Welcome to lecture 40 on Measure and Integration.
In the past two lectures, we had been
looking at the various modes of convergence
for measurable functions.
We will continue the study in today’s lecture
also. We will recall some of the properties
of convergence in measure that we had proved
last time. Then we will go on to what is
called convergence in the pth mean.
.
So, let us just recall what we have been doing;
f and f ns are measurable functions on a
measure space; then, we said that f n converges
to f in measure, if for every epsilon
.bigger than 0, the measure of the set all
x, where f n x minus f x is bigger than or
equal to
epsilon - that measure of that set goes to
0 as n goes to infinity for every (( )). So,
in
some sense, the measure of the set where f
n is away from f by a distance epsilon, that
goes to 0 for every epsilon bigger than 0.
This was called convergence in measure and
we denoted this by saying that f n with an
arrow and the symbol m above f. So, f n converges
to f in measure. So, it is denoted by
this symbol.
.
In the previous lecture, we showed that the
convergence in measure neither implies nor
is
implied by convergence point almost everywhere.
That means if f n converges to f in
measure, then it need not imply that f n converges
pointwise or almost everywhere.
Conversely, if a sequence converges pointwise
or almost everywhere, that need not
imply that it converges in measure.
However, we proved that if the underlying
measure space is finite and f n converges
to f
almost everywhere, then it also converges
in measure. So, that means convergence in
measure is implied by convergence almost everywhere,
if the underlying measure space
is finite.
After that, we looked at what is called almost
uniform convergence for functions. So, we
said a sequence f n converges almost uniformly
to f on a set E, if we can find for every
.epsilon, a subset E epsilon such that the
measure of the set E intersection E epsilon
complement is finite and f n converges uniformly
to f on E. That means except for a
small set of small measure, f n converges
to f uniformly. So, this is called almost
uniform
convergence.
Note: this is different from saying that the
convergence is uniform almost everywhere.
Saying that f n converges to f uniformly almost
everywhere will mean that, except for a
null set, f n converges to f; almost uniform
convergence is saying that outside a set of
measure 0, for every epsilon, there is a set
E epsilon of such that outside that set the
convergence is uniform.
.
In general, we showed that convergence in
measure does not imply convergence almost
everywhere because if convergence almost everywhere
almost uniformly implies
convergence almost everywhere, we want to
conclude that convergence in measure does
not imply convergence almost uniform because
almost uniform convergence implies
almost everywhere convergence.
So, if convergence in measure implies almost
uniform convergence, then it will also
imply a convergence almost everywhere which
is not true, in general. So, convergence in
measure does not imply convergence almost
uniform.
.However, if f n convergence almost uniformly
to f, then it also converges in measure; so,
the converse is always true; namely, if f
n converges almost uniformly to f when f n
converges to f in measure so that the proof
of this is quite simple because f n converges
to f almost uniformly. So, that means for
every delta bigger than 0, we can select a
set E
delta such that outside the measure of the
set, outside E delta is small f n in convergence
uniformly to f on E delta.
So, that is by the property that f n converges
almost uniformly to f; so, for every delta,
we can find a subset. So, let us say we are
analyzing this on the set E. Then measure
of
the set outside E delta is small and f n converges
uniformly to E delta, but what does that
mean?
That means because it is converging uniformly
to f on E delta; that means for every n
integer N bigger than 0, f n must come close
to f for every x in E delta.
.
So, that means for every epsilon bigger than
0, there exists a set n, there exist a natural
number n say that the distance f n minus f
x is less than epsilon, for all n bigger than
N
and for every x belonging to E delta complement.
Then, that means if we look at the set up
points where this is bigger than equal to
epsilon, that will be contained in E delta
and the measure of E delta is less than epsilon.
So, that will prove that for every delta,
we have got a stage n bigger than N naught
such
.that x is f n x minus f x bigger than epsilon
is less than delta; that means f n converges
to
f in measure. So, we have shown that almost
uniform convergence implies convergence
in measure. so these are the basic properties
of relations between convergence in
measure with almost uniform convergence, convergence
pointwise, and so on
Now, we just now said that convergence in
measure need not imply convergence almost
everywhere, in general.
However, one can prove a partial result in
this direction namely, if f n converges to
f in
measure, then there is a subsequence which
converges almost everywhere. This result is
quite useful sometimes when you are analyzing
sequences of measurable functions
which converge in measure.
.
So, this theorem is called Riesz theorem.
