an increasing sequence of nonnegative simple
measurable functions; so, g belongs to
L plus. Also let us observe that each g n
is less than equal to f n ; each g n is less
than or
equal to f n for every n. That is because
g n is the maximum of this (Refer Slide Time:
09:38) and the maximum of each one of them
is less than f 1 is less than f 2 is less
than f n .
The maximum of these g n s is going to be
less than or equal to this f n for every n.
f n is
increasing to f and 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.
.
.t implies that g n is less than or equal
to f n and is less than or equal to f for
every n.
Hence, when g n is increasing to g, that implies
g is less than or equal to f; 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; the claim is that 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,
g n is the maximum of this and so this is
less than
or equal to g n for every n; this is less
than or equal to g n for every n and j between
less
than this.
If we fix and g n is less than or equal to
so so and this is less than or equal to g;
so, s j n is
less than or equal to g n 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; as n goes to infinity, what happens?
This converges to f n . Note that as n goes
to infinity, s j n goes to f j . From this
and this,
these two observations, s j n is less than
or equal to g for every…. If you fix j and
let n go
to infinity, then n is going to cross over
j and s j n as n goes to infinity converges
to f j
.. This implies f j is less than or equal
to g for every j.
.
This implies that f j is less than or equal
to g for every j; f j s are increasing and
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, one observation
from here is hence that g belongs to L
.lus and 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 nonnegative measurable,
then f is also nonnegative measurable.
Now note that integral of f into d mu is same
as integral of g into d mu because f is equal
to g. This is equal to limit n going to infinity
of integral g n into d mu because g n s are
nonnegative simple measurable increasing to
g. By definition this is so, but each g n
is
less than equal to f, if you recall .. Each
g n is less than or equal
to f; so, integral of g n will be less than
or equal to integral of f; so, limit of integrals
of
f n s will be less than or equal to integral
f. This is less than or equal to integral
f into d mu
.. You can even introduce in between; 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 n into d mu
which is less
than or equal to integral f into d mu.
What does this imply? Integral f into d mu
is less than or equal to limit f n integral
of f n
into d mu and that is less than or equal to
f into d mu. That implies that integral of
f into
d mu is equal to limit n going to infinity
integral f n into d mu. That proves the theorem
completely, namely integral of f d mu is equal
to limit n going to infinity integral of f
n d
mu. This is a construction which is quite
useful; this is the kind of analysis one has
to
carry out.
.
Let us go through the proof again so we that
understand what we are doing. Each f j or
f n
is a measurable function. So, I can look at
a sequence s 1 1, s 2 1, s j 1 up to s n 1
which is
.oing to increase to f 1 .. Similarly, the
upper ((.)) is fixed – 2;
so, s 1 2, s 2 2, s j 2 up to s n 2 increases
to f 2 and so on. Each row is increasing to
the
function on the right side and the functions
f 1 , f 2 up to f n s are increasing to the
function
f.
What we do is we look at the maximum of this
column .. What
is this column? This column is the maximum
of the functions s n 1, s n 2, up to s n n.
Call
this as g n ; this function is called g n
. The observation is each g n is a maximum
of
nonnegative simple measurable functions and
so it is nonnegative simple measurable.
Each g n is lesser or equal to f n because
we are going up to this corner only (Refer
Slide
Time.. So, each g n is less than or equal
to… because s n 1 is less than f 1 , s n
2 is
less than f 2 , 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. If we write the limit of g
n to be g,
then g is less than or equal to f by this
simple construction. Also, for any fixed j
let us
look at s j n. Let us look at s n j where
j is fixed and n is going to vary. As n varies,
what
happens to these functions? For every fixed
j, this sequence of functions is going to
be s j
n is less than or equal to g n and g n is
less than or equal to f. g is less than or
equal to f.
Sorry, s j n is less than or equal to g n
for j between 1 and n; that will give us that
f is also
less than or equal to g.
That will prove the theorem that limit of
increasing sequence of nonnegative measurable
functions.... . If f n is a sequence of nonnegative
measurable
functions increasing to f, then integral f
into d mu is equal to limit of n going to
infinity
integral f n into d mu. This is called monotone
convergence theorem, monotone because
we are looking at monotonically increasing
sequences f n s and convergence because we
are looking at the convergence of the integrals
– integral f n into d mu. This proves the
monotone convergence theorem.
