Welcome to lecture 21; on Measure and Integration.
In the previous lecture, we had started looking
at the properties of sequences of integrable
functions and we started proving an important
theorem called Lebesgue dominated convergence
theorem. Let us continue looking at that and
after that we will start looking at the special
case of integration on the real line and that
will give us the notion of Legbsgue integral.
So, let us recall what we had started proving
namely dominated converges theorem which says
that if f n is a sequence of measurable functions;
such that there exists a function g belonging
to l 1; such that all the f n s are dominated
by this integrable function g; almost everywhere
x for all n. And if f n converges to f; then
the limit function is integrable and integral
of f is nothing, but the limit of integrals
of f n’s.
So, the theorem basically says that if f n
is a sequence of measurable functions; all
of them dominated by a single integrable function
g, then all the f n’s become integrable
of course, and if f n’s converges to f;
then f is integrable and integral of f is
nothing, but the limit of integrals of f n’s.
Now, we had proved this theorem in this particular
case, when instead of this almost everywhere
that mod f n’s are dominated by g everywhere
and f n’s converge to f everywhere.
So, to extend this case to almost everywhere;
we have to do on minor modifications.
So, let us look at. let us define the set
N to be the set of all x belonging to X; where
mod f n x is not dominated by g. So, g of
x or union the set of all those points x belonging
to X; such that f n x, does not converge to
the function f of x; so, on N complement;
we have f n x is less than g x and f n x converges
to f of x. So, for every x belonging to N
complement and mu of the set n is equal to
0. Because we are saying that this mod f n
x is less than g x almost everywhere. And
f n x converges to f of x almost everywhere;
so, the set where it does not hold; there
is a set n and that set has n has got measure
0.
So, now let us consider the sequence; indicator
function of N complement times; f n. So, this
is a sequence of functions which are dominated
by; so, this is a sequence of functions for
n bigger than or equal to 1 and they satisfy
the property. So, namely this the indicator
function of N compliment f n mod of that is
less than g; for all x everywhere. And the
functions converged to the indicator function
of N compliment f.
So, by our earlier case; what we get is the
following namely that indicator function of
N complement times f is L 1 is integrable
and integral of f n limit n going to infinity
indicator function of N compliment times f
n; the integral of that converges to the integral
of indicator function of N compliment times
f. So, that is by the earlier case when everything
is true for all points. So that means, so
this is same; but note that mu of N is equal
to 0.
So, that implies that is the integral of f
over N d mu is equal to 0 and we already know
that on N complement; f is integrable. So,
that together with this fact implies integral
of mod integral of mod f d mu is finite; implying
that f belongs to l 1. And this equation which
said that integral of a f n over N complement
converges to integral of f over N complement
and mu of N being 0; together gives us the
condition that integral of f n d mu converges
to integral f d mu.
Because integral of f n d mu is same as integral
of f n over n plus integral over N complement
and integral over N compliment converges to
integral over N complement of f and on n both
are 0, so this gives us the required result.
So, that is how from almost everywhere conditions
are deduced from the fact that something holds
everywhere.
So, this dominated convergence theorem holds
for whenever the sequence f n is dominated
by g and f n converges to f almost everywhere.
Then f limit function is integrable and integral
f converges to integral of f n. So, as I said
this is one of the important theorems which
helps us to interchange; the limit and the
integral sign.
Let us look at some more minor modifications
of this theorem; one thing more we can even
reduce that integral of mod f n minus f d
mu also converges to 0.
So, to deduce that part; we just have to observe
that mod f n minus f is less than or equal
to twice of g and mod f n minus f goes to
0. So, again an application of dominated convergence
theorem; dominated convergence theorem which
we put just now, implies that integral of
mod f n minus f d mu goes to 0. So, that is
a another modification, another consequence
of the dominated convergence theorem.
Let us prove what I call as the series version
of this theorem; namely that if f n is a sequence
of functions which are integrable and integrals
of f n summation 1 to infinity. So, sum of
all the integrals f n mod f n’s are finite;
then the conclusion is that the series f n
x converges almost everywhere. And if you
denote the limit; the sum being f of x, then
that function is integrable and integral f
is equal to summation of integral f n’s.
So, essentially this theorem says that if
the summations of mod f n’s are finite;
then this the series f n x is; I guess is
convergent almost everywhere and integral
of f is equal to integral of summation of
integral f n’s. So, that is again interchange
limit essentially. So, let us see how from
dominated convergence theorem; we can get
this.