So, let us prove Riesz theorem. So, he says,
let
f n be a sequence of measurable functions
converging in measure to a measurable
function f; then there exist a subsequence
f n k such that f n k converges to f everywhere,
to almost everywhere; actually we should be
saying to f n almost.
So, every sequence which converges in measure
is a subsequence which converges
almost everywhere.
.So, let us see a proof of this. So, to prove
this, we have what we are looking at we are
looking at how to construct a subsequence
f n k which converges almost everywhere to
f.
So, that means we have to find f n k such
that…
So, we reformulate the problem as follows:
Look at the set of points. So, f n k we want
with the property that wherever f n k x minus
f of x is bigger than 1 over k, the set of
these points for k equal to union of such
sets for k equal to j to infinity intersection
over j
equal to 1 into infinity - that must be equal
to 0. So, we want a sequence f n k such that
this measure of the set is equal to 0. Why
is that rough?
.
We are saying that f n k will converge to
f almost everywhere if the following: if we
look
at the set intersection j equal to 1 to infinity
union over k equal to j to infinity, the set
of
points x belonging to x such that mod of f
n k x minus f of x bigger than 1 over k -This
set has got measure 0. Why is that? Because
if you take an element x belonging to this,
what will that imply? That if and only if
x belongs to; this is belonging to intersection;
so, x belongs to union k over j to infinity,
x belonging to x such that f n k x minus f
of x
is bigger than or equal to 1 over k, for every
j - That must happen for every j, because
it
belongs to intersection. But what does that
mean?
That is if and only if this belongs to j - that
means for every j there exists k equal to
j, k
equal to j to infinity; so, that means for
every j, there exist k bigger than or equal
to j
.such that, saying that x belongs to this
union means x will belong to at least one
of them;
that means x belongs to for this set, for
some k bigger than or equal to j. so there
is a k
such that bigger than or equal to j such that
x belongs to this such that that means f n
k x
minus f of x is bigger than or equal to k.
For such an x, this must hold. It is bigger
than or equal to k; so, that means it does
not
converge. So, if we show that this set has
got measure 0 ., that
means if x does not belong, then it must converge.
So, what is the meaning of saying? If
you want to say that if x does not belong
to the set - that will mean what? It does
not
belong to the intersection; that means it
does not that means there exist so x belongs
to
this does not belong to intersection; that
means at least it does not belong to one of
them.
So, there exist some j naught such that x
does not belong to this.
If x does not belong to this - for j some
j naught, it does not belong to union. So,
that
means there exist A j naught such that from
j naught onwards, x does not belong to this
.; that means x cannot belong to any one of
them; that means
for every k bigger than or equal to j, there
is A j naught such that for every k bigger
than
or equal to j; bigger than or equal to does
not hold; equal to j; that means this must
happen for less than 1 over k.
So, saying x does not belong to this set will
mean there is a j naught such that every k
bigger than or equal to j naught; the distance
is less than k; that will imply that f n k
x
converges to f of x. So, what do we have?
What we are saying is essentially that this
is
the set of points . where f and k does not
converge to f n. You
want that set to have measure 0.
..
So, once again let me revise. Saying that
f n k does not converge to f x is same as
saying
x does not belong to this set. So, if x does
not belong, so for every x does not belong
to
intersection over j of some sets, that means
there is a set, there is A j naught such that
x
does not belong to j naught. So, it belongs
to this.
.
For some it should be union of k; so, that
is a union here. So, that means for every
k
bigger than j naught, this is less than k.
So, we have to only find a subsequence. So,
we
.have to complete the proof. We have to construct
a subsequence so that measure of this
set is equal to 0.
.
Let us write this set as A. So, A is the set
of points where intersection over j equal
to 1 to
infinity union over k equal to j to infinity,
where f n k x minus f x is bigger than or
equal
to 1 over k. Let us call the inner portion
which is a union set as A j. So, what we want
to
prove is that mu of intersection A j is equal
to mu of… So, we want to prove that mu of
A is 0, but this set A is contained in A j
because this intersection is smaller. So,
A is
subset of A j.
So, showing that mu of the set A is 0, A is
contained in A j, that means mu of A is less
than or equal to mu of A j. So, we will be
through if we can prove that mu of A js
converge to 0 as j goes to infinity. So, we
have to construct a subsequence f n k with
this
property that mu of A js of these sets must
go to 0 as j goes to infinity. So, this will
be
true.
..
We want mu subsequence that mu of A js go
to 0. That will be true if you could find,
say
for example, a sequence f n k such that the
measure of the set where f n k minus f x is
bigger than or equal to k, say it is less
than 1 over 2 to the power k plus 1.
