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Welcome to lecture 39 on Measure and Integration. Today, we will be looking at a special topic
which is called various modes of convergence for measurable functions. So, the topic for
today's discussion is modes of convergence.
Let us recall some of the ways of convergence that you might have already looked into earlier
courses. So, let us take a sequence of functions f n, it is a sequence of functions on measurable
space X S. Now, saying that f n converges to f point wise, it means that f n x is same
as saying that f n x converges to f of x; these are numbers for every x belonging to
X, this is converging point wise. You might have already come across something called
f n converges to f uniformly, so what is the difference between this point wise convergence
and uniform convergence? Let us just write down in terms of our definition of epsilon
delta. Point wise means for every x, for every epsilon
bigger than 0 there is n naught which will depend upon the point x and epsilon such that
f n x minus f of x is less than epsilon for every n bigger than n naught. So that essentially
means that the numbers f n x converges to the number f of x as n goes to infinity, so
the given epsilon is bigger than 0. There is a stage after which f n x is close to f
of x of course, this stage may depend upon the point x and the number epsilon.
Let us look at what is point wise convergence, by saying that f n converges to f uniformly;
that means for every x, for every epsilon is bigger than 0 there is the stage n naught
which does not depend on x which depends only on epsilon such that f n x converges to f
of x; that is, mod f n x minus f of x is less than epsilon for every n bigger than n naught
and the same stage works for every x. You must have already seen that the uniform convergence
implies point wise convergence and the converse need not be true. So the uniform convergence implies point wise
convergence; the converse need not be true that you must have seen in your earlier courses
in analysis, but we are going to look at today is the functions which are defined not only
on measure spaces; that is, actually they are defined on measure spaces not only on
measurable spaces. So, we are going to look at a sequence f n of measurable functions
on a measure space X S mu. We had already looked at some concept called convergence
almost everywhere.
Let us define formally again that f n converges to f almost everywhere if everything is defined
on a measure space X S mu; these are functions defined on the measure space X S mu. We say
f n converges to f almost everywhere, if the set N all x belonging to X such that f n x
does not converge to f of x. If that is a set N then, we want to N should belong to
sigma algebra S and mu of N should be equal to 0, so that is convergence almost everywhere.
Obviously, point wise convergence implies convergence almost everywhere. Obvious examples one can construct to show that this
other way implication need not hold; convergence almost everywhere need not imply point wise
convergence.
Now, let us try to look at a relation between convergences almost everywhere. Just recall
once again what we are saying? We are saying f n convergence to f point wise, if the sequence
of numbers f n x converges to f of x. That is same as saying that for every epsilon bigger
than 0, there exists n naught such that it may depend upon the point x. The number epsilon
is such that absolute value of f n x minus f of x is less than epsilon for every n bigger
than n naught.
So, the numbers f n x converging - the sequence f n x is converging to f of x. By saying f
n converges almost everywhere to f means that the set of points where f n x does not converge
to f of x that set is a set of measure 0, so that is convergence almost everywhere.
Saying that the f n converges uniformly to f that means, for every epsilon there is a
stage n naught which depends only on epsilon n naught on the point x such that for every
x the distance between f n x and f of x is less than epsilon for every n bigger than
or equal to n naught. That stages works for every x, so that is important thing for uniform
convergence.
Instead of writing in English all the time whenever f n converges to a point wise, we
will write f n arrow with the p above indicating it is a point wise converges, f n converges
to f point wise. Convergence almost everywhere will be indicated by f n an arrow pointing
right side to f with a symbol a dot e dot saying almost everywhere to f, but this is
also written as f n converges to f, f n right arrow f, a e indicating and this converges
is almost everywhere. Similarly for uniform convergence, we will
denote it by symbols f n arrow by pointing towards the right and u above the arrow indicating
that it is uniform convergence.
So as I pointed out earlier that f n converges to f, uniformly implies that f n converges
to f point wise and that implies f n converges to f almost everywhere. None of the backward
implications is true, so one can easily construct counter examples. We show that convergence
almost everywhere need not imply convergence point wise and point wise convergence need
not imply convergence which is uniform.
