Tip:
Highlight text to annotate it
X
Welcome to lecture 5 on measure and integration. If you recall, in the previous lectures we
have been looking at the various classes of subsets of a set X with various properties.
We looked at what is an algebra and what is a sigma algebra and a monotone class. Today,
we will start looking at functions defined on classes of subsets of a set X.
We will first look at what are called set functions and then we will look at a very
important example of a step function, namely the length function. Let us start defining
what are called set functions.
Let us start with C-a class of subsets of a set X. Any function mu (this is a Greek
symbol called mu) is defined on the class of subsets C of a set X and taking nonnegative
extended real-valued functions. This interval 0 to plus infinity, both included, denotes
the set of all nonnegative extended real numbers. A function mu defined on this collection C
of subsets of a set X taking values in nonnegative extended real numbers is going to be called
a set function. It is a function whose domain is a collection of sets; that is why it is
called a set function. Next, we will be looking at some special properties;
we will be analyzing such functions. Let us define a set function mu, of course, where
C is a collection of subsets of a set X and 0 to plus infinity the nonnegative extended
real numbers. A set function mu is set to be monotone if it has the following property:
for any two sets A and B in C, mu of A is less than or equal to mu of B whenever A is
a subset of B. It is a monotone property that whenever A is a subset of B and both are in
the collection C, we want that mu of A should be less than or equal to mu of B; this is
called the monotone property.
Next, we look at what is called finite additivity property of a set function mu. A set function
mu is set to be finitely additive, I am emphasizing the point finitely and additive, if it has
the following properties: mu of union of sets Ai, i equal to 1 to n, given any finite collection
of sets A1, A2 up to An in C, mu of the union of the sets is equal to mu of Ais.
Of course, this will be whenever A1, A2, up to An ... This is a finite collection of sets
in C such that their union also belongs to C, for otherwise this number on the left-hand
side of this equation will not be defined. Further, the sets are pairwise disjoint; Ai
intersection Aj is empty for i not equal to j. Once again, let us see what is finite additivity.
Finite additivity means for any finite collection of sets in C, A1, A2, up to An in C, such
that their union is also an element in C. These sets are pairwise disjoint; for any
such finite collection of sets, we want that mu of the union is equal to summation of mu
of the individual Ais. Intuitively, keep in mind that mu in some
sense is denoting the size of a set A and so we are saying mu of the union is equal
to sum of the individual sizes whenever the sets Ais are disjoint; we are requiring it
for any finite collection i equal to 1 to n. If A1, A2, up to An is any finite collection
of sets in C which are pairwise disjoint such that their union is an element in C, mu of
the union is equal to summation of mu of the individual Ais. Such a property is called
finite additivity property of mu or one says mu is finitely additive.
We can extend the generalization of this definition. We will say mu is countably additive, from
finite we are going to countably additive, if mu of union Ans 1 to infinity is equal
to summation of mu of Ans, of course, whenever A1, A2, up to An is a sequence of sets in
C such that the union is also an element of C and they are pairwise disjoint. Countable
additivity is a property about a sequence of sets A1, A2, An and so on in C which are
pairwise disjoint and their union is an element in C; we want that for any such sequence of
pairwise disjoint sets, mu of the union must be equal to summation of mu of Ans, n equal
to 1 to infinity.
There is another notion of called countably subadditive if mu of A is less than or equal
to summation 1 to infinity mu of Ans whenever A is a set in C and A is contained in union
of Ans where Ans are also in C for every n. In some sense, if A is covered by a union
of sets Ans, then we want the size-that is mu of A-to be less than or equal to summation
mu of Ans, n equal to 1 to infinity This is called countable subadditivity because
here we are just saying that mu of A is less than or equal to and we are not requiring
that Ans are pairwise disjoint; this is called countable subadditivity property of the set
function . A set function mu is called a measure on C (C is a collection of subsets) if mu
has the property that it is countably additive (it should be countably additive), the empty
set belongs to C and with the property that mu of empty set is equal to 0. mu is defined
on a collection C of subsets and we want the properties that the empty set belongs to C,
mu of empty set should be 0 and mu on this collection should be countably additive; such
a set function is going to be called a measure on C.
Let us look at some examples of set functions; let us start with a very simple one. Let us
look at a set X which is a countable set; its elements are x1, x2, x3 and so on. X is
equal to xn, n equal to 1, 2, 3 and so on. Let us fix pn-a sequence of nonnegative real
numbers. X is a set which is a countable set with elements x1, x2, x3 and so on and we
are fixing arbitrarily some sequence of nonnegative real numbers.