Suppose we can choose our subsequence in that
way; then what will happen? Then, mu
of A j which is nothing but union of from
k equal to j to infinity of this set; that
is by
countable sub additive property mu of A j
will be less than or equal to summation k
equal to j to infinity mu of these sets and
each mu of these set is less than 1 over 2
to the
power k plus 1. So, that is summation j equal
to infinity. So, this is a geometric series
with common ratio less than half. So, that
will give you 1 over 2 to the power j.
So, we will get mu of A j less than 1 over
2 to the power j. So, as j goes to infinity,
we
will get mu of A j equal to 0 and hence we
will be through.
..
So, we have to only find a subsequence f n
k with this property and that is done by using
the fact that f n convergence to f in measure.
Because it converges in measure, by the
property of convergence in measure, let us
start with epsilon equal to 1. Then
convergence in measure says that mu of set
of those points where f n minus f x is bigger
than 1 will be less than half for every…
that goes to 0. So, that means after some
stage,
the difference of the measure. So, measure
of the set where f n minus f x is bigger than
or
equal to 1 will be small after some stage.
So, that stage we called as n 1.
So, using the fact the convergence in measure,
we can choose n 1 such that measure of
the set where f n 1 minus f x bigger than
1 is less than half.
Now, you proceed inductively. Supposing we
have selected n 1 less than n 2 up to n k
minus 1 have been selected with those required
property, then by fact that is sequence f
n
is converging in a measure for epsilon equal
to 1 over 2 to the power k plus 1.
..
So, let n k a stage bigger than n k minus
1 such that the measure of the set f n k minus
f x
bigger than 1 over k is less than 1 over 2
the power k plus 1. So, by induction, this
existence is instance of a sequence f n k
with this property is complete. Hence, we
will
have that this subsequence converges almost
everywhere to f. So, this is an important
result which is used sometimes to analyze
sequences which converge in measure. So, this
is called Riesz theorem.
So, we have looked at the various properties
of so till now we looked at various
properties of convergence: convergence almost
everywhere, convergence pointwise,
convergence almost uniform, convergence uniform
and convergence in measure, and
relations between such modes of convergence,
There is another mode of convergence which
arises when the functions f ns are in L p
spaces. Recall, we have defined L p spaces
the pth power integrable functions. So, one
would like to know that when sequences converge
in L p, does this convergence have
any relation with convergence in measure - pointwise
convergence, or convergence
almost uniformly? So, we will analyze these
things next.
..
Let us recall what is the meaning of saying
that have a sequence converges in the pth
mean or L p. So, saying that a sequence f
n, there p is bigger than 1 less than infinity
and
f belongs to L p.
So, let us take functions f ns and f in L
p. So, saying that f n converges to f in L
p or
sometimes one also writes this as f n converges
in the pth mean to f, if the pth norm of f
n minus f goes to 0, as n goes to infinity.
This called convergence in the pth mean.
.
.So we would like to find out the relations
between the pth means the and other modes
of
convergence
So the first property is that in general convergence
in the pth mean does not imply
uniform convergence it and it need not imply
almost uniform convergence or
convergence almost everywhere. So convergence
in pth mean need not imply any one of
uniform convergence, almost uniform convergence
or convergence almost everywhere
So, given example of measure space and a sequence
of functions in L p such that f n
converges to f in L p, but f n does not converge
uniformly or almost uniformly or almost
everywhere. So, we will give an example of
a sequence which converges in L p, but does
not converge almost everywhere. So, that example
itself will imply that pth mean cannot
imply uniform convergence because uniform
implies converges almost everywhere.
Similarly, same example will be sufficient
to say that pth mean does not imply almost
uniform
convergence
because
uniform
convergence
implies
almost
uniform
convergence.
So, we want to construct a sequence of functions
in L p such that the sequence converges
in L p, but does not converge almost everywhere.
So, for that, let us look at the measure
space - the interval 0 1, Lebesgue measurable
sets in 0 1, and lambda - the length
function.
let us write, if you recall - we had constructed
a sequence of measurable functions on this
measure space by looking at dividing the interval
into equal parts by using the binary
points, and that was an example which showed
that convergence in measure does not
imply convergence almost everywhere.
..
So, that is the sequence we are looking at
again. So, let us recall that sequence: f
n for
every n bigger than or equal to 1; let n be
represented as k plus 2 to the power m where
this k is between 1 and 2 to the power m.