So, let us define; to show that this series
is convergent almost everywhere; we will actually
show that, so we show is absolutely convergent.
So, for that let us define say g n of x to
be equal to summation mod f k of x; k going
from 1 to n, the partial sums of the absolute
values 1 to n.
So, let us observe that this sequence g n.
So, note g n is a sequence of nonnegative
measurable functions and g n’s are increasing
to some function so, that is there going to
increase to k equal to 1 to infinity; mod
of f k of x; let us call that as g x; they
are increasing to the function g of x. So,
that implies by monotone convergence theorem;
we have integral of g n; d mu must converge
to integral of g d mu.
But integral of g n d mu; so, that is same
as same integral g d mu is equal to limit
n going to infinity of integral g and d mu,
but what is integral of g n; g n is sum of
absolute values of f k 1 to n. So, by linearity
property this is nothing, but limit of n going
to infinity of summation 1 to n; k equal to
1 to n of integral mod f k; d mu and this
limit is nothing, but 1 to infinity and that
is given to be finite. So, let us write that.
So, this is equal to summation k equal to
1 to infinity of integral mod f k d mu; which
is given to be finite. So, hence what do we
get is thus g is integrable; g belongs to
g is a integrable function. So, saying that
g is integrable implies, we call that if a
function is integrable and g is a nonnegative
function. So, g of x is finite almost everywhere;
that is nonnegative function which is a integrable
function must be finite almost everywhere.
So, we get that g is finite almost everywhere
and what is the function g; g is nothing but
the limit of the absolute values of f k x.
So; that means, that prove that the series;
so, hence sigma k equal to 1 to infinity mod
f k of x is finite almost everywhere x. And
once a series is absolutely convergent; it
is also convergent so, that implies that sigma
k equal to 1 to infinity f k x is finite for
almost everywhere x.
So, let us denote this limit by f of x; so,
this is f of x and so, as observed earlier.
So, note; so f of x, we can write also f of
x as the limit n going to infinity of summation
k equal to 1 to n; f k x. And if this functions
are called as something say phi n; then note
that mod phi n, if this is called as phi n;
then what is mod phi n? mod phi n is absolute
value of 1 to n f k x; absolute value of that
and that is less than or equal to summation
1 to n mode of f k 1 to n; and that is nothing,
but our g n which is less than or equal to
g.
So, these partial sums which we called as
phi n’s are all dominated by g and phi n’s
converge to f; so by dominated convergence
theorem. So, what we have got is; so, we have
got.
So, all the phi n’s are less than or equal
to g and phi n’s converge to f almost everywhere.
So, that implies by dominated convergence
theorem; that integral of phi n’s d mu must
converge to integral of f d mu. But this is
nothing, but this phi n’ this is what we
called as phi n that is summation of 1 to
n. So, this is nothing, but summation of 1
to n of integral k equal to 1 to n of integral
f k; d mu must converge to integral f d mu.
And that is same as saying that integral f
d mu is equal to summation k equal to 1 to
infinity; integral f k d mu. So, that proves
the theorem namely if f k is a sequence of
functions, which are integrable and the sum
of the integrals is finite; then the series
f n x; n 1 to infinity itself is convergent
almost everywhere and the limit function is
integrable and integral of the limit function
is equal to summation of integrals of f n’s.
So, this will referred to as the series version
of dominated convergence theorem.
There is another interpretation of the dominated
convergence theorem; when the underlying measure
space is a finite measure space; then one
has that if X S mu is a finite measure space
and f n is a sequence of measurable functions;
such that all of them are dominated by a single
constant M; almost everywhere and f n x converges
to f of x; then integral f n s converges to
integral f.
So, this is a particular case of dominated
convergence theorem when the underlying measure
space is the finite measure space. And the
only thing to observe here is that because
let us see how does this follow from; is a
dominated convergence theorem.
So, we are given that mod f n x is less than
or equal to M; for almost everywhere x. So,
now, look at this; so, look at the constant
function M. So, look at the function g of
X which is equal to M for every x belonging
to X. So, the constant function is measurable;
so, note that g is a nonnegative measurable
function because it is constant function note
that. So, its integral g d mu is equal to
is equal to integral the constant function
M d mu and that is equal to m times the measure
of the whole space x; which is finite.
So, what we are saying is on finite measure
spaces a constant function is always integrable.