.
We have proved the monotone convergence theorem
for when f n is an increasing
sequence .. Naturally, the question arises:
will the similar result
hold if I have a decreasing sequence f n of
nonnegative measurable functions? That result
unfortunately is not true. Here is an example
which says that if f n is a sequence of
functions which are nonnegative measurable
and they decrease to a function f, then
integral of f need not be equal to integral
of f n into d mu. The example is on the
Lebesgue measurable space.
Look at X to be the real line, the sigma algebra
to be the sigma algebra of Lebesgue
measurable sets and mu to be the Lebesgue
measure. Look at the function f n which is
the
indicator function of the interval n to infinity.
This is actually a nonnegative simple
measurable function – each f n ; f n is
decreasing and decreasing to the identity
function
identically equal to 0; that is quite obvious
to see.
.
What is f n ? Here is n and we are looking
at the interval n to infinity. We are looking
at
this interval .. We are looking at the indicator
function of n to
infinity. So, the function is 0 and it is
1 here. This height is 1; this is the function
f n ; it is
0 up to here .. Then, it starts and goes;
that is the function f n .
We take n plus 1; this is n plus 1; so n plus
1 will be 0 here . but
f n is equal to 1 here. 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. The claim is f n decreases to
f of x
which is identically equal to 0 for every
x. If I take any point x on the real line,
then I can
find some integer n, say, n 0 which is on
the right side of it. So, for every x belonging
lo
real line ((.)), I can find a point n 0 – a
positive integer n 0 (of course, it will depend
on x)
such that n 0 of x is bigger than x. That
will imply that the indicator function of
n to
infinity at x is going to be equal to 0 for
every n bigger than equal to n 0 and that
is my f n
of x. So, f n of x is equal to 0 for every
n bigger than n; that means f n of x converges
to f
of x which is equal to 0. So, f n is a sequence
of nonnegative measurable functions which
is decreasing to f identically 0, but if we
look at the integral of each f n , what is
the
integral of each f n ?
.
Integral of f n into d lambda is the integral
of the indicator function n to infinity d
lambda.
That is equal to lambda of n to plus infinity;
that is equal to plus infinity for every n.
Integral of f n is equal to plus infinity
for every n and integral of f into d lambda
is equal
to f is 0 and so it is 0. This implies that
integral f n into d lambda does not converge
to
integral f into d lambda whenever f n is a
decreasing sequence of function nonnegative,
simple nonnegative; even simple function we
have given example here (Refer Slide
Time..
For decreasing sequences, this result does
not hold; that gives the importance to the
monotone convergence theorem; that means whenever
a sequence f n of nonnegative
functions is increasing, then integral f is
equal to limit integral f n into d mu. For
decreasing, this need not hold; this is what
we have shown just now by an example.
.
However, one can prove not an inequality but
some kind of a inequality for a sequence of
nonnegative measurable functions. That is
also an important result
.
Let us prove the result which is called Fatou's
Lemma. It says let f n be a sequence of
nonnegative measurable functions. Then the
integral of limit inferior of f n into d mu
is
less than or equal to limit inferior of the
integrals f n into d mu. This is only an inequality
and it need not be an equality. What we are
saying is if f n is sequence of nonnegative
.easurable functions, then it is always true
that the integral of the limit inferior of
f n s is
less than or equal to limit inferior of the
integrals.
.
Let us give a proof of this theorem. To prove
this theorem, let us just first recall what
is
f n . It is a sequence of nonnegative measurable
functions; each f n is a nonnegative
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
is defined. Limit inferior of f n at a point
x is
defined as you take the infimum from some
stage onwards; so, it is m bigger than or
equal to n of f n of x.
Look at the numbers f m of x for m bigger
than or equal to n; I am looking at the tail
of
the sequence f n of x from m onwards. So,
this number infimum will depend on m (Refer
Slide Time.. Let me take the supremum of this
over all m. First, take the infimum
from some stage onwards and then take the
supremum of these infimums. Let us put a
bracket here. Let me call phi m to be the
infimum from the stage n onwards – 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.
Then, because it is an infimum of a sequence
of functions which are nonnegative
measurable, clearly the observation is that
each phi n is also a nonnegative measurable
function; so, it is a nonnegative measurable
function; that is one. Secondly, we are taking
.he infimum from some stage n onwards; so,
if we increase, the claim is this phi n is
increasing.
