Now, let us look at the additive property
of the integral over the sets. So, this is
just now we have proved that f. So, another
property that if f is integrable, then f must
be a finite number for almost all X; so, a
similar argument as before. So, let us prove
that property also; it says if f is a integrable.
Then this implies that mod f of x is finite
almost everywhere x. So, let us write once
again the idea is; let us write the set x
n to be the set where mod f of x is equal
to plus infinity either. So, f of x is equal
to infinity or equal to minus infinity. So,
together we have put them together in the
set n. So, to show that the set mu of N is
equal to 0; so, if not so; once again, if
not let us write N as x belonging to x such
that such that mod f of x is bigger than say
some quantity is not equal to 0. So, if this
is not equal to 0 and f of x n is the set
where f of x is equal to plus infinity. So,
let us write f of x bigger than N; oh, let
us write, sorry, let us write A n to be the
set where f of x is bigger than n, then each
set a n is in the sigma algebra, right. So,
we want to; sorry; this is not required. So,
let us; this is not required because we want
to just want to show that f is finite.
So, now observe that integral of mod f d mu,
I can write it as integral over N mod f d
mu plus integral over N complement mod f d
mu, right because N and N compliment together
make up the whole space and just now, we observed
that integral of mod f over a set E is a measure.
So, integral of mod f over the whole space
can be written as integral over the set of
mod f over N plus integral of mod f over N
complement and on N, the set is on N mu of
N is equal to 0. So, the first integral is
0. So, it is equal to 0 plus and sorry yes
no let us observe this is.
So, if mu of N is bigger than 0, then what
will happen. So, let us assume mu of N is
bigger than 0, then this integral is equal
to this integral plus integral over N plus
integral over N compliment so; that means,
integral of mod f d mu is always bigger than
integral over integral over N mod f d mu right
because I am just; so, what we are doing.
So, integral of mod f is integral over N plus
integral over N complement, let us just drop
the second term. So, integral of mod f over
the whole space is going to be bigger than
integral over N f d mu.
So, once that is true and on N; the function
takes the value plus infinity. So, this is
going to be equal to plus infinity multiplied
with mu of N. So, if mu of N is bigger than
0, then this will be equal to plus infinity
if mu of N is bigger than 0 and that is a
contradiction that is not possible not true
as f belongs to L 1. So, this integral must
be a finite quantity and here we are saying
in that case it will be equal to infinity.
So, that proves that if a function is integrable,
then it must be finite almost everywhere.
So, now, let us come back to the question
if f is integrable and E belongs to the set
S, then the indicator function of E times
f is integrable that we have just now observed.
So, we let us write nu tilde of E as before
the integral of f over E and we observed that
this number may not be a nonnegative number;
however, is still has the property something
similar to that of countable additive property
for measures namely.
So, let us state that property that if you
take sets E and S in the sigma algebra S which
are pair wise disjoint. So, and E is the union
of this sets, then the claim is that the series
which is integral of E i S f d mu summation
1 to infinity, this series is absolutely convergent
and if we write E as the union, then integral
of f over E is equal to summation of integral
of f over E i S. So, essentially we want to
say that the integral of f of a integrable
function over a set W can be written as in
summation integral of E i S where E i S are
pair wise disjoint and this is; we are saying
its always possible for f to be integrable
function and because why we are saying absolutely
convergent.
So, here is one should note that when E is
equal to union E i; it does not matter whether
you write as E 1 union E 2 union E 3 and so
on or any other order say E 2 union E 1. So,
it does not the union does not depend on the
order in which you write the sequence E i
so; that means, in this series the summation
should not depend upon the order of the terms
and all are nonnegative; that means, we should
prove that these are absolutely convergent
and that is what we want to prove that if
E i S are pair wise disjoint, then the series
integral over E i of f d mu summation 1 to
infinity is a absolutely convergent series
and this integral of f over E is summation
of integral over E i S.
So, let us prove this property. So, let us
we have got E i is a sequence of sets in the
sigma algebra they are pair wise disjoint
is equal to empty set for i not equal to j
and E is equal to union of E i S 1 to infinity.
So, first we want to show. So, first claim
is that the series summation i equal to 1
to infinity of integral over E i of f d mu.