However, we would like to still analyze whether some conclusions can be drawn from the point
wise convergence or almost everywhere convergence in terms of uniform convergence. To analyze,
let us assume that f n converges to f almost everywhere. Let us take a set E which is in
the sigma algebra S such that the measure of this sigma set E is finite and let us be
given with two numbers epsilon and delta arbitrary. Then, we would like to show that there exist
a set E lower epsilon - a measurable set E lower epsilon of S and a positive integer
n naught which will depend upon epsilon and delta such that with the following properties
namely, that E epsilon is a subset of E and the difference E minus E epsilon is small.
For every x belonging to E epsilon, on E epsilon f n x minus f of x is less than delta for
every n bigger than or equal to n naught. That means, for every x the same stage n naught
works, so essentially saying that the difference between f n and f stays less than delta for
all n bigger than n naught and the stage will not depend on epsilon.
So let us prove this result. To prove this, we are given that f n converges to f almost
everywhere. That means, if we define the set N to be the set of all x belonging to X such
that f n x does not converge to f of x then, this set the mu of the set N is equal to 0,
so that is what is given to us. Now, let us look at the set that means, on N complement
on the complement of this set f n x converges to f of x for every x belonging to N complement.
Let us write the set E say m, we are also given delta, so let us write E m delta to
be the set of all points x belonging to X such that f n x such that the difference f
n x minus f of x f n x minus the difference f n of x is less than delta for every n bigger
than or equal to m. Let us look at this set E m delta, this set depends on m and on delta,
so the difference between f n and f is less than delta for every n bigger than or equal
to m. Let us note that these sets E m delta they
belong to the sigma algebra. If the difference between f n and f m f n and f x is less than
delta for all n bigger than m then, that is also going to be true for all n bigger than
n plus one.
That means, the E m delta is an increasing sequence. So, E m delta is a subset of E m
plus one delta further, not only it is increasing because we are given that f n x converges
to f of x for every x, So that means the sequence is increasing and on n complement was that
know the difference is so the sequence is increasing sequence and we can say that n
complement is contained in union of E m delta m equal to 1 to infinity, because on n complement
it converges. So for every x, we can find some stage after which it will be less than
for a given delta.
So, the fact that on n complement f n x converges to f of x that implies that the set n complement
is a subset of the union of E m delta.
Thus, if you look at the sequence E intersection E m complement delta then, this is a decreasing
sequence; it is a decreasing sequence of sets and mu of E being finite implies that limit
n going to infinity mu of E intersection E m complement delta. It must be equal to this,
because E m was increasing sequence and the complements will be decreasing sequence, so
it must converge to intersection of all of them and that is contained in the complement
the e intersection and complement that is equal to 0, because this sequence of sets
decreases to E intersection n complement, so that is equal to 0.
So what does that imply? Saying that this converges to 0 that implies that given there
is the stage. Let us find some E m m naught such that mu of E intersection E complement
m is m naught is less than epsilon. On E intersection E m naught complement, we know that f n x
minus f of x is less than delta for every n greater than or equal to m naught.
So with the given epsilon delta and mu of E finite, we have found a stage m naught and
a set E m naught such that E intersection is nothing but E minus the set that is less
than epsilon and on the set E intersection the difference between f n and f is less than delta .
So that proves the required claim namely, if f n converges to f almost everywhere and
we are given a set E of positive finite measure - mu of E is finite then, for every epsilon
and delta we can find a subset E epsilon inside E such that the measure of on E epsilon f
n x minus f x is less than delta for every n greater than the stage n naught and the
measure of the difference E minus E epsilon is small.
So, this is consequence of the f n converges to f almost everywhere and mu of E has got
finite measure. As a consequence of this, we prove theorem called Egoroff's theorem;
it says that if f n converges to f almost everywhere and E is a set of finite measure
then for given epsilon, there is a set E epsilon such that measure of the set E minus E epsilon
is less than epsilon and f n converges to f uniformly on E epsilon.
We are given that f n converges to f almost everywhere and mu of E is given to be infinite.