. phi n , 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 are taking infimum.
When you take infimum of more numbers,
then infimum can decrease; so, infimum from
the stage n onwards and the infimum from
the stage n plus 1 onwards…. That says that
the infimum from the stage n plus 1
onwards will be bigger than or equal to the
infimum; so, it is increasing; that is, phi
n plus 1
is bigger than or equal to phi n of x for
every n. It is an increasing sequence of
nonnegative measurable functions and its limit
is nothing but the limit inferior; so, it
is
increasing.
.
It is increasing and limit n going to infinity
of phi n is equal to limit inferior of f n
, n going
to infinity. The stage is set – perfect
– for application of monotone convergence
theorem
.. phi n is a sequence of nonnegative measurable
functions and
phi n s are increasing. We can apply monotone
convergence theorem. It implies that by
monotone convergence theorem the integral
of limit n going to infinity of phi n into
d mu
is equal to limit integral phi n into d mu,
n going to infinity.
The left-hand side is nothing but integral
of limit inferior n going to infinity of f
n into d
mu; that is equal to limit of integral f n
s. Now, let us look at what is phi n (Refer
Slide
.ime.. phi n is the infimum from the stage
n onwards; each phi n is less than or
equal to f n ; that is the observation from
here by the definition of phi n (Refer Slide
Time:
28:50). We have that 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 ; it
will be less than or equal to limit inferior
of n going
to infinity.
What we are observing here is because each
phi n is less than or equal to f n , this
implies
(this is what we are using here) that if phi
n is less than or equal to f n , then the
integrals of
phi n s are increasing; so, its limit exists;
so, it is limit n going to infinity integral
phi n into
d mu is less than or equal to… However,
integral of f n s may not exist. We can say
that it
will be less than or equal to limit inferior
of integral f n s into d mu; this is what
is being
used in this conclusion.
That proves the theorem called Fatou's Lemma.
So, we have two important results for
nonnegative .. We have got two important results
for sequences
of nonnegative measurable functions. One of
them is called the monotone convergence
theorem which says if f n is an increasing
sequence of nonnegative measurable functions
increasing to a function f, then integrals
of f n s will converge to integral of f. Keep
in
mind that the monotone convergence theorem
is for a nonnegative sequence of
nonnegative measurable functions which is
increasing to f.
In case the sequence of nonnegative measurable
functions is not increasing, then we have
Fatou's Lemma which says that for any sequence
f n of nonnegative measurable
functions, the integral of the limit inferior
of f n s is going to be less than or equal
to, is
always less than or equal to, limit inferior
of the integrals. These are the two important
theorems which help us to relate the limit
of the integrals with the integral of the
limits;
we will see applications of this in the rest
of our course.
With this, we conclude the section on definition
of an integral for nonnegative
measurable functions. Let us just recall what
we have done. We started with defining the
integral of nonnegative simple measurable
functions, the functions which look like linear
combinations of indicator functions sigma
a i indicator functions of A i and for them
we
defined the integral to be nothing but summation
of a i times mu of A i . We showed that it
is independent of the representation and we
proved various properties of the integral
for
nonnegative simple measurable functions.
.hen we looked at the class of nonnegative
measurable functions. Since every
nonnegative measurable function is a limit
of some sequence of nonnegative simple
measurable functions increasing to that function
f, we defined integral of the nonnegative
simple measurable functions to be nothing
but the limit of the integrals of that sequence
of nonnegative simple measurable functions
increasing to it. We showed this integral
is
independent of the limit of the sequence s
n you select which increases to f; then we
proved various properties including monotone
convergence theorem and Fatou's Lemma.
Now, let us look at how we can define the
integral for a function which is not necessarily
nonnegative.
.
For that, we will do the following. Let us
look at a function f. Keep in mind we have
got
a measure space X as mu which is complete;
f is a function which is defined on X taking
extended real values; we want to define integral
f into d mu – we want to know what it
should look like; of course, we would like
this integral to have nice properties; we
would
like to have it to be a function integral
to be a linear operation.
Let us recall the function f can be written
as f plus minus f minus. We can split it into
two parts – the positive part and the negative
part of the function – where f plus n f
minus are both nonnegative functions and 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.