This is absolutely convergent absolutely convergent.
Let us observe; what is absolute convergent
means; that means, absolute values of this
terms that this is a series of nonnegative
terms that must converge, but. So, for that
let us note that absolute value of integral
E over E i of f d mu is less than or equal
to integral of mod f over E i d mu, right.
So, that is the property of integrable functions
that absolute value of the integral is less
than or equal to integral of the absolute
value and for nonnegative function. So, mod
f is a nonnegative function and E is a disjoint
union of sets. So, that implies that integral
over E i of mod f d mu if i sum it up i equal
to 1 to infinity that is same as integral
over E of mod f d mu and f being integrable
that is a finite number. So, here we are used
2 things; one for non negative measurable
functions integral over a set is a measure.
So, integral of mod f d mu over E i summation
one to infinity is equal to integral of absolute
value of f over E and f being integrable this
is finite. So, that proves that the series
integral f d mu over E i is absolutely convergent,
right because this sum is less than or equal
to sum. So, from these 2 it implies that this
series is absolutely convergent.
So, once the series is absolutely convergent
its sum is equal to sum of the partial sums.
So, now, we can easily write. So, this implies
that the claim holds.
So, the series is absolutely convergent hence
integral over E f d mu is equal to limit n
going to infinity of the partial sums. So,
i equal to 1 to n integral of f over E i d
mu and that is nothing, but partial sums and
that is same as right saying that sigma i
equal to 1 to infinity integral over E i of
f d mu. So, that proves that the integral
of a integrable function over a set need not
be a measure.
But it has we can say this is a countably
additivity property of this integral that
integral over E is equal to summation of integrals
over E i S whenever E is a union of pair wise
disjoint sets E i. So, this is the property
that we are just now proved. So, these were
some of the properties that we have proved
about the integral of integrable functions
and now we want to prove an important property,
we want to analyze this sequences of integrable
functions to we want to analyze if the property
that if f n is a sequence of integrable functions
and it converges to a function f can we say
that f is integrable and can we say integral
of f n s will converge to integral of f.
We have seen that this need not be true even
for non negative functions, but they under
some suitable condition we can say that integral
of f n s will converge to integral of f and
that is an important theorem called Lebesgue
dominated convergence theorem. So, let us
prove interchange of integral with the limits
and look at the theorem called Lebesgue dominated
convergence theorem.
.
So, it says let f n be a sequence of measurable
functions such that there exists a function
g which is integrable with the property that
integral of mod f n s are less than or equal
to g for almost all x nu for all n, then the
claim is if f n s converges to f almost everywhere
then the limit function is integrable one
and. Secondly, integral of the limit function
is equal to limit of the integrable functions
So, let us observe once again what is given
and what is true we are saying f n is a sequence
of measurable functions and all the f n s
are dominated by a single function g which
is integrable and this dominance could be
almost everywhere. So, all the f n s are dominated
by a single function g and. So, the conclusion
is if f n s converges to f then f is integrable
and integral of f is equal to limit of integral
of f n s. So, this is what is called dominated
convergence theorem and it is an important
theorem. So, let us prove this theorem.
So, we are given that all the f n s are measurable
for n bigger than or equal to one mod f n
x is less than or equal to g of x for almost
all x and for every n and we are given that
f n x converges to f of x almost everywhere.
So, to prove the required claim for the time
being, let us assume that this almost everywhere
is everywhere that is the proof is not going
to change much. So, we will see that improver.
So, let us assume for the time being f n x
is less than mod f n x is less than or equal
to g of x where g is L one is a integrable
function. So, that implies that integral of
mod f n x d mu x is less than or equal to
g x d mu x and g is integrable. So, that is
finite. So, that implies that each f n is
a integrable function also.
Mod f n converges to mod f right because f
n converges to f and each mod f n is less
than or equal to g. So, that implies. So,
this implies this implies that mod f n x is
mod f n x is less than or equal to g and that
converges. So, that implies that mod f x is
also less than or equal to g of x for every
x. So, that once again implies integral mod
f d mu is less that integral g d mu which
is finite. So, this again implies that f is
in L 1 of mu.