For any subset A contained in X, let us define mu of the empty set to be equal to 0; for
the set A if it is nonempty, let us define mu of A to be equal to summation over those
pis such that xi belongs to A. A is a subset of X and so sum of the xis will belong to
A. Look at those indices i such that xi belongs to A; pick up those pis from the given sequence
pn and add them up; that is called mu of A. mu of A is defined as summation over those
pis such that xi belongs to A. We want to check that this is a measure on
the collection of all subsets of the set X. That is quite obvious because mu of empty
set is defined to be equal to 0. Let us observe that if A is a singleton set, then mu of the
singleton set is going to be the number pi; if a set A is a countable disjoint union of
sets, let us check that this mu is a measure.
We are defining mu of A to be equal to summation pi where i is such that xi belongs to A. mu
is countably additive. Let us take a set A which is union of Ais, i equal to 1 to n and
Ais is a subset of A subset of where Ai is any subset of X. We have to check that mu
of A is equal to... We want Ai intersection Aj to be empty . We want this to be equal
to mu of Ais, i equal to 1 to infinity. Let us observe; it is enough to check when
each Ai is a singleton xi; let us check that case first. What is A? This is not xi because
x itself is x1, x2, up to xn and so this will be the whole space. Let us look at the special
case when Ai is equal to some xki, I bigger than or equal to 1 . Then, the set mu of A
is going to be equal to summation p of ki, i equal to 1 to infinity. This can be written
as limit n going to infinity i equal to 1 to infinity pi up to n. These are nonnegative numbers and so this
sum is nothing but the limit of the partial sums. These are nonnegative and so there is
no problem in writing it that way.
That means mu of A is equal to limit n going to infinity of sigma i equal to 1 to n of
pi. That means what we want to check? It is summation of mu of each Ai; so, this is limit
n going to infinity of summation i equal to 1 to n mu of Ai because each one is pi. It
is p of ki, sorry ; this summation p of ki. This is ki and this is mu of Ai.
That is equal to i equal to 1 to infinity mu of Ai. mu of A is equal to summation mu
of Ais whenever Ai is a singleton set; if not, it is a finite set; each finite is a
union of finite sets. For a nonnegative series, you can add it anyway you like; it is easy
to check that mu of A is equal to summation mu of Ai, i equal to 1 to infinity whenever
Ais are contained in X and Ai intersection Aj is empty. That says that this set function
mu that we have defined is countably additive . This is what is called a discrete measure
because it is given by a sequence and pi is called the mass at the point xi.
As we observed, mu of the singleton xi is equal to pi for every i and mu of the whole
space is equal to summation mu of the singletons is pi; mu of X is equal to summation of pis.
The obvious consequence of this is that mu of X is finite whenever this series is convergent.
One says this discrete measure mu is finite; that is, mu of X is less than infinity; mu
of the whole space is finite if and only if summation mu of pis is less than infinity.
If this summation of pis-the series pi is convergent and its sum is equal to 1, then
this measure mu is called a discrete probability distribution on the set X which is x1, x2,
up to xn.
This is a very special case which plays an important role in the theory of probability
and so on. X is the set of the numbers 0, 1, 2 and so on. Let us fix any number p which
is between 0 and 1 and define pk to be equal to n k (this is the binomial coefficient n
k) p to the power k into 1 minus p raised to the power n minus k, k between 0 and n.
This is called the binomial distribution because of this binomial coefficient appearing in
the definition of pk. It is quite easy to check that the summation
of these pks is equal to 1. That is because summation of these pks is summation k equal
to 0 to n and this side is nothing but p plus 1 minus p raised to power n and that is equal
to 1. This is a distribution which plays a very important role in probability; this is
a probability distribution. Supposing you have got a coin and you are tossing a coin
with probability p for head appearing, then this pk represents the probability that in
n tosses you will get k heads. Another special case of this discrete distribution
is called the Poisson distribution which is characterized by the definition that pk is
equal to lambda to the power k into e raised to the power minus lambda divided by k factorial.
This is called Poisson distribution; this is another important distribution in the theory
of probability. Finally, when we take only a finite number of points 0, 1 up to n and
pk is 1 over k and each point is given the same mass 1 over k, then this is called the
uniform distribution. There are special cases of discrete probability distributions.