So, we showed that, for every n can be written
uniquely in the form k plus 2 to the power
m where m is a positive integer and k is
between 1 and 2 to the power m.
So, whenever m n is represented this way,
we define f n to be the indicator function
of
the interval k by 2 to the power m and k plus
1 by 2 to the power m. So, look at the k th
interval of length 1 over 2 to the power m
and define f n to be the indicator function
of
this interval.
..
So, we showed earlier that this sequence of
measurable functions does not converge
almost everywhere and this converges in measure
to the function identically equal to 0.
So, this sequence of measurable functions
converges in measure to f identically 0, but
it
does not converge to f at any point in 0 1.
So, this does not converge almost everywhere
or at any point, actually.
Let us just prove that this sequence of measurable
functions is in L p because there is an
indicator function of an interval. So, the
function takes the value 1 on this interval
and
interval is finite. So, obviously, it implies
this is an L p integrable function. So, indicator
functions of a sub interval; so, f n belongs
to L p. So, we have got a sequence of
functions in L p.
Let us look at the what is the L p norm of
f n because it is the indicator function of
an
interval of a length 1 over 2 to the power
m. So, it is precisely equal to 1 over 2 to
the
power m; the function takes the value 1 on
the interval of length 1 over 2 to the power
m;
so, its L p norm is 1 over 2 to the power
raise to the power 1 over p.
So, this implies that the norm of f n minus
f which is identically 0 goes to 0 as n goes
to
infinity because as n goes to infinity, m
also goes to infinity.
..
So, this is the sequence of functions in L
p which converges to the function identically
0,
but the sequence does not converge almost
everywhere.
So, that proves our claim namely, convergence
in the pth norm does not imply
convergence uniform or convergence almost
uniform or convergence pointwise. So,
convergence in the pth mean does not imply
this; so, as we indicated, this cannot imply
uniform or almost uniform because both of
them imply convergence almost everywhere.
.
.Next, let us look at fact that convergence
in the pth mean always implies convergence
in
measure. So, let us take a sequence f n which
converges to f in L p and let epsilon greater
than 0 be arbitrary. So, what we have to show?
So, let us just look at what we have to
show - f n converges f in L p.
.
We have given that f n converges to f in pth
mean. To show - 
f n converges to f in
measure; so, what does it mean? That is for
every epsilon bigger than 0, we have to look
at the measure of the set x belonging to x
such that f n of x minus f of x bigger than
or
equal to epsilon, limit n going to infinity;
that must be equal to 0. So, this is what
we
have to show.
So, let us call E to be the set x belonging
to x where f n x minus f of x is bigger than
or
equal to epsilon. Now, we know that f n goes
to f in L p; that is equivalent to saying
that
f n minus f norm of that; here the norm goes
to 0. Let us take the power p also. what will
that mean is the following.
..
So, saying that in L p, let us look at the
norm f n minus f p b; this is equal to integral
mod
f n minus f to the power p d mu. So, this
goes to 0, but this is equal to integral over
E of
the same thing - f n minus f to the power
p d mu plus integral over E complement of
the
same thing; so, mod f n minus f to the power
p d mu.
Now, on the set E, f n is bigger than or equal
to epsilon; so, on the set E, this difference
is bigger than epsilon. So, it is bigger than
or equal to epsilon to the power p into mu
of
the set E plus integral over E complement,
but this is a non-negative function; so, even
if
I drop this term, it will still remain bigger
than or equal to mu of E. So, that implies
that
mu of E is less than or equal to mod mu of
E is norm of f n minus f to p divided by
epsilon point to power p.
So, the inequality that we get for this is
actually an important inequality in the theory
of
probability and this is called Chebyshev's
inequality; very simple inequality, but you
see,
it gives the powerful outcome. So, that is
following because f n converges to f n L p,
so
this goes to 0 .; that means that mu of E
equal to 0.
So, here f n is bigger than f by distance
epsilon mu of that it depends on epsilon;
so that
goes to 0 for every epsilon. So, that will
prove that, if f n converges to f in the pth
mean,
then it converges in the measure.
..
.
So, as we looked just now, the proof is simple
look at. The set where f n minus f x is
bigger than or equal to epsilon, the integral
mod of f n minus f n, norm of f n minus f
to
the power p can be separated into two parts:
One part where the function f n minus f is
bigger than epsilon is this and remaining
part will drop. So, inequality still stays;
so, mu
of E is less than or equal to this, which
goes to 0.
So, that proves that convergence in the pth
mean implies convergence in measure.
..
However, in general, the convergence almost
everywhere does not imply convergence in
the pth mean.