This is the clue how we should go about it
.; f can be written as
. difference of two nonnegative measurable
functions if f is measurable and integral
of
nonnegative functions is defined; so, integral
of f plus is defined and integral of f minus
is defined.
.
If our integration is going to be linear,
it is all but necessary that our integral…
We
should define integral f into d mu; whatever
be the way we define it, it should have the
property f plus into d mu minus integral of
f minus into d mu; this is what we would like
to have. This is defined and this is defined
.. Now, the question
is: is the difference defined?
The difference will be defined if both of
these quantities are finite numbers. It is
defined
if f plus into d mu is finite and integral
f minus into d mu is finite. That means whenever
f is a measurable function, the integral f
plus into d mu is finite and integral f minus
into
d u is finite, we can define its integral
to be equal to integral of f plus into d mu
minus
integral f minus into d mu.
.
With that, let us define what is called an
integrable function. A measurable function
f
defined on X taking extended real values is
said to be mu integrable (of course mu is
the
measure underlying the space, which is fixed)
if both integral of f plus into d mu… f
plus
is a nonnegative measurable function and f
minus is a nonnegative measurable function;
so, by our earlier discussion, both these
numbers f plus into d mu and f minus into
d mu
are defined.
If they are both finite, in that case we say
that the function f is integrable and its
integral
is defined as integral f plus minus integral
f minus. Integral of f is defined as integral
of
the positive part of the function minus the
integral of the negative part of the function.
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 and
in that case we define the integral of f (we
have
written as f into d mu) to be integral of
f plus minus integral of f minus.
We will denote by the symbol capital L lower
1 of X, S, mu to be the class of all muintegrable
functions. In case it is clear what is X,
what is mu and what is S, we can
sometimes simply write it as L1 of X or simply
L 1 of mu. If we understand what is the
underlying measure space, the space of integrable
functions, either it will be explicitly
written as L 1 of X, S, mu or sometimes simply
as L1 of X or L 1 of mu. This is the class
of all integrable functions; that means all
functions f such that integral f plus is finite
and
integral f minus is finite; in that case,
integral of f is defined as integral f plus
minus
.ntegral of f minus. So for all integrable
functions f belonging to L 1 of X, we have
integral f into d mu.
.
We will now study the properties of this integral.
The first property is let us fix functions
f and g which are integrable and a and b to
be real numbers. If f and g are measurable
functions and mod f of x is less than g of
x for almost all x and g belongs to L1 , then
the
claim is f is in L1 .
.
.his is a very simple property we want to
check. f and g are measurable functions on
x
and we are given that mod f of x is less than
or equal to g of x almost everywhere mu.
From here, the first claim is that if g is
L1 of x, then that implies that f is in L1
of x. To
prove that, let us observe. Note that we are
given f of x is less than or equal to g of
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; here, this property is not true.
Then, we know that N belongs to the sigma
algebra and mu of N is equal to 0 n of n is
equal to 0. Now note that because mod of f
is nonnegative, mod f belongs to L plus; g
is
nonnegative measurable and so g belongs to
L plus. It is almost everywhere and so mod
f
of x is less than or equal to g of x on N
complement; that is what is given to us. This
implies mod f of x into d mu x which we can
write as integral over N mod f of x into d
mu of x plus integral over N complement mod
f of x into d mu. Now, let us observe that
the set N has got measure 0; so this part
is 0 . and on N
complement, mod f is less than or equal to
g.
.
This is equal to 0 plus integral over N complement
of mod f of x into d mu of x and on N
complement, f is less than or equal to g.
This is less than or equal to integral over
N
complement of g of x into d mu x. That is
less than or equal to integral over the whole
space g of x into d mu x which is finite.
What we have shown is that in case 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
(Refer
Slide Time..
Hence, integral mod f into d mu is finite.
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,
the
positive part is less than or equal to mod
f; the negative part also is less than or
equal to
mod f. That implies that integral f plus into
d mu and integral f minus into d mu – both
of
them are less than or equal to integral mod
f which is finite.
We have shown that the integral of f plus
and integral of f minus are both finite
whenever mod of f is less than or equal to.
What we have shown is this property is true
..
.
This implies that integral of f plus and integral
of f minus both are finite. That implies f
belongs to L 1 .
.