Next, we give an important example of a measure which is defined on the collection of all
intervals in the real line. To do that, let us fix our notations. We will denote by I
the collection of all intervals on the real line. For an interval I with end points a
and b (the left end point being a and the right end point being b), we will write it
as I of a comma b; a will denote the left end point and b will denote the right end
point. We are not saying that this is an open interval
a comma b; we are just saying that it is an interval with left end point a and right end
point b where the left or the right may or may not be or both may or may not be included
in that interval. It is just an interval with end points a and b; the left end point is
a and the right end point is b. On this collection of all intervals, we are going to define a
function. For example, recall that the open interval
a comma a is the empty set. In the interval 0 to plus infinity, the square brackets indicate
that we are including 0 and we are including plus infinity. This is a closed interval in
R star, x belongs to R star-the extended real numbers, x bigger than or equal to 0; this
is same as the open interval, closed on the left 0 and open on the right infinity in the
real line, union the special symbol plus infinity that we had added in the extended real numbers.
With these notations, we define what is called the length function on the class of intervals.
It is a set function lambda defined on I taking values in 0 to infinity and is defined by
take any interval I with left end point a and right end point b. We define it as the
absolute value of b minus a if a and b are both real numbers. That means if the interval
I is a finite interval with end points a and b, then its length is defined as b minus a
and we define it equal to plus infinity in case either the left end point a is minus
infinity or the right end point b is equal to plus infinity or both; length of I for
an unbounded interval is defined as plus infinity. This function is called a length function
on the class of all intervals. This length function is going to play an important role
in our subject; let us study its properties.
Next, we will be studying properties of this length function. The first property is that
the length function has the property that lambda of the empty set is 0 because empty
set is an open interval with left end point, say, a and right end point a; it is an open
interval a comma a which is the empty set; by the very definition, that is equal to a
minus a which is equal to 0. Next, let us check that this is a monotone set function,
namely, length of I is less than or equal to length of J if I is a subset of J.
We want to check that whenever we have got intervals I comma J and I is a subset of J,
this should imply that length of I is less than or equal to length of J. Since the intervals
are characterized by the end points... Case I: let us say I is infinite, say I is equal
to minus infinity to a. Since I is a subset of J, obviously J has to start with minus infinity and can
go up to some point c where c is bigger than or equal to a. Essentially, what we are saying
is if this is a and on this side all of it is the interval I and if J is to contain I,
then J must be ending somewhere here-that is c . Clearly, both are infinite and so length
of I is equal to plus infinity is length of J; that case is obvious.
Let us look at the next case. I is a subset of J; J is infinite; whether I is infinite
or not does not matter because the length of I is always less than or equal to plus
infinity which is equal to length of J. J being infinite, its length is always going
to be plus infinity and so this is obvious if this is the case. Finally, let us look
at the case when both I and J are finite. Let us say I has got the end points a and
b. Now, I is subset of J and that means the end
points of J have to be somewhere here and here . If I is with end points a, b and J
is with end point c, d, then we should have c is less than or equal to a less than or
equal to b is less than or equal to d. This implies this . That is same as saying that
d minus c is bigger than or equal to b minus a and that is saying that the length of J
is bigger than or equal to length of I. The monotone property is checked . The length
function lambda is a monotone property; if I is a subset of J whenever an interval I
is contained in another interval J, the length of I is less than or equal to length of J.
Next, let us look at another property; it is called the finite additivity property.
What should be finite additivity property? Whenever an interval I is a disjoint union
of some other intervals, then the length of I should be summation of length of Js. What
we are saying is that if an interval I is written as a finite union of intervals Jis,
i equal to 1 to n where these Jis are pairwise disjoint, then we want length of I to be equal
to summation length of Jis; that is going to be called finite additivity property.
Let us check the finite additivity property. If I is equal to union of Jis where all are
intervals where Ji intersection Jk is empty, then that should imply that length of I is
equal to summation length of Jis, i equal to 1 to n. Let us assume that if I is infinite
and I is equal to union of Jis 1 to n, then that implies at least one of Jis is infinite
because if all of them are finite intervals, their union will be again a finite interval.
So, I infinite implies I equal to union of Jis implies at least one of these Jis has
to be infinite . That implies lambda of I is equal to plus infinity is equal to summation
lambda of Jis, i equal to 1 to n because one of them is plus infinity. So the case when
I is infinite is okay. Let us look at the case when I is finite.