So by the result, it just now proved for every delta equal to 1 over n; let us apply the
previous result for epsilon equal to delta equal to 1 over n there exists a set E n contained
in E such that mu of E minus E n is less than 1 over n. On E n on the set let us write E
m here just for the sake of clarity, because we are going to apply it for every n. For
every E m there is a set and a stage n m such that f n of x minus f of x is less than 1
over m for every n greater than that stage n m.
By the result, we have proved that whenever f n converges to f almost everywhere and E
is a set of finite measure then with delta equal to 1 over n, sorry, this is less than
there is a set E m with that this is less than epsilon; for every epsilon, let us do
it for epsilon two to the power m . So, we are applying the previous result with
delta equal to 1 over m by replacing epsilon by epsilon divided by 2 to the power m. We
will say that is set, so that the difference between E and E m is less than epsilon by
2 to the power m and on the set E m the difference f n minus f x is less than 1 over m for every
n bigger than or equal to E m.
Let us define the set E epsilon equal to intersection of E m, m equal to 1 to infinity. With that let
us compute that this is the required set. So mu of E minus E epsilon is equal to mu
of union of E minus E m, m equal to 1 to infinity, because E epsilon is intersection, so minus
that will make it union, which is less than or equal to sigma m equal to 1 to infinity
mu of measure of E minus E m which is less than epsilon by 2 m; so which is less than
or equal to m equal to 1 to infinity of epsilon by 2 to the power m which is less than or
equal to epsilon. So, we get that measure of the set E minus
E epsilon is less than epsilon. Also if we look at x belongs to E epsilon that is equal
to intersection of E m, m equal to 1 to infinity.
What does that imply; that means, because it belongs to intersection it implies that
x belongs to E m for every m; mod of f n x minus f of x will be because it belongs to
E m for every m it is less than 1 over m for every n bigger than or equal to n m. So that
means for every m we can find a stage after which the difference f n x minus f x is less
than 1 over m for every x; so that implies that f n converges to f uniformly on E epsilon
that converges to E epsilon uniformly. So, this proves Egoroff's theorem namely,
which says that if f n converges to f almost everywhere then with given any set E of finite
measure, we can find a part of it. So that on the part of that set E f n converges to
f uniformly and the measure of the remaining that is E minus E epsilon is small is less
than epsilon. So for every epsilon we can find a set E epsilon contained in E such that
mu of E minus E epsilon is less than epsilon and on E epsilon f n converges f uniformly,
so this is what is called Egoroff's theorem.
So as a particular case of Egoroff's theorem; let us define and make new definition for
measurable functions f n and f, and a set E in the sigma algebra. One says, f n converges
almost uniformly to f on E if for every epsilon there is a set E epsilon belonging to the
sigma algebra such that the measure of the difference E minus E epsilon is small. On
E epsilon f converges a uniformly, such convergence is called almost uniform convergence essentially,
saying it is uniform except on a set of measure small that is what almost uniform is to be
understood.
So, Egoroff's theorem can be restated as if f n converges to f almost everywhere on a
set E of finite measure then, f n converges to f almost uniformly on E. So almost everywhere
convergence implies almost uniform convergence on every set of finite positive measure. In
particular case, when mu of x is finite this will imply f n converges to f almost everywhere
on the set x, so this will imply that f n converges to f almost uniformly on x.
On finite measure spaces almost everywhere convergence implies convergence which is almost
uniform. In fact converse of this proposition is also true when we have got the measure
space finite. So, the converse says that if mu of x is finite and f n converges to f almost
uniformly on x then f n converges to f almost everywhere.
So, the proof of that is almost obvious because by the given hypothesis for every integer
n, we can find a set f n in S in the sigma algebra such that the measure of the set f
n is small and f n converges to f uniformly on the complement of it. Let us define the
set F which is intersection of all this F n's then measure of the set f is less than
or equal to measure of each f n which is less than 1 over n.
The set f is a set of measure 0. Outside this if f if a point x belongs to f complement
then that mean it belongs to some f n complement for some n and on that convergence is uniform,
so f n converges to f in particular. That will imply that on f complement f n x converges
to f of x, so that proves the converse part of the Egoroff's theorem for finite measure
spaces.