So, for that, let us look at a simple example.
Look at the Lebesgue measure space R
Lebesgue measurable sets lambda. So, that
is the Lebesgue measure, the Lebesgue
measure space and look at the function f n
which is defined as n raise to power minus
1
by p into the indicator function of the interval
0 to n.
So, first of all, let us observe that this
function belongs to L p because integral of
f n raise
to power p is just n to the power minus 1
to the power p. So, that is 1 over n and into
the
measure of the interval 0 and that is n. So,
each f n is a L p function; its norm is equal
to
L p norm is 1 and this convergence uniformly
to f. That is obvious because the values of
the function that is taken on a larger and
larger interval is 1 over n; so, becoming
smaller
and smaller. So, we can always find for every
x, we can find a stage after which the
distance will be less than 0. So, that means
f n converges. So, it is easy to check. Let
us
just verify that f n converges to f uniformly.
.
.However, we just now said that the L p norm
of each f E is 1; that does not converge to
f
in the pth norm. So, what we are saying is
convergence almost everywhere does not
imply convergence in the pth mean. This is
convergence almost everywhere.
Here is another observation that none of uniform
convergence or almost uniform
convergence or convergence in measure imply
convergence in the pth mean - that means
none of these; so, that means in general,
uniform convergence does not apply
convergence in the pth mean or convergence
almost uniform does not imply, and
similarly, convergence in measure need not
imply convergence in the pth mean.
Because obviously both uniform convergences
are almost uniform convergence, apply
convergence almost everywhere. So, if this
were true, then we will have a contradiction;
so, this is true.
..
Next, we want to say that convergence in the
measure need not imply convergence in the
pth mean. So, for that, let us look at the
example of the measure space 0 1, Lebesgue
measure space. Let us define g n of x equal
to n to the power 1 over p; that is something
similar to the previous one; instead of minus
1 over p, it is n to the power p and into
the
indicator function of 0 to 1 over n.
So, here, the interval where function is nonzero
is shrinking, but the value is increasing.
In the previous one, the value was decreasing
and the length was increasing. So, it is
other way round of this.
So, look at this function - g ns all of this
functions. g n are in L p because integral
mod g
n will be equal to the power p will be n into
to indicator function; so, integral is equal
to
1. So, each one of them has got an L p with
integral equal to 1. It is obvious that the
sequence g n converges in measure; it converges
in measure; that is obvious because the
set where it is going to be bigger than or
equal to epsilon is going to be shrinking.
It is 0
to 1 over n; so, that will go to 0. So, converges
in measure and identically equal to 0, and
its integrals L p norms are equal to 1.
..
Hence, g n does not converge to f identically
0 in the pth mean. That means, we as
produce a sequence of functions which is convergence
in measure, but does not imply
convergence in L p space.
.
However, if the underlying measure space is
finite, then the uniform convergence does
imply convergence in the pth mean. So, that
is quite obvious to verify.
.Let us just look at look at mu of a measure
space so that mu of x is finite and let f
n be a
sequence of functions in L p, converging uniformly
to a measurable function f. We want
to show that f n converges to f in L p also.
What is uniform convergence? uniform convergence
means for every epsilon bigger than
0, there is a stage after which… given every
epsilon bigger than 0, you can find a stage
n
naught such that the distance between f n
and f is less than epsilon, for every n bigger
than or equal to n 0 and for every x. So,
uniform means for every x, the same stage
satisfies the required property. So, for every
x, there is a single stage n naught such that
f
n x minus f of x is less than epsilon.
So, now, let us look at the absolute value
of the function mod f to the power p. See
we
have just given that f is converging uniformly
and each f n is in L p, but we do not know
whether f is in L p or not.
.
We have to first prove that this is in L p,
but look at the absolute value of f to the
power
p. So, by triangle inequality, I can write
it as mod of f n minus f plus. So, it should
be
always less than or equal to; by triangle
inequality add and subtract f n to the power
p.
Now, absolute value of a plus b is always
less than or equal to 2 times the maximum
of
the 2 values. So, this is less than or equal
to 2 times the maximum value f n minus f and
absolute value of f n; of course, everything
raise to power p. But that is same as less
than
.or equal to 2 to the power p and the maximum
of this to the power p. Now, 2 to the
power p and this maximum will always be less
than or equal to the sum. So, we can write
this is less than or equal to 2 to the power
p mod of f n minus f to the power p plus mod
of f n to the power p.