I is finite. I is equal to union of Jis and Jis are pairwise disjoint. Now, let us say
the interval I has got end points a and b; let us say the left end point is a and the
right endpoint is b; here is a and here is b. We want to compute the length of I. Note:
the length of I is same as the length of the closed interval a comma b. I can include the
end points in the interval I because the length depends only on the values of the end points;
it does not matter whether the end points are inside or not. What we are saying is:
without loss of generality, let I be equal to a comma b. This is the interval a comma
b . Now, I is equal to union of Jis. The point
a belongs to this union and so it should belong to one of the intervals Jis. It belongs to
one of the interval Jis and actually it has to be end point of one of the intervals of
Jis because the interval cannot start somewhere else. Sorry, the interval J1 is starting at
a and ending somewhere let us call it as b1; the end point may or may not be included.
The first interval I1 we can assume it starts here and ends somewhere here; that is, b1
. Now, the point b1 is again in that union . So, either it is already included in the
interval I1 or it should be an end point of another interval in the union J1, J2 up to
Jns. The second one must start here and end somewhere here; that is, b2 . What we are
saying is I2-some other interval; you can rename it as I2; it should start somewhere
again at b1 and end somewhere here and so on. Here will be the last one an and that
should be bn. What we are saying is this is going this way; we can arrange the end points
of J1, J2 and Jn such that such that a is same as a1 less than or equal to b1 is equal
to a2 less than or equal to b2 and so on; so, an less than or equal to bn which is equal
to b. You can rearrange the end points of these intervals because this is a union and
that is a disjoint union ; this is what is possible for us to arrange.
That clearly says that b minus a is equal to bn minus a1; that is equal to summation
bi minus ai, i equal to 1 to n, adding and subtracting these terms in between; that is
same as i equal to 1 to n lambda of Ji. Whenever i is a finite interval, i is equal to union
of Jis (they are pairwise disjoint), we have got that the length of this b minis a is the
length of the interval I is equal to summation length of Jis. That means that the length
function lambda is finitely additive . This is the property of lambda-the length function
being finitely additive; if an interval I is a finite union of pairwise disjoint intervals,
then length of the interval I is equal to summation length of Jis.
Next, let us look at another property. Supposing I is a finite interval such that I is contained
in union 1 to n Iis where a finite union of the intervals..., we are no longer saying
that they are disjoint, then the claim is that length of I must be less than or equal
to summation length of these intervals Iis. If you drop the condition that these are pairwise
disjoint, we are saying if an interval I is covered by a finite union of intervals, then
the length of I must be less than or equal to summation of length of these intervals
Iis. Let us look at the proof of this. The proof of this is once again similar to the
earlier properties.
We are saying I is contained in union of Iis, i equal to 1 to n. Obviously, if one of Ii
is infinite, clearly this implies length of I is less than or equal to summation length
of Iis; that is obvious because one of these terms on the right-hand side in the summation
is plus infinity which is always greater than or equal to length of I, whatever be I. Let
us suppose so that each Ii is finite. This is a finite union and that implies I is finite.
As before, without loss of generality, we assume that I is equal to a comma b; here
is once again the same picture; here is a and here is b . The point a belongs to the
interval I; this is my interval I and a belongs to I; that means it belongs to this union
. It will belong to at least one of the intervals Iis. Let us name any one of them which contains
the point a to be I1 and let us say the end points of that are a1 and b1. The point a
belongs to one of the intervals Iis because it is in the union and so it will belong to
one of them, say I1; let us say the end points of I1 are a1 and b1. Here is the end point
a1 and here is the end point b1. Now the possibility is this b1 is on the right side of b; one
possibility is it is on the right side of b.
Let us write either b1 is bigger than or equal to b; that means my picture looks like this;
here is a1, here is a, here is b and here is b1. Then, length of I which is equal to
b minus a is less than or equal to b1 minus a1, which is equal to length of I1 and is
obviously less than or equal to summation length of Iis, i equal to 1 to n. In case
b1 is on the right side, we are obviously through by this case .