Let us draw a picture and try to see what this implication mean. We have got convergences;
one is convergence point wise, so this is point wise convergence. We have got convergence
almost everywhere, convergence which is uniform and convergence which is almost uniform.
We know that the point wise convergence implies convergence which is almost everywhere. Convergence
uniform implies which is point wise convergence. So convergence uniform implies convergence
point wise and point wise implies almost everywhere. The other way around inequality is this is
not true and this implication is also not true in general .
We just now showed that convergence almost everywhere obviously does not imply this convergence.
Convergence uniform obviously implies convergence almost uniform, because almost uniform means
outside a set of measure small, so this is uniform implies that convergence also almost
uniform . So, almost everywhere convergence or point
wise convergence cannot imply this because it does not imply this . This implication
convergence almost everywhere obviously does not imply in general this, but what we can
say is that when it is finite almost this implies this when mu of X is finite and of
course, this also implies the other way round when this is finite.
So this is what one way, almost everywhere implies convergence almost uniform that is
Egoroff's theorem and the converse always hold, so these two are same if underlying
measure spaces is a finite measure space. These are the implications that we can say
as true. Next, let us point out the fact that Egoroff's
theorem says that almost everywhere convergence on finite measure spaces gives rise to almost
uniform convergence. This can be use to prove a important theorem called Luzin's theorem,
which says that if f is a real valued function on reals which is measurable and epsilon greater
than 0 is arbitrary then, there exists a continuous function g such that where f differs from
g is a set of measure small. So, what it says that if every measurable
function is almost continuous then, you can look at it this way. We will not prove this
result. Basically, the idea is on every interval one tries to apply Egoroff's theorem and then
try to patch it up. Those who are interested should look at the text book, refer for this
result by saying that Egoroff's theorem has an application namely every measureable - real
valued measurable function on reals is almost equal to a continuous function; that means,
the set of we can find for given any epsilon one can find a continuous function such that
the difference f x not equal to g x, it is the measure of that set which is small.
So, these are the relationships between point wise convergence, convergence almost everywhere,
uniform convergence and almost uniform convergence, but keep in mind almost uniform convergence
is not uniform almost everywhere, so these two are different things. Almost uniform means
that except outside a set of measure small the convergence is uniform, but if you say
uniform almost everywhere that means, outside a set of measure 0 the convergence is uniform.
So almost uniform is not same as uniform almost everywhere, so keep that in mind.
Next we would like to discover another important mode of convergence called convergence in
measure. This mode of convergence is very useful in the theory of probability analyzing
what are called convergence of random variables. We will not be go to the probabilistic aspect
of this; we will just look at the measure thyroidic aspect of what is convergence in
measure is.
So, a sequence of measurable functions f n is set to converge in measure to a measurable
function f on a measure space excess mu. If for every epsilon, look at the set, where
f n minus f x is bigger than or equal to epsilon, this is the kind of set where f n is not going
to; the difference remains bigger than epsilon. So, collect all such points and if we look
at the measure of this set, if the measure of this set goes to 0 and it becomes smaller
and smaller as n goes to infinity, then we say f n converges to f in measure. So saying
f n convergence to f in measure is for every epsilon, this is important for every epsilon
bigger than 0. Look at the measure of the set where f n minus
f is bigger than or equal to epsilon, measure of that set limit n going to infinity should
be equal to 0; this is called convergence in measure for function. This will write as
f n with an arrow f and above the arrow, we will put the symbol m indicating that f n
converges to f in measure.
Let us look at some examples to understand convergence in measure, these are the almost
everywhere convergence.
Let us look at the sequence of functions f n which is defined on the Lebesgue measure
space R that is real line Lebesgue measurable sets and the length function. Let us take
the function f n to be equal to the indicator function of n to n plus 1. So, what we are
doing is f n it is the indicator function of the interval n to n plus 1, if this is
a line, this is 1, 2, 3, n and n+1. What is f 1? f 1 is the indicator function
of 1 to 2, f 1 is nothing but the function, so this is f 1. What is f 2? f 2 is the indicator
function from 2 to 3, so this is f 2, this is f 3 and so on and this is f n . So what
is f n? f n is the function, it nothing but the constant function one on the interval
n to n plus 1, this graph is shifting as you go.