So, what does that imply? We can integrate
both sides with respect to mu. So, integral
will be less than or equal to 2 to the power
p into f n naught minus f to the power p d
mu
plus 2 to the power p of f n naught - we just
now showed this is less than epsilon. So,
this
will be less than 2 to the power p epsilon
to the power p mu of x plus 2 to the power
p
norm of f n to the power p; each of them being
finite says that the function f is in L p.
.
So, what we have shown is - if the underlying
space is having finite measure and f ns
belongs to f, f ns belongs to L p and converges
uniformly to f that the limit function is
also in L p; so that is what we have shown.
Now, look at the difference. So, the difference
of f n minus f to the pth norm is by
definition integral of mod f n minus f to
the power p raise to power 1 over p; but for
n
bigger than n naught, this difference is less
than epsilon to the epsilon to the power p
raise to the power 1 over p; so that is epsilon
into mu x raise to the power 1 over p. So,
as
epsilon goes to 0, this will go to 0. So,
that proves that f n converges to f in the
pth mean.
..
Let us also analyze what happens: The relations
between L p spaces and other modes of
convergence and the underlying spaces of finite
measure. So, the result says, even if the
underlying convergence space is finite, then
none of uniform convergence or
convergence almost everywhere need imply convergence
in the pth mean. So, this
condition is not good enough to ensure that
almost uniform convergence that will imply.
We have just looked at the function g n x
is equal to n to the power 1 over p. So, these
are all L p functions and their L p norm equal
to 1. So, they can all converge in L p, but
we know that this converges in measure and
hence almost uniform. Also, the
convergence in the pth mean need not imply
almost uniform convergence or
convergence almost everywhere. Other way round,
inequality - even when this is finite,
the convergence in pth mean need not imply
almost uniform or convergence.
..
So, that earlier example of the measure space
0 on 0 1, Lebesgue measurable sets, and f
n
to be the indicator function of the interval
I k m, where I k m is the interval of length
to
the power m - just now we consider this example.
So, these examples of functions they
converge to f uniformly. So, it converges
to f identically 0 in the pth mean, but we
know
that does not converge almost everywhere.
Hence, it can also converge almost uniformly
because almost uniform convergence implies
converges almost everywhere.
So, these are the various ways of looking
at modes of convergence and analyzing them.
Let me just look at all the implications put
together in the form of a diagram. So, here
is
the diagram; look at this.
..
So, here, we have got the vision of uniform
convergence uniform almost everywhere in
the pth mean here is pointwise, pointwise
almost everywhere, in measure and almost
uniform. As we all know that uniform convergence
is the strongest one. So, uniform we
have already seen earlier; the uniform implies
pointwise; of course, pointwise implies
pointwise almost everywhere, and this other
way round - implications need not hold;
simple examples.
Uniform will imply uniform almost everywhere;
actually uniform is uniform almost
everywhere with the set to be empty set. So,
uniform implies uniform almost
everywhere. Just now we have proved that mu
of x is finite; so, this green arrow
indicates that implication under the condition
that mu of the whole space is finite. So,
uniform convergence almost everywhere and
underlying space finite implies
convergence in the pth mean - that we saw
just now.
We have already seen uniform implies pointwise
apply pointwise almost everywhere.
and we also saw that pointwise converges pointwise
almost everywhere; in general
linearly implied convergence in measure, but
when an underlying measure space is
finite, pointwise will imply convergence in
measure. Of course, pointwise almost
everywhere need not imply almost uniform,
but we showed today that if the underlying
space is finite measure, then the pointwise
implies almost uniform. Obviously, almost
uniform we have shown implies pointwise almost
everywhere.
.Finally, we also showed today that almost
uniform implies in measure and when the
underlying space is finite, in measure implies
convergence in the pth mean; we use
Chebyshev’s inequality. Just now we said
that converges always in the pth mean always
implies convergence in measure. So, this is
the overall picture of various modes of
convergence.
With this, we come to the end of the basic
concepts of measure theory; whatever we have
not done is essentially looking at various
ways of necessary and sufficient conditions
under which a sequence f n converges to f
in L p - that is one thing we have not done,
but
the limited scope of lectures.
We wanted to cover in almost 40 lectures the
basic concepts of measure theory and that
we have covered. Another thing that we have
not proved is looking at the change of
variables formula for R n; that again requires
a bit of work and technical things. So, we
have not covered that in our discussions.
So, with that, we come to an end of this course
video lecture on Measure Theory. In the
next lecture, I will just try to give an overall
view of the things that we have done in our
course.
Thank you.
.