What is the other possibility? Case two. This is the picture . We have got a, we have got
b, here is a1 and b1 is not on the right side but on the left side of b; let us take that
as the picture. In that case, the point b1 belongs to that union. b1 is in the interval
a, b and so it will belong to that union. b1 belongs to I and so it belongs to the union
. It will belong to one of the intervals in the Iis. Let us call that as some interval
I2. b1 belongs to I2; that means a2 must start
here and b2 will either be somewhere here or it will be on the right side. If it is
on the right side of it, that means what? Let us say I is on the right side; here is
b2; instead of here, let us say b2 is here . Then, the length of the interval I which
is equal to b minus a is less than or equal to b2 minus a1 which is less than or equal
to b2 minus a2 plus b1 minus a1. So, b2 minus a1 is less than or equal to b2 minus a2 plus
b1 (we are adding something bigger) and then a1.
In that case, length of I will be less than or equal to length of I1 plus length of I2;
that is anyway less than or equal to summation length of Iis, i equal to 1 to n. If you go
on repeating this process, what does that mean? What is the other possibility? b1 is
inside; that means here is a and here is b . If it is not outside, then it must be inside;
that means here is a1; here was our b1; here is a and somewhere here is b2; it is not on
the right side; it is on the left side. Once again, b2 belongs... and then we can
proceed in the same way. At some stage we will be through; if not, then we will have
a1 is less than equal to a is less than or equal to a2 less than or equal to b1 less
than or equal to b2 less than or equal to so on less than or equal to an less than or
equal to b less than or equal to bn. What we are saying is either will be through at
some finite stage or we can rearrange eventually after n stages the end points in that way.
In that case again, lambda of I which is equal to b minus a, here is a and here is b , is
less than or equal to same idea bn minus a1; go on adding and subtracting; it is less than
or equal to bn minus an plus bn minus 1 minus a n minus 1 and so on plus b1 minus a1; that
is equal to sigma lambda of Ij, j equal to 1 to n. Whenever we are in a finite stage,
the end points can be rearranged nicely and we get this property; the length function
is having the property that whenever a interval I is covered by a finite union of intervals,
then the length of I is less than or equal to summation length of Iis.
Let us look at an extension of this property. Supposing I is a finite interval such that
I is covered by a union of intervals Iis 1 to infinity, that means the interval I is
covered by a countable union of intervals Iis; then again, the claim is length of I
is less than or equal to summation length of Iis. Let us prove this property. Keep in
mind that here we are assuming our interval I is a finite interval.
Interval I is contained in union of intervals Iis, i equal to 1 to infinity; these are intervals;
I is finite. This implies length of I is less than or equal to summation length of Iis,
i equal to 1 to infinity; this is what we want to prove. Obvious case: if any one of
the terms on this side-lambda of Ii-is infinite, then we are through. Note: if Ii is infinite
for some I, then what will happen? Length of Ii will be equal to plus infinity which
is bigger than or equal to length of I, whatever it may be-whether finite or infinite. It implies
sigma length of Ij, j equal to 1 to infinity is also bigger than or equal to lambda because
one of them is infinite; that case is obvious. Let us assume that not only is I finite but
all the intervals Iis are also finite; we want to check this property .
What we want to check is the following. I is finite with end points a and b' we can
assume it is a closed interval because the length of I is not going to change. Each Ij
is finite with left end point aj and right end point bj. We are not saying that we are
assuming these Ijs are open or closed or anything' we are just naming the end points. We are
saying I looks like this-a and b; each Ij is aj, bj . We are not saying that these end
points are included. We are given that I which is a comma b is contained in union of Ij,
j equal to 1 to infinity. If this was finite, then we already know how
to manipulate that; that we have already done earlier in the previous case. The idea is:
from that infinite union, bring it to a finite union. Here is a closed bounded interval contained
in an infinite union and we want to say this is going to be contained in a finite union.
Somewhere, the compactness property of the interval a to b is going to be used, but for
that we need the intervals to be open. Let us make these intervals Ijs open but,
of course, the lengths will change. Let epsilon greater than 0 be fixed. Select an open interval,
say, we call it as Jj such that this Jj includes our interval Ij and does not change the length
much. Length of this Jj is equal to say length of Ii plus epsilon; so, slightly increase.
What we are saying in this picture is take an interval from here to here the open interval
from here to here ; call that as Jj. Each Ij which was from aj to here (bj) is enclosed
in an open interval slightly bigger but the length portion that you had is at the most
equal to epsilon.