Further, let us calculate what is integral
of mod f into d mu. Mod f, if you recall,
is
nothing but f plus 
plus f minus. That means this is equal to
integral d mu plus integral of
f minus into d mu. So, integral of mod f is
nothing but integral of f plus plus integral
of f
minus into d mu. Both of them are finite;
that we have already observed. We wanted to
.heck that integral of mod f is less than
or equal to integral of integral g d mu, which
we
have already checked.
We have already checked that integral of f
plus which is less than integral of mod f
is
less than… So mod f is less than integral
g implies this and we do not have to do this
.. It is less than or equal to integral g
d mu; that follows directly
from that mod f. We have shown it is integrable
and its integral is finite; so, this is less
than or equal to integral of g.
. This proves the first property: if f and
g are measurable
functions and mod f of x is less than or equal
to g of x for almost all x and if g is
integrable, then f is integrable. 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 property that if f
and g are equal almost everywhere and f is
integrable, then g is integrable and integrals
of the two are equal. That property is
something similar to what we have just now
shown; a similar analysis will work.
.
We have two functions f and g; f of x is equal
to g of x almost everywhere mu. Let us
write the set N – x belonging to X where
f of x is not equal to g of x. Then, by the
given
condition, mu of N not equal to that is equal
to 0; f of x is equal to g of x for every
x
belonging to N complement.
.
.e are given that the function f is belonging
to L 1 . We want to show that g belongs to
L 1 and that is because if f of x is equal
to g of x almost everywhere, then that implies
that
mod f of x is also less than or equal to mod
g of x almost everywhere. The sets are not
equal; wherever they are equal, mod x is equal
to g x because on the N complement that
will happen; that is ((.)).
That implies integral of f of x is equal to
g of x. Sorry, we should say that f x is equal
to
g x almost everywhere .. That implies f of
x is equal to g of x;
mod f x is equal to mod g of x. Just now we
showed that whenever f and g are equal
almost everywhere, integral mod f d mu is
equal to integral mod g d mu. Either of them
finite implies the other is finite. We are
given this is finite and so this implies mod
g d
mu is finite. It implies once again that g
is in L 1 . g is in L 1 and 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; that
means 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, integral g is equal to
integral f whenever f and g are equal almost
everywhere. .
These are simple properties of integrable
functions that we have looked at. 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.
If f is in L1 , then we want to check that
alpha
is also in L1 and alpha f of d mu is equal
to alpha times integral of f into d mu.
.
To check that property, let us just observe
one thing. Just now we looked at this kind
of
analysis: if f belongs to L 1 of x, it is
same as if and only if mod f belongs to L
1 of x.
Why is that? Once again, let us do this because
this we are going to use again and again.
Saying that f belongs to L 1 this implies
integral of f plus d mu is finite and integral
of f
minus d mu is finite. What is mod f? mod f
is equal to f plus plus f minus. That implies
.ntegral f plus d mu plus integral f minus
d mu is finite and this is equal to integral
of
mod f d mu. So, f belonging to L1 implies
integral of mod f is finite. Let us look at
the
converse part – if mod f integral is finite.
.
This is given to us: 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. All are nonnegative measurable
functions; that implies integral of f plus
d mu is less than integral mod f d mu which
is finite and integral f minus d mu is less
than integral mod f d mu which is finite;
that implies that f belongs to L 1 . Saying
that
. a function is integrable is equivalent to
saying that mod f
which is a nonnegative measurable function
has got a finite integral. This property will
be used again and again. Let us see how this
property is used in our proposition.
.
a belongs to the real line and f is L1 . Look
at mod of a f. Mod of a f is less than or
equal
to mod a into mod f. All are nonnegative 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
implies that a f is an integrable function.
Now, not only is it integrable but the integral
of
a f d mu you can write as integral of a f
plus d mu minus integral of a f minus d mu.
Now, the possibility is either a is equal
to 0; in this case, a f will be 0 and everything
is 0;
so, no problem. 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.
This a comes out because of the property for
integral of nonnegative measurable functions;
so, this will be finite. In case a is less
than
0, this becomes a of minus negative part and
again that thing is okay and similarly for
a
minus. That proves the property that if f
is integrable and a is a real number, then
a f is
integrable and a comes out ..
We will continue looking at the properties
of integrable functions. In the next lecture,
we
will show that this integral is a linear operation
on the space of integrable functions and
various other properties of this space of
integrable functions and integral on it. We
will
continue this study in the next lecture. Thank
you.