Clearly, f n x converges to f x for every x and that is obvious because, if we are given
a point x somewhere, here is a point x then, after some stage we can find n which is bigger
than this. So we can find something - some n naught which is bigger than this, then f
n naught of x will be equal to 0 . That means for every x, we can find a stage
n naught so that this is equal to 0; that will happen for every n bigger than that;
that means, the f n x converges to f of x which is identically equal to 0 . So, the
sequence of functions f n x point wise converges to the function f of x which is identically
0, so f n converges to f identically 0 point wise.
Let us look at the measure of the set, where f n minus f is bigger than or equal to say
1, because f n gives the value 1 or 0 only. The set where f n x differs from f of x is
precisely from the interval n to n plus 1. Lebesgue measure of the set where f n x minus
f of x is bigger than or equal to epsilon equal to 1 is equal to 1 and that never goes
to 0. This is the sequence of measurable functions which converges point wise, but it does not
converge measure to the function f identically equal to 0.
So, point wise convergence need not imply convergence in measure. Point wise convergence
- convergence almost everywhere in particular also does not imply convergence in measure.
Let us look at another example which is easy to describe pictorially. So, what we are going
to look at is, look at the interval 0, 1 divided into two equal parts that is 0, half and 1.
In the next stage divided into four equal parts 1, 2; that is 0, 1 by 4, half, 3 by
4, 1 and so on. So define f 1 to be the indicator function
of the whole interval 0 to 1; define f 2 to be the indicator function of the interval
0 to half, so this is f 2 which is the indicator function of 0 to half. This is f 3 which is
equal to indicator function of half to 1, so that is f 3. Similarly, this is f 4, this
is f 5, this is f 6, f 7 and so on. So each f n is nothing but the indicator function
of interval with n points which are obtained by dividing the interval into equal parts
at every stage. So, it is a indicator function of a interval
I k n, where the length of I k n is equal to 1 over 2, it is going to depend on which
stage? It is something like interval of length 1 over 2 to the power n or something like
this, because it is going to be shrinking .
If this is how the sequence f n is obtained then, it is clear that the set on which f
n is 1 is becoming smaller and smaller. So, guess is this f n converges to f identically
0 in measure, but this sequence f n does not converge to f point wise. For that the reason
is given any point x we can always find some f n like this, where the value of the function
f n is going to be equal to 1, because of that what is happening is f 1 is the indicator
function of the whole interval, f 2 it is this .
So, the set over which the function is non 0 is taking the value 1, it is fluctuating
but it is moving, at that most stage it becomes 0. Actually at every point x there will be
some f n which will be decreasing the value 1, so this will not converge almost everywhere
or point wise. This will be an example of a sequence which converges in measure but
not point wise. One can formally write this as follows. To
write this more rigorously, let us look at this. For every n, first of all choose unique
integer m such that n lies between 2 to the power m, n 2 to the power m plus 1 that is
obvious. That for any n you can find an integer with this property .
Next look at an integer k, for any integer k between 0 and 2 to the power m this number
n can be written as 2 to the power m plus k, because n to n plus 1 can be divided into
intervals of length 1 over 2 to the power m. Once, you do that you can find a k such
that this n is equal to 2 to the power m plus k.
So, what we are saying for every natural number n? It can be expressed uniquely as in the
form. To every n, you can associate a pair namely, small m comma k where m is such that
n lies between 2 to the power m less than or equal to and less than 2 to the power m
plus 1 and k is between 0 and 2 to the power m. This correspondence n with this pairs is
unique. Obviously, as n goes to infinity, this m will also go to infinity and conversely
as m goes to infinity, n goes to infinity .