Now, what happens is the following. a, b is contained in the union of Ijs; each Ij is
contained in the union of Jjs; each Jj is an open interval; we had taken an open interval
. We have got an open cover of the closed bounded interval a, b. Heine-Borel property
of the real line which says that whenever a closed bounded interval is covered by a
collection of open intervals implies there exist some n such that a finite number of
them will cover it; so a, b will be contained in union of j equal to 1 to n Jjs; a finite
number of them will cover it. This implies by our earlier case that length
of I, this was my interval I , is less than or equal to sigma length of Jjs, 1 to n. Each
one of them is less than or equal to sigma j equal to 1 to n length of Ij plus epsilon.
Now, we want to separate out this summation and let it go to infinity to infinity but
the problem will come because of the summation epsilon added n times. That summation will
tend to become very very large; we do not want that to happen. What we do is we revise
our construction.
For a given epsilon, select an open interval Jj says that this holds. So, instead of epsilon
for the interval Ij, let us divide it by 2 to the power j. Instead of having this extra
length to be equal to same length as epsilon for every interval Ij, for Ij we want this
extra length to be equal to epsilon by 2 to the power j .
Once we do that, we are in a better shape because now this will be 2 to the power j.
That means it is less than or equal to summation j equal to I can put it 1 to infinity because
this is less than or equal to lambda of Ij plus summation epsilon by 2 to the power j,
j equal to 1 to infinity. Now, this series is convergent because it is a geometric series
with common ratio 1 by 2, which is less than 1. This term is equal to epsilon .
What we are saying is length of I is less than or equal to summation length of Ijs plus
a number epsilon but epsilon was arbitrary. Let epsilon go to 0. We will get length of
I is less than or equal to summation length of Ijs. What we are saying is that the countable
property that we looked at namely length of I is less than or equal to summation length
of Iis whenever an interval I which is finite is covered by any countable union, then the
length of I is less than or equal to length of Iis. We have extended that earlier property;
whenever a finite covering is there, we have extended it to a countable infinite covering
but only for finite intervals. We would like to extend this to even arbitrary intervals
which are not necessarily finite.
For that, we will have to do a little bit of more work. Let us look at the next property
which says the following. Let I be a finite interval such that I is equal to union 1 to
infinity In where Ins are pairwise disjoint. Then, at least we can conclude that the length
of I is equal to summation length of Ins. Whenever a finite interval is a countable
union of pairwise disjoint intervals, then the length of I is equal to summation length
of Ins. Let us prove this property.
What we have got is I is equal to union of Ijs, j equal to 1 to infinity; Ijs are pairwise
disjoint; I is finite implies length of I is equal to summation length of Ijs. Note
that we have already proved, just now, that if an interval is written as this-a union
of countable disjoint union, the length of I (we have just now shown) is less than or
equal to length of Ijs added up, j equal to 1 to infinity; call it (1). Length of I is
less than or equal to this ; this we proved just now for finite intervals. We need only
to show that length of I is bigger than or equal to summation j equal to 1 to infinity
length of Ij; only this is to be shown.
Here is the interval a to b. I is finite; here is the finite interval I. I1 is a subset
of I; it should be somewhere inside; somewhere is a1 and somewhere is b1 . Similarly, I2
is also inside I; somewhere it has to be; either it has to be a2 here and b2 here or
it could be here somewhere and so on. For every n, let us consider the end points an, bn of Ins. We can arrange
them, there are only finitely many of them, such that here is a, here is a1, here is b1,
here is a2, here is b2 and so on and here is an and here is bn and here is b.
That means we can arrange them in such a way that a is less than or equal to a1 less than
or equal to b1 which is less than or equal to a2 less than or equal to b2 and so on less
than or equal to an less than or equal to bn which is less than or equal to b. This
implies by simple algebra that length of I is equal to b minus a. This is b and this
is a and I am going to make it shorter bn and a1; this is bigger than or equal to bn
minus a1 which is bigger than or equal to bn minus an plus bn minus 1 (the next one
here) minus an minus 1 and so on plus b1 minus a1.
This put together is nothing but equal to sigma i equal to 1 to n length of Ii. What
we are saying is for every n, the end points of the intervals I1, I2 up to In can be rearranged
in this fashion. Hence by looking at the ordering of this, the length of I is bigger than this
. This happens for every n; that implies length of I is bigger than or equal to sigma I equal
to 1 to infinity because this is happening for every n, I can let it go to infinity,
length of I. The other way around inequality is also proved.
That means we have proved that whenever I is a finite interval which is written as a
countable union of pairwise disjoint intervals, then length of I is equal to sigma length
of Ins . With that, we prove an important property of the length function for finite
intervals. We will continue our study of the length function in the next lecture. Thank
you.