So, under the given conditions we are shown
if f n s are dominated by a integrable function
and f n s converge to f, then f is a integrable
function. So, to look at the limits let us
observe. So, note f n converges to f right
and mod f n is less than equal to mod g. So,
this implies that look at the sequence f n.
So, look at the sequence f n minus g. So,
look at the sequence f n minus g. So, look
at the sequence this is a sequence of measurable
functions of course, they are mod a f n is
less than or equal to g. So, this will be
negative.
So, let us look at we want nonnegative. So,
let us look at g minus f n look at this sequence
instant. So, this is a sequence of nonnegative
measurable functions because mod g is bigger
than mod f n is bigger than or equal to g
that is given. So, g minus f n; so, is a sequence
of measurable functions and let us observe
g minus f n is bigger than or equal to 0 because
g is bigger than or equal to f n. So, this
is a sequence of nonnegative measurable functions;
and so, let us write that g minus f n is a
sequence of nonnegative measurable functions
and it converges to g minus f because f n
converges to f.
So, now we can apply Fatou’s lemma. So,
implies by Fatou’s lemma that integral limit
inferior of g minus f n d mu will be less
than or equal to limit inferior of integral
of g minus f n d mu. So, that is application
of Fatou’s lemma. So, recall we had Fatou’s
lemma which was applicable for functions which
are not necessarily sequence of functions
is not necessarily increasing. So, look at
this now let us compute both sides. So, what
is the left hand side? So, this is equal to
integral limit infimum of g minus f n is g
minus limit inferior plus limit inferior of
minus f n right. So, that is the left hand
side d mu and that is equal to integral of
g.
So, look at that is equal to integral of g
d mu and what can you say about this f n is
a all are integrable functions. So, everything
is finite. So, this limit inferior of minus
f n is equal to minus limit superior of f
n s d mu. So, this is a property of limit
superior and limit inferior that limit inferior
of minus f n is equal to minus of limit superior.
So, this is equal to. So, this is equal to
integral of g d mu minus integral limit superior
of f n s. So, limit f n is convergent. So,
limit superior is same as f of x f of x d
mu d mu x. So, that is a left hand side and
let us see; what is the right hand side once
again. So, limit inferior of integral. So,
this is equal to limit inferior of integral
integral of g minus f n is integral g and
that does not depend upon limit. So, is integral
g d mu and then limit inferior of minus. So,
that will be minus limit superior of integral
f n d mu.
So, from these 2; so, this is less than or
equal to this. So, what does that imply? So,
that implies that integral g d mu Minus integral
f d mu, right.
So, I am just writing integral; this is less
than or equal to integral g d mu minus limit
superior of integral f n d mu, right. So,
everything is finite. So, I can cancel out
this and negative sign gives you other way
inequality. So, it implies integral f d mu
is bigger than or equal to limit superior
of integral f n d mu. So, looking at the sequence
g minus f n we got this is that g minus f
n is nonnegative converges to g minus f gives
us this.
Similarly, if I look at the sequence g plus
f n that is again a sequence of nonnegative
measurable functions and application of Fatou’s
lemma. So, Fatou’s lemma will give me that
integral; integral of f d mu is less than
or equal to limit inferior of integral f n
d mu. So, a similar application of Fatou’s
lemma to this sequence will give me this.
So, 1 and 2; so, 1 plus 2 together imply that
integral f d mu is bigger than limit superior
that is always bigger than limit inferior
and that is bigger than integral f d mu implies
that the sequence. So, limit integral f n
d mu exists this limit exist and is equal
to integral f d mu. So, that proves dominated
convergence theorem.
So, the proof of dominated convergence theorem
is essentially very simple it is just a straight
forward application of Fatou’s lemma because
mod f n is less than or equal to g implies
g minus f n and g plus f n both are sequences
of nonnegative measurable functions. So, apply
Fatou’s lemma and we have the conclusion
that integral of f is equal to limit of integrals
of f n s. So, we have proved this under the
conditions that f n converges to f n x converges
to f of x and f n x is dominated by g of x
for every x the modification for this for
almost everywhere things is simple and we
will do it next time.
Thank you very much.