How the function f n is defined? So define f n for any n, represent it uniquely as k
plus 2 to the power m and take f n to be the indicator function of I k m. This is the interval
starting at k by 2 to the power m and going to k plus 1 2 to the power m; that means,
it is a interval of length 1 over 2 to the power m. f n is 1 on this interval and you
see that this interval is shifting but never becoming smaller and smaller, and never vanishing.
So, f n is the sequence of some measurable functions because of the indicator functions
of intervals. For every x we can find a stage; for any n, if n is equal to this we can find
a stage, so that x will also belong to an interval of the starting with l 0 with base
as 2 to the power m plus 1. We mean that for every n there is a stage and dash such that
f n n dash of f n dash at the point x is also equal to 1
So, this will prove that the sequence f n does not converge to the function identically
0 point wise. On the other hand, it is quite obvious that the set of points where this
is bigger than epsilon for every epsilon bigger than 1, it is empty set and less than or equal
to 1, it is the interval which is becoming smaller and smaller.
So, it says that this is a sequence of functions that will go to 0. This is a sequence of functions
which converges in measure, but not point wise. What we have shown is that the convergence
in measure is neither implied by point wise convergence or convergence almost everywhere
and neither implies convergence in measure. So, convergence in measure and point wise
convergence are neither implied nor implied by each other. This is the concept of convergence
in measure which is quite different from point wise convergence.
However, when the underlying space is with finite measure one can draw some conclusions,
because in the theory of probability the underlying measure space has got total mass 1; so we
want to look at this concept of convergence in measure when the underlying measure space
has got finite measure.
We want to prove that if mu of x is finite and f n converges to f almost everywhere then,
f n converges also in measure to the set f. What we want to show is that convergence for
finite measure spaces convergence almost everywhere implies convergence in a measure.
Let us look at a proof of this fact; recall that saying f n converges in measure to f
is same as said; this is what we want to show? We want f n converges to f in measure, so
what we have to show? We have to show that for every epsilon, if we look at the set of
those points where f n minus f is bigger than or equal to epsilon then, the measure of this
set goes to 0 as n goes to infinity for every epsilon.
Let us call this set x belonging to X f n x minus f of x bigger than or equal to epsilon;
a set, it depends on epsilon and it depends on n, let us call this set as A n epsilon.
So A n epsilon is equal to set of points where f n minus f is bigger than or equal to epsilon
and we want to show that the measure of this set goes to 0 as n goes to infinity.
Now, let us observe that A n epsilon is obviously a subset of the union of the sets A m epsilons
from m equal to n, because this is one of the sets in the union. So this is showing
that mu of A n epsilon goes to 0 is same as enough if we can show that measure of this
union goes to 0.
Look at this unions A m epsilon m bigger than or equal to n, this is a sequence of sets
as n increases this union is going to become smaller and smaller. This is a sequence of
sets which is decreasing, so it is a decreasing sequence of sets and it decreases to; where
it will decrease? It will decrease to the intersection of the sets n equal to 1 to infinity
of these sets; mu of the set x being finite, this is a sequence of sets which is decreasing
to this set and mu of x being finite implies mu of the limit is equal to limit of the mu
of the sets. So, limit n going to infinity mu of this is equal to this.
Let us observe that mu of the set which is intersection n equal to 1 to infinity, union
m equal to 1 to infinity A m epsilon, what is this set? If x belongs to this set that
means, for every n there is m such that x belongs to A m for some m bigger than or equal
to n, which is same as the saying. So x does not converge f of x because A m epsilon is
the set where it is bigger than epsilon. What it says? It says that this is a set where
f n does not converge to f of x and we are given that f n converges to f almost everywhere.
So this set has got measure 0, this limit is equal to measure 0. So that proves the
fact that this is of measure 0 that proves the limit of A n epsilon is 0 that means,
f n converges to f n measure. So, we had started looking at the convergence
in measure and we have proved, gave examples to illustrate that neither point wise convergence
implies convergence in measure nor convergence in measure implies point wise convergence.
However, when the underlying space is finite convergence almost everywhere does imply convergence
in measure. We will continue this study of convergence
in measure and also look at its relation with what is called convergence in mean or convergence
in L P spaces. We will do that in next lecture, thank you.