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.
.
.o, 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
.onvergence. 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.
.et 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
.omplement 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
.he 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
.ess 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.
.
.o, 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
.onverges 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
.hen 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
.psilon 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.
.o, 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
.ise 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 .
37:12).
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
. 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
.lmost 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.
.
.et 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.