So, implies g is L 1; so, f n x bounded by
M. So, that is constant function and that
is a integrable function; once we have that
and f n x converges to f of x almost everywhere.
So, now, by dominated convergence theorem
is applicable and that implies integral fd
mu is equal to integral f n d mu limit n going
to infinity.
So, the main thing is on finite measure spaces
a constant function becomes integrable because
of this reason. So, this is what is called
bounded convergence theorem and it is quite
useful; when underlying measure space is a
finite measure space. So, let us look at what
we have proved till now; we have looked at
the space of integrable functions and proved
linearity property and an important theorem
called dominated convergence theorem.
So, if you recall for nonnegative measurable
functions; we had two theorems one was monotone
convergence theorem namely that was theorem;
when f n is a sequence of nonnegative measurable
functions increasing to a function f, then
integral of f is equal to limit of integrals.
So; that means, interchange of limit and integration
is possible by monotone convergence theorem
whenever the sequence f n is monotonically
increasing and sequence of nonnegative measurable
functions.
The second theorem which involved sequences
of measurable functions was again for nonnegative
measurable functions and that was called fatous
lemma. So, there we do not emphasize; we do
not require the sequence f n be nonnegative
and measurable; we only want the sequence
f n to be a sequence of nonnegative measurable
functions; they need not be increasing. So,
for such a sequence we had that integral of
the limit inferior of the sequence f n is
less than or equal to limit inferior of the
integrals of f n; so, that was Fatou’s lemma.
And now we have the third theorem; dominated
convergence theorem which again helps you
to interchange the notion of integral and
the limiting operation under the conditions
that all the f n’s dominated by a single
integrable function.
So, these are the three important theorem
which help us to interchange limit and the
integral signs. Let us at this stage emphasize
one more point about this technique of integration
and so, basically for integral; we started
with simple functions and then we go to a
nonnegative functions and then we defined
it for integrable functions.
So, this process of step by step defining
the integral can be is useful in proving many
results and I call it as the simple function
technique. So, this is a technique which is
used very often to prove some results about
integrable functions and non negative measurable
functions. So, what is the technique; let
me outline that and then I will give a illustration
of this.
Suppose, you want to show that a certain property,
say let us all that property as star holds
for all integrable functions. So, to prove
that the property holds for all integrable
function; the technique is as follows basically
that show that this property star holds for
all non negative simple measurable functions.
So, if you want to show a property holds for
all integrable functions; first show that
it holds for the class of nonnegative simple
measurable functions. And next show that star
holds for non negative measurable; integrable
function by using the fact that non negative
measurable functions are limits of increasing
the limit of simple measurable functions.
So, and use their one use is normally the
monotone convergence theorem. So, using monotone
convergence theorem; one extends the property
star from simple measurable functions to nonnegative
measurable functions or nonnegative integrable
functions.
And then keeping in mind that for a function
f it can be split into positive part and negative
part. So, f can be written as f plus minus
f minus and if a property holds for nonnegative
functions. So, about integral; so, f for plus
that will hold for f minus that will hold
and then conclude from there that it holds
for f also. So, this is what I call as the
simple function technique to prove results
about integrable functions.
To give a illustration of this; let us look
at the following result. Let us take a measurable
space X S mu; which is sigma finite measure
space and let us look at a function f which
is integrable on this measure space and is
nonnegative. So, we have got a sigma finite
measure space and f is a nonnegative integrable
function on this measure space. Let us define
nu of E, for every set in the sigma algebra;
let us define nu of e to be integral of fd
mu over the set E integral of f over the set
E is denoted by nu of E; for every set e in
the sigma algebra S. Then we had already shown
that this nu, the set function nu is in fact,
a finite measure on S. So, this we have already
proved.
But what we want to prove now is that further
if g is any integrable function within on
the measure space X S nu; this nu is the new
measure. So, if g is integrable on X with
respect to nu, then the product function f
into g is integrable with respect to mu and
this relation holds integral fd mu. So, integral
of f with respect to nu is equal to integral
of f into g with respect to mu.
So, what we have done is by fixing a function
f; which is nonnegative, we have defined a
new measure on the measurable space by nu
of e to be equal to integral of efd mu. And
we are saying; if you want to integrate a
function with respect to a function g, with
respect to this new measure; then it is same
as integrating the function; the product function
f g with respect to the old measure mu.
Thanks
